Similarity

Given,

(i) In △ABC and △PQR

$$\begin{aligned}\frac{AB}{PQ} = \frac{3.2}{4} \\[1em] = \frac{32}{40} = \frac{4}{5} \\[1em] \frac{AC}{PR} = \frac{3.6}{4.5} \\[1em] = \frac{36}{45} = \frac{4}{5} \\[1em] \frac{BC}{QR} = \frac{3}{5.4} \\[1em] = \frac{30}{54} = \frac{5}{9}\end{aligned}$$

Since, the ratio of the sides of both the triangle are not same. Hence, they are not similar.

(ii) In △DEF and △LMN

∠ E = ∠ N = 40°

$$\begin{aligned}\frac{DE}{LN} = \frac{4}{2} = \frac{2}{1} \\[1em] \frac{EF}{MN} = \frac{4.8}{2.4} = \frac{2}{1} \\[1em]\end{aligned}$$

Thus, by SAS rule of similarity △DEF ~ △LMN.

Given two right-angled triangles,

One angle of both triangles will be equal to 90°.

Given, acute angle of both triangles are equal let it be a°.

The third angle of both triangles will be [180° - (90 + a)°].

Hence, by AA rule of similarity both triangles are similar.

Given, △ABC ~ △EDF

$$\begin{aligned}\therefore \frac{AB}{DE} = \frac{AC}{EF} = \frac{BC}{DF}. \\[1em] \text{Consider, } \frac{AB}{DE} = \frac{AC}{EF} \\[1em] \Rightarrow \frac{5}{12} = \frac{7}{EF} \\[1em] \Rightarrow EF = \frac{84}{5} \\[1em] \Rightarrow EF = 16.8 \text{ cm.}\end{aligned}$$

Now consider,

$$\begin{aligned}\frac{AB}{DE} = \frac{BC}{DF} \\[1em] \Rightarrow \frac{5}{12} = \frac{BC}{15} \\[1em] \Rightarrow BC = \frac{5 \times 15}{12} \\[1em] \Rightarrow BC = \frac{75}{12} \\[1em] \Rightarrow BC = 6.25.\end{aligned}$$

Hence, EF = 16.8 cm and BC = 6.25 cm.

Given △ABC ~ △DEF, so

$$\begin{aligned}\therefore \frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF} \\[1em] \text{Consider, } \frac{AB}{DE} = \frac{AC}{DF} \\[1em] \frac{4}{6} = \frac{AC}{12} \\[1em] AC = \frac{48}{6} \\[1em] AC = 8.\end{aligned}$$

Now consider,

$$\begin{aligned}\frac{AB}{DE} = \frac{BC}{EF} \\[1em] \frac{4}{6} = \frac{BC}{9} \\[1em] BC = \frac{36}{6} \\[1em] BC = 6\end{aligned}$$

Perimeter of △ABC = AB + BC + AC = 4 + 6 + 8 = 18 cm.

Hence, perimeter of △ABC = 18 cm.

Given, △ABC ~ △PQR

$$\begin{aligned}\therefore \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} \\[1em]\end{aligned}$$

Since triangles are similar so the ratio of perimeter will be equal to ratio of sides.

$$\begin{aligned}\Rightarrow \frac{\text{Perimeter of △ABC}}{\text{Perimeter of △PQR}} = \frac{AC}{PR} \\[1em] \frac{32}{48} = \frac{AC}{6} \\[1em] AC = \frac{192}{48} \\[1em] AC = 4.\end{aligned}$$

Hence, length of AC = 4 cm.

Let △ABC ~ △DEF in which shortest side of △ABC be BC = 6cm.
△DEF, DE = 8cm, EF = 4cm and DF = 7cm.

Since, △ABC ~ △DEF so,

$$\begin{aligned}\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \\[1em] \text{Consider, } \frac{AB}{DE} = \frac{BC}{EF} \\[1em] \frac{AB}{8} = \frac{6}{4} \\[1em] AB = \frac{48}{4} \\[1em] AB = 12.\end{aligned}$$

Now consider,

$$\begin{aligned}\frac{BC}{EF} = \frac{AC}{DF} \\[1em] \frac{6}{4} = \frac{AC}{7} \\[1em] AC = \frac{42}{4} \\[1em] AC = 10.5.\end{aligned}$$

Hence, the other sides of triangle are 10.5cm and 12cm.

In the given figure,

AB ∥ DE, AC = 3cm, CE = 7.5cm, BD = 14cm.

From the figure,

∠ACB = ∠DCE [Vertically opposite angles]
∠BAC = ∠CED [Alternate angles]

Then, by AA rule of similarity, △ABC ~ △CDE.

So,

$$\begin{aligned}\Rightarrow \frac{AC}{CE} = \frac{BC}{CD} \\[1em] \Rightarrow \frac{3}{7.5} = \frac{BC}{CD} \\[1em] \Rightarrow 7.5BC = 3CD \qquad \text{[....Eq 1]} \\[1em]\end{aligned}$$

∵ BD = 14 cm, Let BC = x cm

∴ From fig, CD = (14 - x) cm.

Putting these values of BC and CD in equation 1 we get,

$$\begin{aligned}\Rightarrow 7.5x = 3(14 - x) \\[1em] \Rightarrow 7.5x = 42 - 3x \\[1em] \Rightarrow 7.5x + 3x = 42 \\[1em] \Rightarrow 10.5x = 42 \\[1em] \Rightarrow x = \frac{42}{10.5} \\[1em] \Rightarrow x = 4.\end{aligned}$$

∴ x = 4 and 14 - x = 10.

Hence, CB = 4cm and DC = 10cm.

Considering △ACO and △BDO,

∠ AOC = ∠ BOD [Vertically opposite angles]
∠ A = ∠ B [Alternate angles]

Then, by AA rule of similarity, △AOC ~ △BOD.

So,

$$\begin{aligned}\Rightarrow \frac{OA}{OB} = \frac{OC}{OD} = \frac{AC}{BD} \\[1em] \Rightarrow \text{Consider, } \frac{AC}{BD} = \frac{OA}{OB} \\[1em] \Rightarrow \frac{3.6}{2.4} = \frac{OA}{3.2} \\[1em] OA = \frac{3.6 \times 3.2}{2.4} \\[1em] OA = \frac{11.52}{2.4} \\[1em] OA = 4.8.\end{aligned}$$

Now, consider

$$\begin{aligned}\frac{AC}{BD} = \frac{OC}{OD} \\[1em] \frac{3.6}{2.4} = \frac{OC}{4} \\[1em] OC = \frac{14.4}{2.4} \\[1em] OC = 6.\end{aligned}$$

Hence, OA = 4.8 cm and OC = 6 cm.

In △RPQ and △RTS.

∠R = ∠R ( Common )
∠P = ∠RTS (Given)

Hence, by AA rule of similarity △RPQ ~ △RTS.

In △ABC and △ADC

∠C = ∠C (Common angle for both triangle)

∠BAC = ∠ADC (Given)

Then, by AA rule of similarity, △BAC ~ △ADC.

So,

$$\begin{aligned}\frac{CA}{DC} = \frac{BC}{CA} \\[1em] CA^2 = DC \times BC.\end{aligned}$$

Hence proved.

(i) Given, AP = 2PB and CP = 2PD.

$\therefore \frac{AP}{PB} = \frac{2}{1} \text{ and } \frac{CP}{PD} = \frac{2}{1}.$

∠ APC = ∠ BPD [Vertically opposite angles]

So by SAS rule of similarity △ACP ~ △BDP.

Since, triangles are similar,

∴ ∠ CAP = ∠ PBD.

Since, these angles are alternate angles therefore, AC ∥ BD.

Hence, proved that △ACP ~ △BDP and AC ∥ BD.

(ii) Since triangles are similar. So,

$$\begin{aligned}\frac{AP}{PB} = \frac{AC}{BD} \\[1em] \frac{AC}{BD} = \frac{2}{1} \\[1em] BD = \frac{AC}{2} \\[1em] BD = \frac{4.5}{2} \\[1em] BD = 2.25.\end{aligned}$$

Hence, BD = 2.25 cm.

(i) From the given figure,

∠ ADE = ∠ ACB (Given)
∠ A = ∠ A. (Common)

Hence, by AA rule of similarity △ABC ~ △AED.

(ii) Since triangles are similar,

$$\begin{aligned}\therefore \frac{BC}{DE} = \frac{AB}{AE} = \frac{AC}{AD} \\[1em]\end{aligned}$$

From figure:
AD = AB - BD = 6 - 1 = 5 cm.

Consider,

$$\begin{aligned}\frac{AB}{AE} = \frac{AC}{AD} \\[1em] \frac{6}{3} = \frac{AC}{5} \\[1em] AC = \frac{30}{3} \\[1em] AC = 10.\end{aligned}$$

Hence, the length of AC = 10 cm.

Given, ∠ PQR = ∠ PRS.

∠ P = ∠ P (common for both the triangles).

By AA rule of similarity, △PQR ~ △PRS.

Then,

$$\begin{aligned}\frac{PQ}{PR} = \frac{PR}{PS} = \frac{QR}{SR} \\[1em] \text{Considering, } \frac{PQ}{PR} = \frac{PR}{PS} \\[1em] \frac{PQ}{8} = \frac{8}{4} \\[1em] PQ = \frac{64}{4} \\[1em] PQ = 16.\end{aligned}$$

Hence, the length of PQ = 16 cm.

Consider △ABC

Given, AB = AC

∠ B = ∠ C [Angles opposite to equal sides (Property of isosceles triangle)]

Considering △BMP and △CNP

∠ M = ∠ N = 90°.

∠ B = ∠ C.

So, by AA rule of similarity △BMP ~ △CNP.

As triangles are similar,

$$\begin{aligned}\Rightarrow \frac{BM}{CN} = \frac{MP}{NP} \\[1em] \Rightarrow BM \times NP = CN \times MP.\end{aligned}$$

Hence proved.

Let two similar triangles be △ABC and △PQR.

We know that when triangles are similar ratio of corresponding sides are equal.

$\therefore \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}.$

By property of ratio i.e.,

if $\frac{a}{b} = \frac{b}{c} = \frac{d}{e},$ then each ratio = $\frac{\text{sum of antecedents}}{\text{sum of consequents}}$.

So,

$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{AB + BC + AC}{PQ + QR + PR}$

Since, AB + BC + AC = Perimeter of △ABC and PQ + QR + PR = Perimeter of △PQR. So,

$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{\text{Perimeter of △ABC}}{\text{Perimeter of △PQR}}.$

Hence, proved.

Consider triangle AOB and COD,

∠ AOB = ∠ COD [Vertically opposite angles]

∠ OAB = ∠ OCD [Alternate angles]

So, by AA rule of similarity △AOB ~ △COD.

As triangles are similar, ratio of sides will be similar,

$\frac{OA}{OC} = \frac{OB}{OD}$

Putting values of sides from question in equation,

$$\begin{aligned}\frac{3x - 19}{x - 3} = \frac{x - 4}{4} \\[1em] 4(3x - 19) = (x - 3)(x - 4) \\[1em] 12x - 76 = x^2 - 4x - 3x + 12 \\[1em] x^2 - 7x + 12 = 12x - 76 \\[1em] x^2 - 7x - 12x + 12 + 76 = 0 \\[1em] x^2 - 19x + 88 = 0 \\[1em] x^2 - 11x - 8x + 88 = 0 \\[1em] x(x - 11) - 8(x - 11) = 0 \\[1em] (x - 8)(x - 11) = 0 \\[1em] x - 8 = 0 \text{ or } x - 11 = 0 \\[1em] x = 8 \text{ or } x = 11.\end{aligned}$$

Hence, the value of x = 8 or 11.

(i) Consider △EFG and △CGD

∠ EGF = ∠ CGD [Vertically opposite angles]

∠ FEG = ∠ GCD [Alternate angles are equal]

So, by AA rule of similarity △EFG ~ △CGD.

Then,

$$\begin{aligned}\frac{EG}{GC} = \frac{EF}{CD} \\[1em] \frac{5}{10} = \frac{EF}{18} \\[1em] EF = \frac{90}{10} \\[1em] EF = 9.\end{aligned}$$

Hence, the length of EF = 9 cm.

(ii) Consider △ABC and △EFC

∠ C = ∠ C [Common angles]

∠ ABC = ∠ EFC [Alternate angles are equal]

So, by AA rule of similarity △ABC ~ △EFC

Then,

$$\begin{aligned}\frac{AC}{EC} = \frac{AB}{EF} \\[1em] \frac{AC}{EG + GC} = \frac{15}{9} \\[1em] \frac{AC}{5 + 10} = \frac{15}{9} \\[1em] \frac{AC}{15} = \frac{15}{9} \\[1em] AC = \frac{225}{9} \\[1em] AC = 25.\end{aligned}$$

Hence, the length of AC = 25 cm.

Consider △AEF and △CED

∠AEF = ∠CED [Vertically opposite angles]

∠F = ∠C [Alternate angles are equal]

So, by AA rule of similarity △AEF ~ △CED

Then,

$$\begin{aligned}\frac{AF}{CD} = \frac{AE}{ED} \\[1em] \frac{7.5}{4.5} = \frac{y}{3} \\[1em] y = \frac{22.5}{4.5} \\[1em] y = 5.\end{aligned}$$

Consider △ABE and △ACD

∠A = ∠A [Common angles]

∠ABE = ∠ACD [Alternate angles are equal]

So, by AA rule of similarity △ABE ~ △ACD.

Then,

$$\begin{aligned}\frac{EB}{CD} = \frac{AE}{AD} \\[1em] \frac{x}{4.5} = \frac{y}{AE + ED} \\[1em] \frac{x}{4.5} = \frac{y}{y + 3} \\[1em] x = \frac{4.5 \times 5}{8} \\[1em] x = \frac{22.5}{8} \\[1em] x = \frac{225}{80} = \frac{45}{16} \\[1em] x = 2\frac{13}{16}.\end{aligned}$$

Hence, the value of x = $2\frac{13}{16}$ cm and y = 5 cm.

Given ∠A = 90°,

or, ∠BAD + ∠DAC = 90° .....(i)

Now, consider △ADC

∠ADC = 90°

or, ∠DCA + ∠DAC = 90° .....(ii)

From equation (i) and equation (ii)

We have,

∠BAD + ∠DAC = ∠DCA + ∠DAC

∠BAD = ∠DCA .....(iii)

So, from △BDA and △ADC

∠BDA = ∠ADC = 90°
∠BAD = ∠DCA [From equation (iii)]

So, by AA rule of similarity △BDA ~ △ADC.

Since, corresponding sides of similar triangles are proportional,

$$\begin{aligned}\therefore \frac{BD}{AD} = \frac{AD}{DC} = \frac{AB}{AC} \\[1em] \text{Considering, } \frac{BD}{AD} = \frac{AD}{DC} \\[1em] AD^2 = BD \times CD \\[1em] AD^2 = 2 \times 8 \\[1em] AD^2 = 16 \\[1em] AD^2 - 16 = 0 \\[1em] AD^2 - 4^2 = 0 \\[1em] (AD - 4)(AD + 4) = 0 \\[1em] AD - 4 = 0 \text{ and } AD + 4 = 0 \\[1em] AD = 4 \text{ and } AD = -4.\end{aligned}$$

Since, length cannot be negative hence, AD ≠ -4.

Hence, the length of AD = 4 cm.

Let AB be tower and CD be pole.

BE = Shadow of tower and
DE = Shadow of telephone pole.

Considering △ABE and △CDE

∠ABE = ∠CDE (Both are equal to 90°)
∠AEB = ∠CED [Common angles]

So, by AA rule of similarity △ABE ~ △CDE. Hence, the ratio of corresponding sides will be equal.

$$\begin{aligned}\therefore \frac{AB}{CD} = \frac{BE}{DE} \\[1em] \Rightarrow \frac{15}{CD} = \frac{24}{16} \\[1em] \Rightarrow CD = \frac{15 \times 16}{24} \\[1em] \Rightarrow CD = \frac{240}{24} \\[1em] \Rightarrow CD = 10.\end{aligned}$$

Hence, the height of telephone pole is 10 m.

Let AB be the pole and DE be the woman as shown in the figure below:

Height of pole (AB) = 6 m
and height of a woman (DE) = 1.5 m

Here shadow EF = 3 m

Let BE(Distance of woman from pole) = x meters.

Considering △ABF and △EFD

∠ABF = ∠DEF (Both are equal to 90°)
∠F = ∠F [Common angles]

So, by AA rule of similarity △ABF ~ △EFD. Hence, the ratio of corresponding sides will be equal.

$$\begin{aligned}\therefore \frac{BF}{EF} = \frac{AB}{DE} \\[1em] \Rightarrow \frac{3 + x}{3} = \frac{6}{1.5} \\[1em] \Rightarrow \frac{3 + x}{3} = \frac{60}{15} \\[1em] \Rightarrow 15(3 + x) = 180 \\[1em] \Rightarrow 45 + 15x = 180 \\[1em] \Rightarrow 15x = 180 - 45 \\[1em] \Rightarrow 15x = 135 \\[1em] \Rightarrow x = \frac{135}{15} \\[1em] \Rightarrow x = 9.\end{aligned}$$

Hence, woman is 9 m away from the pole.

(i) Considering △ABC and △ADE,

∠ A = ∠ A (Common angle)
∠ ADE = ∠ABC (Alternate angle)

So, by AA rule of similarity △ABC ~ △ADE.

$$\begin{aligned}\therefore \frac{AE}{AC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{AE}{AE + EC} = \frac{AD}{AD + BD} \\[1em] \Rightarrow \frac{AE}{AE + EC} = \frac{3}{3 + 4} \\[1em] \Rightarrow 7AE = 3(AE + EC) \\[1em] \Rightarrow 7AE = 3AE + 3EC \\[1em] \Rightarrow 7AE - 3AE = 3EC \\[1em] \Rightarrow 4AE = 3EC \\[1em] \Rightarrow \frac{AE}{EC} = \frac{3}{4}.\end{aligned}$$

Hence, AE : EC = 3 : 4.

(ii) Since, △ABC ~ △ADE

$$\begin{aligned}\therefore \frac{DE}{BC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{AD}{AD + BD} \\[1em] \Rightarrow \frac{DE}{5} = \frac{3}{3 + 4} \\[1em] \Rightarrow DE = \frac{15}{7} \\[1em] \Rightarrow DE = 2\frac{1}{7}. \\[1em]\end{aligned}$$

Hence, DE = $2\frac{1}{7}.$

Considering △ABC and △PBQ,

∠B = ∠B (Common angle)
∠BPQ = ∠BAC (Corresponding angle are equal)

So, by AA rule of similarity △ABC ~ △PBQ.

$$\begin{aligned}\therefore \frac{BQ}{BC} = \frac{BP}{AB} \\[1em] \Rightarrow \frac{BQ}{8} = \frac{6}{AP + PB} \\[1em] \Rightarrow \frac{BQ}{8} = \frac{6}{4 + 6} \\[1em] \Rightarrow BQ = \frac{48}{10} \\[1em] \Rightarrow BQ = 4.8\end{aligned}$$

CQ = BC - BQ = 8 - 4.8 = 3.2 .

Hence, BQ = 4.8 cm and CQ = 3.2 cm.

Considering △PQR and △PXY,

∠P = ∠P (Common angle)
∠PXY = ∠PQR (Corresponding angles)

So, by AA rule of similarity △PQR ~ △PXY. Since ratio of corresponding sides is same,

$$\begin{aligned}\therefore \frac{PX}{QX} = \frac{PY}{YR} \\[1em] \Rightarrow \frac{1}{3} = \frac{PY}{4.5} \\[1em] \Rightarrow PY = \frac{4.5}{3} \\[1em] \Rightarrow PY = 1.5 \\[1em]\end{aligned}$$

As triangles are similar so ratio of corresponding sides is same,

$$\begin{aligned}\therefore \frac{XY}{QR} = \frac{PX}{PQ} \\[1em] \Rightarrow \frac{XY}{9} = \frac{1}{1 + 3} \\[1em] \Rightarrow XY = \frac{9}{4} \\[1em] \Rightarrow XY = 2.25\end{aligned}$$

Hence, the value of PY = 1.5 cm and XY = 2.25 cm.

(i) Considering △ABC and △ADE,

∠A = ∠A (Common angle)
∠ADE = ∠ABC (Corresponding angles)

So, by AA rule of similarity △ABC ~ △ADE. Since ratio of corresponding sides is same,

$$\begin{aligned}\therefore \frac{AD}{AB} = \frac{AE}{AC} \\[1em] \Rightarrow \frac{AD}{AD + DB} = \frac{AE}{AE + EC} \\[1em] \Rightarrow \frac{x}{x + x - 2} = \frac{x + 2}{x + 2 + x - 1} \\[1em] \Rightarrow \frac{x}{2x - 2} = \frac{x + 2}{2x + 1} \\[1em] \Rightarrow x(2x + 1) = (x + 2)(2x - 2) \\[1em] \Rightarrow 2x^2 + x = 2x^2 - 2x + 4x - 4 \\[1em] \Rightarrow 2x^2 - 2x^2 + x + 2x - 4x = 4 \\[1em] \Rightarrow -x = -4 \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, the value of x = 4.

(ii) Since, triangles are similar so, ratio of corresponding sides is same,

$$\begin{aligned}\therefore \frac{AD}{AB} = \frac{AE}{AC} \\[1em] \Rightarrow \frac{AB - DB}{AB} = \frac{AC - EC}{AC} \\[1em] \Rightarrow \frac{2x - (x - 3)}{2x} = \frac{2x + 3 - (x - 2)}{2x + 3} \\[1em] \Rightarrow \frac{x + 3}{2x} = \frac{x + 5}{2x + 3} \\[1em] \Rightarrow (x + 3)(2x + 3) = 2x(x + 5) \\[1em] \Rightarrow 2x^2 + 3x + 6x + 9 = 2x^2 + 10x \\[1em] \Rightarrow 2x^2 - 2x^2 + 9 = 10x - 9x \\[1em] \Rightarrow x = 9 \\[1em]\end{aligned}$$

Hence, the value of x = 9.

(i) So, according to question two triangles will be formed PQR and PEF.

Comparing ratios of corresponding side we get,

$$\begin{aligned}\Rightarrow \frac{PE}{PQ} \text{ and } \frac{PF}{PR} \\[1em] \Rightarrow \frac{PE}{PE + EQ} \text{ and } \frac{PF}{PF + FR} \\[1em] \Rightarrow \frac{3.9}{3.9 + 3} \text{ and } \frac{8}{8 + 9} \\[1em] \Rightarrow \frac{3.9}{6.9} \text{ and } \frac{8}{17} \\[1em] \Rightarrow \frac{39}{69} \text{ and } \frac{8}{17}.\end{aligned}$$

Since, both ratios are different hence, EF and QR are not parallel.

(ii) The triangles are shown in the figure below:

Comparing ratios of corresponding side we get,

$$\begin{aligned}\Rightarrow \frac{PE}{PQ} \text{ and } \frac{PF}{PR} \\[1em] \Rightarrow \frac{0.18}{1.28} \text{ and } \frac{0.36}{2.56} \\[1em] \Rightarrow \frac{18}{128} \text{ and } \frac{36}{256} \\[1em] \Rightarrow \frac{9}{64} \text{ and } \frac{9}{64}.\end{aligned}$$

Since, both ratios are same hence by converse of basic proportionality theorem triangles are similar and so EF and QR are parallel.

Checking the ratios in order to check the similarity of triangles PAB and PQR.

$$\begin{aligned}\Rightarrow \frac{PA}{PQ} \text{ and } \frac{PB}{PR} \\[1em] \Rightarrow \frac{PA}{PQ} \text{ and } \frac{PB}{PB + BR} \\[1em] \Rightarrow \frac{5}{12.5} \text{ and } \frac{4}{4 + 6} \\[1em] \Rightarrow \frac{50}{125} \text{ and } \frac{4}{10} \\[1em] \Rightarrow \frac{2}{5} \text{ and } \frac{2}{5} \\[1em]\end{aligned}$$

Since, both ratios $\Big(\frac{PA}{PQ} = \frac{PB}{PR}\Big)$ are same hence by converse of basic proportionality theorem triangles are similar and so AB and QR are parallel.

Considering △ABC and △BDE,

∠B = ∠B (Common angle)
∠DEB = ∠ACB (Corresponding angles)

So, by AA rule of similarity △ABC ~ △BDE. Since ratio of corresponding sides is same,

$$\begin{aligned}\Rightarrow \frac{BE}{BC} = \frac{BD}{BA} \\[1em] \Rightarrow \frac{BE}{BE + EC} = \frac{BD}{BA} \\[1em] \Rightarrow \frac{4}{4 + 2} = \frac{BD}{AB}\\[1em] \Rightarrow \frac{4}{6} = \frac{BD}{AB} \qquad \text{[....Eq 1]} \\[1em]\end{aligned}$$

Considering △ABL and △BDC,

∠B = ∠B (Common angle)
∠BDC = ∠BAL (Corresponding angles)

So, by AA rule of similarity △ABL ~ △BDC. Since ratio of corresponding sides is same,

$$\begin{aligned}\Rightarrow \frac{BD}{BA} = \frac{BC}{BL} \\[1em] \Rightarrow \frac{BD}{BA} = \frac{BE + EC}{BE + EC + CL} \\[1em] \Rightarrow \frac{BD}{BA} = \frac{4 + 2}{4 + 2 + CL} \\[1em] \Rightarrow \frac{BD}{BA} = \frac{6}{6 + CL} \qquad \text{[....Eq 2]} \\[1em]\end{aligned}$$

Solving Eq 1 and Eq 2 we get,

$$\begin{aligned}\Rightarrow \frac{6}{6 + CL} = \frac{4}{6} \\[1em] \Rightarrow 36 = 4(6 + CL) \\[1em] \Rightarrow 24 + 4CL = 36 \\[1em] \Rightarrow 4CL = 36 - 24 \\[1em] \Rightarrow 4CL = 12 \\[1em] \Rightarrow CL = 3.\end{aligned}$$

Hence, the length of CL is 3 cm.

Given, ∠D = ∠E

So, AD = AE [Sides opposite to equal angles]

Given, $\frac{AD}{DB} = \frac{AE}{EC} \qquad \text{[....Eq 1]}$

Hence, by basic proportionality theorem, DE is parallel to BC.

As AD = AE so in order to satisfy Eq 1, DB = EC.

AB = AD + DB = AE + EC

and AC = AE + EC.

Hence, AB = AC which means ABC is an isosceles triangle.

Consider △POQ

AB || PQ ....[ Given ]

So, By basic proportionality theorem,

$\frac{OA}{AP} = \frac{OB}{BQ} \qquad \text{[....Eq 1]}$

Then consider △OPR

AC || PR ....[ Given ]

So, By basic proportionality theorem,

$\frac{OA}{AP} = \frac{OC}{CR} \qquad \text{[....Eq 2]}$

Comparing Eq 1 and Eq 2 we get,

$\Rightarrow \frac{OB}{BQ} = \frac{OC}{CR}$

Hence, by basic proportionality theorem BC || QR.

Trapezium ABCD is shown in the figure below:

Consider △OAB and △OCD,

∠AOB = ∠COD [Vertically opposite angles are equal]
∠OBA = ∠ODC [Alternate angles are equal]
∠OAB = ∠OCD [Alternate angles are equal]

Therefore, by AA rule of similarity △OAB ~ △OCD,

$$\begin{aligned}\Rightarrow \frac{AO}{CO} = \frac{BO}{DO} \\[1em] \Rightarrow \frac{AO}{BO} = \frac{CO}{DO} \text{ (On cross-multiplication)}\end{aligned}$$

Hence, proved that $\frac{AO}{BO} = \frac{CO}{DO}.$

We know that,

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

$\therefore \frac{BD}{DC} = \frac{AB}{AC}$.

Putting values in above equation we get,

$$\begin{aligned}\Rightarrow \frac{3}{DC} = \frac{6}{4} \\[1em] \Rightarrow DC = \frac{3 \times 4}{6} \\[1em] \Rightarrow DC = \frac{12}{6} \\[1em] \Rightarrow DC = 2 \text{ cm}.\end{aligned}$$

BC = BD + DC = 3 + 2 = 5 cm.

Hence, the length of BC = 5 cm.

(i) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △PQR}} = \frac{(1)^2}{(3)^2} \\[1em] = \frac{1}{9} \\[1em] = 1 : 9.\end{aligned}$$

Hence, the ratio of area of △ABC to △PQR = 1 : 9.

(ii) Let the corresponding sides be in ratio x : y.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △PQR}} = \frac{x^2}{y^2} \\[1em] \Rightarrow \frac{25}{36} = \frac{x^2}{y^2} \\[1em] \Rightarrow \Big(\frac{5}{6}\Big)^2 = \Big(\frac{x}{y}\Big)^2 \\[1em] \Rightarrow \frac{x}{y} = \frac{5}{6} \\[1em] \Rightarrow x : y = 5 : 6.\end{aligned}$$

Hence, the ratio of corresponding sides of △ABC and △PQR = 5 : 6.

Let the length of EF be x cm.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △DEF}} = \frac{BC^2}{EF^2} \\[1em] \Rightarrow \frac{9}{16} = \frac{(2.1)^2}{x^2} \\[1em] \Rightarrow \frac{9}{16} = \frac{2.1 \times 2.1}{x^2} \\[1em] \Rightarrow \frac{9}{16} = \frac{4.41}{x^2} \\[1em] \Rightarrow x^2 = \frac{4.41 \times 16}{9} \\[1em] \Rightarrow x^2 = \frac{70.56}{9} \\[1em] \Rightarrow x^2 = 7.84 \\[1em] \Rightarrow x = \sqrt{7.84} \\[1em] \Rightarrow x = 2.8\end{aligned}$$

Hence, the length of EF = 2.8 cm.

Let the area of △DEF be x sq. cm

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △DEF}} = \frac{BC^2}{EF^2} \\[1em] \Rightarrow \frac{54}{x} = \frac{3^2}{4^2} \\[1em] \Rightarrow \frac{54}{x} = \frac{9}{16} \\[1em] \Rightarrow x = \frac{54 \times 16}{9} \\[1em] \Rightarrow x = 6 \times 16 \\[1em] \Rightarrow x = 96.\end{aligned}$$

Hence, the area of △DEF = 96 sq. cm.

Let the length of altitude of other △ be x cm

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding altitudes.

$$\begin{aligned}\therefore \frac{\text{Area of first △}}{\text{Area of second △}} = \Big(\frac{\text{Altitude of first △}}{\text{Altitude of second △}}\Big)^2 \\[1em] \Rightarrow \frac{36}{25} = \Big(\frac{2.4}{x}\Big)^2 \\[1em] \Rightarrow \frac{36}{25} = \frac{5.76}{x^2} \\[1em] \Rightarrow x^2 = \frac{5.76 \times 25}{36} \\[1em] \Rightarrow x^2 = \frac{144}{36} \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x^2 - 4 = 0 \\[1em] \Rightarrow (x - 2)(x + 2) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 2 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -2.\end{aligned}$$

Since length cannot be negative so, x ≠ -2.

Hence, the length of altitude of other triangle = 2 cm.

Considering △QOA and △POB,

∠ QOA = ∠ POB (Vertically opposite angles are equal)
∠ QAO = ∠ PBO (Both are equal to 90°)

Hence, by AA axiom △QOA ~ △POB.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

Let the area of △QOA be x cm².

$$\begin{aligned}\therefore \frac{\text{Area of △QOA}}{\text{Area of △POB}} = \frac{QO^2}{PO^2} \\[1em] \Rightarrow \frac{x}{120} = \frac{9^2}{6^2} \\[1em] \Rightarrow \frac{x}{120} = \frac{81}{36} \\[1em] \Rightarrow x = \frac{120 \times 81}{36} \\[1em] \Rightarrow x = \frac{9720}{36} \\[1em] \Rightarrow x = 270.\end{aligned}$$

Hence, the area of △QOA is 270 cm²

(i) Considering △OAB and △OCD,

∠ AOB = ∠ COD (Vertically opposite angles are equal)
∠ BAO = ∠ OCD (Alternate angles are equal)

Hence, by AA axiom △OAB ~ △OCD.

(ii) Since triangles are similar hence ratio of corresponding sides are equal,

$$\begin{aligned}\therefore \frac{AO}{OC} = \frac{AB}{CD} \\[1em] \Rightarrow \frac{10}{5} = \frac{6.5}{CD} \\[1em] \Rightarrow CD = \frac{6.5 \times 5}{10} \\[1em] \Rightarrow CD = \frac{32.5}{10} \\[1em] \Rightarrow CD = 3.25 \text{ cm}.\end{aligned}$$

Similarly,

$$\begin{aligned}\frac{AO}{OC} = \frac{OB}{OD} \\[1em] \Rightarrow \frac{10}{5} = \frac{OB}{2.8} \\[1em] \Rightarrow OB = \frac{2.8 \times 10}{5} \\[1em] \Rightarrow OB = \frac{28}{5} \\[1em] \Rightarrow OB = 5.6 \text{ cm.}\end{aligned}$$

Hence, the length of CD = 3.25 cm and OB = 5.6 cm.

(iii) In part (i) we have proved that △OAB ~ △OCD.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △OAB}}{\text{Area of △OCD}} = \frac{AO^2}{OC^2} \\[1em] = \frac{10^2}{5^2} \\[1em] = \frac{100}{25} \\[1em] = \frac{4}{1} \\[1em] = 4 : 1.\end{aligned}$$

Hence, the ratio of area of △OAB and △OCD is 4 : 1.

Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

Let the area of △ABC be x cm².

$$\begin{aligned}\therefore \frac{\text{Area of △ADE}}{\text{Area of △ABC}} = \frac{DE^2}{BC^2} \\[1em] \Rightarrow \frac{28}{x} = \frac{6^2}{9^2} \\[1em] \Rightarrow \frac{28}{x} = \frac{36}{81} \\[1em] \Rightarrow x = \frac{28 \times 81}{36} \\[1em] \Rightarrow x = \frac{2268}{36} \\[1em] \Rightarrow x = 63.\end{aligned}$$

Hence, area of △ABC is 63 cm².

Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

Given AD : DB = 1 : 2.

$$\begin{aligned}\Rightarrow \frac{AD}{AB - AD} = \frac{1}{2} \\[1em] \Rightarrow 2AD = AB - AD \\[1em] \Rightarrow 2AD + AD = AB \\[1em] \Rightarrow 3AD = AB \\[1em] \Rightarrow \frac{AD}{AB} = \frac{1}{3} \\[1em] \Rightarrow AD : AB = 1 : 3.\end{aligned}$$

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ADE}}{\text{Area of △ABC}} = \frac{AD^2}{AB^2} \\[1em] \Rightarrow \frac{\text{Area of △ADE}}{\text{Area of △ADE + Area of ⏢DBCE}} = \frac{1^2}{3^2} \\[1em] \Rightarrow \frac{\text{Area of △ADE}}{\text{Area of △ADE + Area of ⏢DBCE}} = \frac{1}{9} \\[1em] \Rightarrow 9 \times \text{Area of △ADE} = \text{Area of △ADE + Area of ⏢DBCE} \\[1em] \Rightarrow 8 \text{ Area of △ADE} = \text{ Area of ⏢DBCE } \\[1em] \Rightarrow \frac{\text{Area of △ADE}}{\text{ Area of ⏢DBCE }} = \frac{1}{8} = 1 : 8.\end{aligned}$$

Hence, the ratio of the areas of △ADE and trapezium DBCE is 1 : 8.

(i) Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

(ii) Given AD = $\frac{1}{2}$BD

$$\begin{aligned}\Rightarrow AD = \frac{1}{2}(AB - AD) \\[1em] \Rightarrow 2AD = AB - AD \\[1em] \Rightarrow 2AD + AD = AB \\[1em] \Rightarrow 3AD = AB \\[1em] \Rightarrow \frac{AD}{AB} = \frac{1}{3} \\[1em] \Rightarrow AD : AB = 1 : 3.\end{aligned}$$

Since triangles ADE and ABC are similar so, ratio of their corresponding sides will be equal

$$\begin{aligned}\therefore \frac{AD}{AB} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{1}{3} = \frac{DE}{4.5}\\[1em] \Rightarrow DE = \frac{4.5}{3} \\[1em] \Rightarrow DE = 1.5\end{aligned}$$

Hence, the length of DE = 1.5 cm.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ADE}}{\text{Area of △ABC}} = \frac{AD^2}{AB^2} \\[1em] \Rightarrow \frac{\text{Area of △ADE}}{\text{Area of △ABC}} = \frac{1^2}{3^2} \\[1em] \Rightarrow \text{Area of △ADE} = \frac{1}{9} \times \text{ Area of △ABC} \\[1em] \Rightarrow \text{Area of △ADE} = \frac{1}{9} \times 18 \\[1em] \Rightarrow \text{Area of △ADE} = 2 \text{ cm}^2\end{aligned}$$

Area of trapezium DBCE = Area of △ABC - Area of △ADE = (18 - 2) cm² = 16 cm².

Hence, the area of trapezium DBCE = 16 cm².

(i) Considering △DEC and △ABC,

∠ C = ∠ C (Common angles)
∠ ABC = ∠ DEC (Both angles are equal to 90°)

Hence, by AA axiom △DEC ~ △ABC.

(ii) Since △DEC ~ △ABC, so, ratio of their corresponding sides will be equal

$$\begin{aligned}\therefore \frac{AB}{DE} = \frac{AC}{CD} \\[1em] \Rightarrow \frac{6}{4} = \frac{15}{CD} \\[1em] \Rightarrow CD = \frac{15 \times 4}{6} \\[1em] \Rightarrow CD = 10 \text{ cm}\end{aligned}$$

Hence, the length of CD = 10 cm.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △DEC}} = \frac{AB^2}{DE^2} \\[1em] \Rightarrow \frac{\text{Area of △ABC}}{\text{Area of △DEC}} = \frac{6^2}{4^2} \\[1em] \Rightarrow \frac{\text{Area of △ABC}}{\text{Area of △DEC}} = \frac{36}{16} \\[1em] \Rightarrow \frac{\text{Area of △ABC}}{\text{Area of △DEC}} = \frac{9}{4} \\[1em].\end{aligned}$$

Hence, the ratio of the area of △ABC : area of △DEC = 9 : 4.

(i) We need to find $\frac{AD}{AB}$,

Given,

$$\begin{aligned}\frac{AD}{DB} = \frac{3}{2} \\[1em] \Rightarrow \frac{AD}{AB - AD} = \frac{3}{2} \\[1em] \Rightarrow 2AD = 3(AB - AD) \\[1em] \Rightarrow 2AD = 3AB - 3AD \\[1em] \Rightarrow 5AD = 3AB \\[1em] \Rightarrow \frac{AD}{AB} = \frac{3}{5}.\end{aligned}$$

Hence, the ratio $\frac{AD}{AB} = \frac{3}{5}$.

Considering △ADE and △ABC,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

Since triangles ADE and ABC are similar so the ratio of corresponding sides will be equal.

$$\begin{aligned}\therefore \frac{DE}{BC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{3}{5}.\end{aligned}$$

Hence, the ratio $\frac{DE}{BC} = \frac{3}{5}$.

(ii) Considering △DEF and △CBF,

∠ DFE = ∠ BFC (Vertically opposite angles)
∠ EDF = ∠ FCB (Alternate angles are equal)

Hence, by AA axiom △DEF ~ △CBF.

Since triangles are similar hence the ratio of the corresponding sides will be equal,

$$\begin{aligned}\therefore \frac{EF}{FB} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{EF}{FB} = \frac{3}{5}.\end{aligned}$$

Hence, the value of $\frac{EF}{FB} = \frac{3}{5}.$

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △DEF}}{\text{Area of △CBF}} = \frac{EF^2}{FB^2} \\[1em] \Rightarrow \frac{\text{Area of △DEF}}{\text{Area of △CBF}} = \frac{3^2}{5^2} \\[1em] \Rightarrow \frac{\text{Area of △DEF}}{\text{Area of △CBF}} = \frac{9}{25}.\end{aligned}$$

Hence, the ratio of the area of △DEF : area of △CBF = 9 : 25.

(i) Considering △PMN and △PQR,

∠ P = ∠ P (Common angles)
∠ PMN = ∠ PQR (Corresponding angles are equal)

Hence, by AA axiom △PMN ~ △PQR.

Given,

$$\begin{aligned}\frac{PM}{MQ} = \frac{2}{3} \\[1em] \Rightarrow \frac{PM}{PQ - PM} = \frac{2}{3} \\[1em] \Rightarrow 3PM = 2(PQ - PM) \\[1em] \Rightarrow 3PM = 2PQ - 2PM \\[1em] \Rightarrow 3PM + 2PM = 2PQ \\[1em] \Rightarrow 5PM = 2PQ \\[1em] \Rightarrow \frac{PM}{PQ} = \frac{2}{5}.\end{aligned}$$

Since triangles are similar hence the ratio of the corresponding sides will be equal,

$\therefore \frac{MN}{QR} = \frac{PM}{PQ} = \frac{2}{5}.$

Hence, $\frac{MN}{QR} = \frac{2}{5}$

(ii) Considering △OMN and △ORQ,

∠ MON = ∠ QOR (Vertically opposite angles are equal)
∠ OMN = ∠ ORQ (Alternate angles are equal)

Hence, by AA axiom △OMN ~ △ORQ.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \frac{MN^2}{QR^2} \\[1em] \Rightarrow \frac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \frac{2^2}{5^2} \\[1em] \Rightarrow \frac{\text{Area of △OMN}}{\text{Area of △ORQ}} = \frac{4}{25}.\end{aligned}$$

Hence, the ratio of the Area of △OMN : Area of △ORQ = 4 : 25.

(i) Given,

$$\begin{aligned}\frac{AP}{PB} = \frac{2}{3} \\[1em] \Rightarrow \frac{AP}{AB - AP} = \frac{2}{3} \\[1em] \Rightarrow 3AP = 2(AB - AP) \\[1em] \Rightarrow 3AP = 2AB - 2AP \\[1em] \Rightarrow 3AP + 2AP = 2AB \\[1em] \Rightarrow 5AP = 2AB \\[1em] \Rightarrow \frac{AP}{AB} = \frac{2}{5}.\end{aligned}$$

Considering △APO and △ABC,

∠ A = ∠ A (Common angles)
∠ APO = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △APO ~ △ABC.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APO}}{\text{Area of △ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{\text{Area of △APO}}{\text{Area of △ABC}} = \frac{2^2}{5^2} \\[1em] \Rightarrow \frac{\text{Area of △APO}}{\text{Area of △ABC}} = \frac{4}{25}.\end{aligned}$$

Hence, the ratio of the area of △APO : area of △ABC = 4 : 25.

(ii) In parallelogram PBCQ opposite sides are equal,

so, PB = QC. Hence, $\frac{AP}{QC} = \frac{AP}{PB} = \frac{2}{3}$.

Considering △APO and △CQO,

∠ AOP = ∠ QOC (Vertically opposite angles)
∠ OAP = ∠ OCQ (Alternate angles are equal)

Hence, by AA axiom △APO ~ △CQO.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APO}}{\text{Area of △CQO}} = \frac{AP^2}{QC^2} \\[1em] \Rightarrow \frac{\text{Area of △APO}}{\text{Area of △CQO}} = \Big(\frac{AP}{QC}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of △APO}}{\text{Area of △CQO}} = \Big(\frac{2}{3}\Big)^2 = \frac{4}{9}.\end{aligned}$$

Hence, the ratio of the area of △APO : area of △CQO = 4 : 9.

Given, AB || DC and AB = 2CD.

$\therefore \frac{AB}{CD} = \frac{2}{1}$.

Considering △AOB and △COD,

∠ AOB = ∠ COD (Vertically opposite angles)
∠ OCD = ∠ OAB (Alternate angles are equal)

Hence, by AA axiom △AOB ~ △COD.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △AOB}}{\text{Area of △COD}} = \frac{AB^2}{CD^2} \\[1em] \Rightarrow \frac{\text{Area of △AOB}}{\text{Area of △COD}} = \frac{2^2}{1^2} \\[1em] \Rightarrow \frac{\text{Area of △AOB}}{\text{Area of △COD}} = \frac{4}{1}.\end{aligned}$$

Hence, the ratio of the area of △AOB : area of △COD = 4 : 1.

(i) Given, AM = 6 cm and AN = 10 cm and area of parallelogram ABCD is 45 cm².

Area of parallelogram = base x height = CD x AM = BC x AN.

$$\begin{aligned}\therefore AM \times CD = 45 \\[1em] \Rightarrow 6 \times CD = 45 \\[1em] \Rightarrow CD = \frac{45}{6} \\[1em] \Rightarrow CD = \frac{15}{2} \\[1em] \Rightarrow CD = 7.5\end{aligned}$$

In parallelogram AB = CD = 7.5 cm.

Hence, the length of AB = 7.5 cm.

(ii) Given, AM = 6 cm and AN = 10 cm and area of parallelogram ABCD is 45 cm².

Area of parallelogram = base x height = CD x AM = BC x AN.

$$\begin{aligned}\therefore AN \times BC = 45 \\[1em] \Rightarrow 10 \times BC = 45 \\[1em] \Rightarrow BC = \frac{45}{10} \\[1em] \Rightarrow BC = 4.5\end{aligned}$$

Hence, the length of BC = 4.5 cm.

(iii) Considering △ADM and △ABN,

∠ ADM = ∠ ABN (Opposite angles of a parallelogram are equal)
∠ AMD = ∠ ANB (Both angles are equal to 90°)

Hence, by AA axiom △ADM ~ △ANB.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ADM}}{\text{Area of △ANB}} = \frac{AM^2}{AN^2} \\[1em] \Rightarrow \frac{\text{Area of △ADM}}{\text{Area of △ANB}} = \frac{6^2}{10^2} \\[1em] \Rightarrow \frac{\text{Area of △ADM}}{\text{Area of △ANB}} = \frac{36}{100} \\[1em] \Rightarrow \frac{\text{Area of △ADM}}{\text{Area of △ANB}} = \frac{9}{25}\end{aligned}$$

Hence, the ratio of the area of △ADM : area of △ANB = 9 : 25.

(i) Considering △ADB and △EFB,

∠ B = ∠ B (Common angles)
∠ DAB = ∠ FEB (Corresponding angles are equal)

Hence, by AA axiom △ADB ~ △EFB.

We know that when triangles are similar the ratio of the corresponding sides are equal,

$$\begin{aligned}\therefore \frac{AB}{BE} = \frac{AD}{EF} \\[1em] \Rightarrow \frac{EF}{AD} = \frac{EB}{AB} \\[1em]\end{aligned}$$

Given, $\frac{AE}{EB} = \frac{2}{3}$.

$$\begin{aligned}\Rightarrow \frac{AB- EB}{EB} = \frac{2}{3} \\[1em] \Rightarrow 3(AB - EB) = 2EB \\[1em] \Rightarrow 3AB - 3EB = 2EB \\[1em] \Rightarrow 3AB = 2EB + 3EB \\[1em] \Rightarrow 3AB = 5EB \\[1em] \Rightarrow \frac{EB}{AB} = \frac{3}{5}.\end{aligned}$$

Since, $\Rightarrow \frac{EF}{AD} = \frac{EB}{AB} = \frac{3}{5}.$

Hence, EF : AD = 3 : 5.

(ii) Considering △ABD and △BEF,

∠ B = ∠ B (Common angles)
∠ DAB = ∠ FEB (Corresponding angles are equal)

Hence, by AA axiom △ADB ~ △BEF.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △BEF}}{\text{Area of △ABD}} = \frac{EF^2}{AD^2} \\[1em] \Rightarrow \frac{\text{Area of △BEF}}{\text{Area of △ABD}} = \frac{3^2}{5^2} \\[1em] \Rightarrow \frac{\text{Area of △BEF}}{\text{Area of △ABD}} = \frac{9}{25}.\end{aligned}$$

Hence, area of △BEF : area of △ABD = 9 : 25.

(iii) From (ii) we get,

$\therefore \frac{\text{Area of △ABD}}{\text{Area of △BEF}} = \frac{25}{9}$

⇒ 9 x Area of △ABD = 25 x Area of △BEF
⇒ 9 x Area of △ABD = 25 x (Area of △ABD - Area of trapezium AEFD)
⇒ 9 x Area of △ABD = 25 x Area of △ABD - 25 x Area of trapezium AEFD
⇒ 16 x Area of △ABD = 25 x Area of trapezium AEFD

$\therefore \frac{\text{Area of △ABD}}{\text{Area of trapezium AEFD}} = \frac{25}{16}$

Hence, area of △ABD : area of trapezium AEFD = 25 : 16.

(iv) Considering △FEO and △OBC,

∠ FOE = ∠ BOC (Vertically opposite angles are equal)
∠ FEO = ∠ OCB (Alternate angles are equal)

Hence, by AA axiom △FEO ~ △OBC.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \frac{\text{Area of △FEO}}{\text{Area of △OBC}} = \Big(\frac{EF}{BC}\Big)^2$

Since, in parallelogram opposite sides are equal so, BC = AD.

$$\begin{aligned}\Rightarrow \frac{\text{Area of △FEO}}{\text{Area of △OBC}} = \Big(\frac{EF}{AD}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of △FEO}}{\text{Area of △OBC}} = \frac{3^2}{5^2} \\[1em] \Rightarrow \frac{\text{Area of △FEO}}{\text{Area of △OBC}} = \frac{9}{25}.\end{aligned}$$

Hence, the ratio of the area of △FEO : area of △OBC = 9 : 25.

(i) Draw QN ⊥ CB as shown in the figure below:

$$\begin{aligned}\Rightarrow \frac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \frac{\frac{1}{2} BP \times QN}{\frac{1}{2} PC \times QN} \\[1em] \Rightarrow \frac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \frac{BP}{PC} \\[1em] \Rightarrow \frac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \frac{1}{2} \\[1em] \therefore \text{Area of △BPQ} = \frac{1}{2}\text{Area of △CPQ} = \frac{1}{2} \times 20 = 10 \text{cm}^2.\end{aligned}$$

Hence, the area of △BPQ = 10 cm².

(ii) Considering △CDP and △BQP,

∠CPD = ∠QPB (Vertically opposite angles are equal)
∠PDC = ∠PQB (Alternate angles are equal)

Hence, by AA axiom △CDP ~ △BQP.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △CDP}}{\text{Area of △BQP}} = \frac{PC^2}{BP^2} \\[1em] \Rightarrow \frac{\text{Area of △CDP}}{\text{Area of △BQP}} = \frac{2^2}{1^2} \\[1em] \Rightarrow \frac{\text{Area of △CDP}}{\text{Area of △BQP}} = \frac{4}{1}\end{aligned}$$

∴ Area of △CDP = 4 × Area of △BQP = 4 × 10 = 40 cm².

Hence, the area of △CDP = 40 cm².

(iii) Area of ||gm ABCD = 2 Area of △DCQ (As △DCQ and ||gm ABCD have same base and are between same parallels)

$$\begin{aligned}= 2(\text{Area of △CDP + Area of △CPQ}) \\[1em] = 2(40 + 20) \\[1em] = 2 \times 60 \\[1em] = 120 \text{ cm}^2.\end{aligned}$$

Hence, the area of ||gm = 120 cm².

Given, ratio of the areas of △ADE and trapezium DBCE = 4 : 5.

$$\begin{aligned}\therefore \frac{\text{Area of △ADE}}{\text{Area of trapezium DBCE}} = \frac{4}{5} \\[1em] \Rightarrow \frac{\text{Area of △ADE}}{\text{Area of △ABC - Area of △ADE}} = \frac{4}{5} \\[1em] \Rightarrow 5 \text{ Area of △ADE} = 4 (\text{Area of △ABC - Area of △ADE}) \\[1em] \Rightarrow 9 \text{ Area of △ADE} = 4 \text{Area of △ABC} \\[1em] \Rightarrow \frac{\text{Area of △ADE}}{\text{Area of △ABC}} = \frac{4}{9}.\end{aligned}$$

Considering △ABC and △ADE,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ADE}}{\text{Area of △ABC}} = \frac{DE^2}{BC^2} \\[1em] \Rightarrow \frac{4}{9} = \Big(\frac{DE}{BC}\Big)^2 \\[1em] \Rightarrow \frac{DE}{BC} = \sqrt{\frac{4}{9}} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{2}{3}.\end{aligned}$$

Hence, the ratio of DE : BC is 2 : 3.

(i) Given, AB = 2DC or, $\frac{AB}{DC} = \frac{2}{1}.$

Considering △AEB and △EDC.

∠E = ∠E (Common angles)

∠EDC = ∠EAB (Corresponding angles are equal)

Hence, by AA axiom △AEB ~ △EDC.

Since triangles are similar, hence the ratio of the corresponding sides will be equal

$$\begin{aligned}\therefore \frac{AE}{ED} = \frac{AB}{DC} \\[1em] \Rightarrow \frac{AD + ED}{ED} = \frac{2}{1} \\[1em] \Rightarrow \frac{3 + ED}{ED} = \frac{2}{1} \\[1em] \Rightarrow 3 + ED = 2ED \\[1em] \Rightarrow ED = 3.\end{aligned}$$

Hence, the length of ED = 3 cm.

(ii) Since, △AEB ~ △EDC. Hence the ratio of the corresponding sides will be equal

$$\begin{aligned}\therefore \frac{BE}{EC} = \frac{AB}{DC} \\[1em] \Rightarrow \frac{BC + EC}{EC} = \frac{2}{1} \\[1em] \Rightarrow \frac{4 + EC}{EC} = \frac{2}{1} \\[1em] \Rightarrow 4 + EC = 2EC \\[1em] \Rightarrow EC = 4.\end{aligned}$$

BE = BC + EC = 4 + 4 = 8 cm.

Hence, the length of BE = 8 cm.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △EDC}}{\text{Area of △AEB}} = \frac{DC^2}{AB^2} \\[1em] \Rightarrow \frac{\text{Area of △EDC}}{\text{Area of △AEB}} = \frac{1^2}{2^2} \\[1em] \Rightarrow \frac{\text{Area of △EDC}}{\text{Area of △AEB}} = \frac{1}{4} \\[1em] \Rightarrow \frac{\text{Area of △EDC}}{\text{Area of △EDC + Area of ⏢ABCD}} = \frac{1}{4} \\[1em] \Rightarrow 4 \text{Area of △EDC} = \text{Area of △EDC + Area of ⏢ABCD} \\[1em] \Rightarrow 4 \text{Area of △EDC} - \text{Area of △EDC} = \text { Area of ⏢ABCD} \\[1em] \Rightarrow 3 \text{ Area of △EDC} = \text { Area of ⏢ABCD} \\[1em] \Rightarrow \frac{\text{ Area of △EDC}}{\text { Area of ⏢ABCD}} = \frac{1}{3}.\end{aligned}$$

Hence, the ratio of area of △EDC : area of trapezium ABCD = 1 : 3.

(i) Considering △APB and △CPD.

∠APB = ∠CPD (Vertical opposite angles are equal)

∠PAB = ∠PCD (Alternate angles are equal)

Hence, by AA axiom △APB ~ △CPD.

Since triangles are similar, hence the ratio of the corresponding sides will be equal

$$\begin{aligned}\therefore \frac{BP}{PD} = \frac{AB}{CD} \\[1em] \Rightarrow \frac{BP}{BD - BP} = \frac{9}{6} \\[1em] \Rightarrow \frac{BP}{12 - BP} = \frac{9}{6} \\[1em] \Rightarrow 6 BP = 9(12 - BP) \\[1em] \Rightarrow 6BP = 108 - 9BP \\[1em] \Rightarrow 15 BP = 108 \\[1em] \Rightarrow BP = \frac{108}{15} \\[1em] \Rightarrow BP = 7.2 \text{ cm}.\end{aligned}$$

Hence, the length of BP = 7.2 cm

(ii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APB}}{\text{Area of △DPC}} = \frac{AB^2}{CD^2} \\[1em] \Rightarrow \frac{\text{Area of △APB}}{\text{Area of △DPC}} = \frac{9^2}{6^2} \\[1em] \Rightarrow \frac{\text{Area of △APB}}{\text{Area of △DPC}} = \frac{81}{36} = \frac{9}{4}.\end{aligned}$$

Hence, area of △APB : area of △DPC = 9 : 4.

(i) Considering △ACD and △BCA.

∠C = ∠C (Common angles)

∠ABC = ∠DAC (Given)

Hence, by AA axiom △ACD ~ △BCA.

(ii) Since triangles are similar, hence the ratio of corresponding sides will be equal

$$\begin{aligned}\Rightarrow \frac{AC}{BC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{4}{BC} = \frac{5}{8} \\[1em] \Rightarrow BC = \frac{4 \times 8}{5} \\[1em] \Rightarrow BC = \frac{32}{5} \\[1em] \Rightarrow BC = 6.4 \text{ cm}.\end{aligned}$$

Similarly,

$$\begin{aligned}\Rightarrow \frac{CD}{CA} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{CD}{4} = \frac{5}{8} \\[1em] \Rightarrow CD = \frac{4 \times 5}{8} \\[1em] \Rightarrow CD = \frac{20}{8} \\[1em] \Rightarrow CD = 2.5 \text{ cm}.\end{aligned}$$

Hence, the length of BC = 6.4 cm and CD = 2.5 cm.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ACD}}{\text{Area of △ABC}} = \frac{AD^2}{AB^2} \\[1em] \Rightarrow \frac{\text{Area of △ADC}}{\text{Area of △ABC}} = \frac{5^2}{8^2} \\[1em] \Rightarrow \frac{\text{Area of △ADC}}{\text{Area of △ABC}} = \frac{25}{64}.\end{aligned}$$

Hence, the ratio of area of △ACD : area of △ABC = 25 : 64.

(i) Considering △PQR and △PST.

∠P = ∠P (Common angles)

∠PQR = ∠PST (Both are equal to 90°)

Hence, by AA axiom △PQR ~ △PST.

(ii) $\frac{\text{Area of △PQR}}{\text{Area of △PST}} = \frac{PQ^2}{PS^2} = \frac{5^2}{2^2} = \frac{25}{4}$

or,

$\Rightarrow \frac{\text{Area of △PQR}}{\text{Area of △PQR - Area of SRQT}} = \frac{25}{4}$

⇒ 4 Area of △PQR = 25 Area of △PQR - 25 Area of SRQT
⇒ 25 Area of SRQT = 25 Area of △PQR - 4 Area of △PQR
⇒ 25 Area of SRQT = 21 Area of △PQR

$\Rightarrow \frac{\text{Area of △PQR}}{\text{Area of SRQT}} = \frac{25}{21}$

Hence, area of △PQR : area of quadrilateral SRQT is 25 : 21.

(i) Considering △ADE and △ACB.

∠A = ∠A (Common angles)

∠AED = ∠ABC (Both are equal to 90°)

Hence, by AA axiom △ADE ~ △ACB.

(ii) △ABC is a right angled triangle.

By pythagoras theorem,

$$\begin{aligned}AC^2 = AB^2 + BC^2 \\[1em] AB^2 = AC^2 - BC^2 \\[1em] AB^2 = 13^2 - 5^2 \\[1em] AB^2 = 169 - 25 \\[1em] AB^2 = 144 \\[1em] AB = \sqrt{144} \\[1em] AB = 12 \text{ cm}.\end{aligned}$$

Since triangles are similar hence the ratio of their corresponding sides are equal.

$$\begin{aligned}\therefore \frac{AE}{AB} = \frac{AD}{AC} \\[1em] \Rightarrow \frac{4}{12} = \frac{AD}{13} \\[1em] \Rightarrow AD = \frac{4 \times 13}{12} \\[1em] \Rightarrow AD = \frac{13}{3} \\[1em] \Rightarrow AD = 4\frac{1}{3}.\end{aligned}$$

Similarly,

$$\begin{aligned}\therefore \frac{AE}{AB} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{4}{12} = \frac{DE}{5} \\[1em] \Rightarrow DE = \frac{4 \times 5}{12} \\[1em] \Rightarrow DE = \frac{5}{3} \\[1em] \Rightarrow DE = 1\frac{2}{3}.\end{aligned}$$

Hence, the length of AD = $4\frac{1}{3}$ cm and of DE = $1\frac{2}{3}$ cm.

(iii) Area of a right angled triangle is given by

$\frac{1}{2} \times \text{Base} \times \text{Height}$.

Area of △ADE = $\frac{1}{2} \times AE \times DE$

$$\begin{aligned}= \frac{1}{2} \times 4 \times \frac{5}{3} \\[1em] = \frac{10}{3} \text{ cm}^2.\end{aligned}$$

Area of quadrilateral BCED = Area of △ABC - Area of △ADE

$$\begin{aligned}= \frac{1}{2} \times BC \times AB - \frac{10}{3} \\[1em] = \frac{1}{2} \times 5 \times 12 - \frac{10}{3} \\[1em] = 30 - \frac{10}{3} \\[1em] = \frac{90 - 10}{3} \\[1em] = \frac{80}{3} \text{ cm}^2\end{aligned}$$

Hence, area of △ADE : area of quadrilateral BCED is

$$\begin{aligned}= \frac{\frac{10}{3}}{\frac{80}{3}} \\[1em] = \frac{10 \times 3}{80 \times 3} \\[1em] = \frac{1}{8} \\[1em] = 1 : 8.\end{aligned}$$

Hence, area of △ADE : area of quadrilateral BCED is 1 : 8.

Let their be two isosceles triangles ABC and DEF.

∠A = ∠D (Given, vertical angles are equal)

Since, triangles are isosceles so,

∠B = ∠C = $\frac{180 - ∠A}{2}$ and ∠E = ∠F = $\frac{180 - ∠D}{2}$.

Since, ∠A = ∠D so, we can say

∠B = ∠E.

Hence, by AA axiom △ABC ~ △DEF.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △DEF}} = \frac{(\text{Height of △ABC})^2}{(\text{Height of △DEF})^2} \\[1em] \Rightarrow \frac{7}{16} = \Big(\frac{\text{Height of △ABC}}{\text{Height of △DEF}}\Big)^2 \\[1em] \Rightarrow \frac{\text{Height of △ABC}}{\text{Height of △DEF}} = \sqrt{\frac{7}{16}} \\[1em] \Rightarrow \frac{\text{Height of △ABC}}{\text{Height of △DEF}} = \frac{\sqrt{7}}{4}.\end{aligned}$$

Hence, the ratio of their corresponding heights is $\sqrt{7} : 4$.

(i) Since, the model of the triangular plot is made to the scale of 1 : 250000.

∴ K (Scale factor) = 250000.

Actual length of AB = k × (the length of AB in model) = 250000 × 3 = 750000 cm.

1 cm = 10^-5 km.

∴ 750000 cm = 750000 × 10^-5 km = 7.5 km.

Hence, the length of AB is 7.5 km.

(ii) Since the plot is a right angled triangle.

We know area of right angled triangle is given by

$\Rightarrow \frac{1}{2} \times \text{ base } \times \text{ height}.$

Hence, area of the model is,

$$\begin{aligned}= \frac{1}{2} \times AB \times BC \\[1em] = \frac{1}{2} \times 3 \times 4 \\[1em] = 6 \text{ cm}^2.\end{aligned}$$

We know that 1 cm² = 10^-10 km².

Hence, area of model = 6 × 10^-10 km².

Area of the actual plot = k² × (Area of the model)

Putting values in above equation,

$$\begin{aligned}= (250000)^2 \times 6 \times 10^{-10} \\[1em] = 625 \times 10^{8} \times 6 \times 10^{-10} \\[1em] = 3750 \times 10^{-2.} \\[1em] = 37.5 \text{ km}^2.\end{aligned}$$

Hence, the area of the plot is 37.5 km².

Since map is drawn to a scale of 1 : 50000.

∴ k (Scale factor) = 50000.

Length of the diagonal AC of the rectangle can be given by pythagoras theorem i.e. $\sqrt{AB^2 + BC^2}$.

Putting values we get,

$$\begin{aligned}\Rightarrow AC = \sqrt{AB^2 + BC^2} \\[1em] \Rightarrow AC = \sqrt{6^2 + 8^2} \\[1em] \Rightarrow AC = \sqrt{36 + 64} \\[1em] \Rightarrow AC = \sqrt{100} \\[1em] \Rightarrow AC = 10 \text{ cm}.\end{aligned}$$

Actual length of diagonal = k × length of diagonal in model.

$$\begin{aligned}= 50000 \times 10 \\[1em] = 500000 \text{ cm} \\[1em] = 500000 \times 10^{-5} \text{ km} \\[1em] = 5 \text{ km}\end{aligned}$$

Hence, actual length of diagonal = 5 km.

(ii) Area of the model ABCD = AB × BC = 6 × 8 = 48 cm².

Area of the actual plot = k² × (Area of the model)
= (50000)² x 48
= 25 x 10⁸ x 48
= 1200 x 10⁸
= 12 x 10¹⁰ cm²

We know that 1 cm² = 10^-10 km².

∴ Actual area of plot = 12 × 10¹⁰ × 10^-10 km² = 12 km².

Hence, the actual area of the plot is 12 km².

(i) Since, the model of the square plot is constructed with scale of 1 : 25000.

k (Scale factor) = 25000.

Area of the actual plot = k² × (Area of the model of the plot)

Given, area of the model = 72 cm². Putting values in above equation,

$$\begin{aligned}= (25000)^2 \times 72 \\[1em] = 625000000 \times 72 \\[1em] = 45000000000 \\[1em] = 45 \times 10^9 \text{ cm}^2.\end{aligned}$$

We know that 1 cm² = 10^-10 km².

∴ Actual area of plot = 45 × 10⁹ × 10^-10 km² = 4.5 km².

Hence, the actual area of the plot is 4.5 km².

(ii) We know that,

$\frac{1}{2}$ (length of diagonal)² = area of square.

Putting value of area of square plot = 4.5 km² in above equation we get,

$$\begin{aligned}\Rightarrow \frac{1}{2} \text{ (Length of diagonal)}^2 = 4.5 \\[1em] \Rightarrow \text{(Length of diagonal)}^2 = 9 \\[1em] \Rightarrow \text{Length of diagonal} = \sqrt{9} \\[1em] \Rightarrow \text{ Length of diagonal} = 3.\end{aligned}$$

Hence, the length of diagonal in the actual plot of land is 3 km.

(i) Since, the model of the building is constructed with scale 1 : 30.

∴ k (Scale factor) = 30

Height of building = k × Height of model of the building = 30 × 80 = 2400 cm = $\frac{2400}{100}$ m = 24 m.

Hence, the height of building is 24 m.

(ii) Volume of the tank = k³ × (the volume of the model)

Given, volume of tank = 27 m³. Let volume of model be x m³. Putting value in above equation we get,

$$\begin{aligned}\Rightarrow 27 = (30)^3 \times x \\[1em] \Rightarrow x = \frac{27}{30 \times 30 \times 30} \\[1em] \Rightarrow x = \frac{27}{27000} \\[1em] \Rightarrow x = \frac{1}{1000}.\end{aligned}$$

∴ x = $\frac{1}{1000} \text{ m}^3 = \frac{1}{1000} \times (100 \text{ cm})^3 = \frac{1000000}{1000} \text{ cm}^3 = 1000 \text{ cm}^3$.

Hence, the volume of the model is 1000 cm³.

(i) Given, the height of the model = 0.8 m

Since, the model of the building is constructed with scale 1 : 50.

∴ k (Scale factor) = 50

⇒ Height of building = k × Height of model of the building = 50 × 0.8 = 40 m

Hence, height of building = 40 m.

(ii) The floor area of a flat = k² x the floor area of a model flat

⇒ 20 = 50² x Floor area of a model flat

⇒ 20 = 2500 x Floor area of a model flat

⇒ Floor area of a model flat = $\frac{20}{2500}$ = 0.008 m²

Hence, the floor area of that in the model = 0.008 m².

(i) Since, the model of the ship is made to the scale of 1 : 200.

∴ K (Scale factor) = 200.

Actual length of the ship = k × (the length of model) = 200 × 4 = 800 m.

Hence, the length of the ship is 800 m.

(ii) Area of the deck of the ship = k² × (Area of the deck of the model)

Let area of deck of model be x m².

$$\begin{aligned}\Rightarrow 160000 = (200)^2 \times x \\[1em] \Rightarrow x = \frac{160000}{200 \times 200} \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, the area of the deck of the ship is 4 m².

(iii) Volume of the ship = k³ × (the volume of the model)

Given, volume of model = 200 litres = $\frac{200}{1000} m^3 = 0.2 m^3$.

Putting value in above equation we get,

$$\begin{aligned}= (200)^3 \times 0.2 \\[1em] = 8000000 \times 0.2 \\[1em] = 1600000\end{aligned}$$

Hence, the volume of the model of the ship is 1600000 m³.

Given, △ABC ~ △QPR

∴ ∠A = ∠Q, ∠B = ∠P and ∠C = ∠R

∠C = 180° - (70° + 50°) = 180° - 120° = 60°.

∴ ∠R = 60°.

Hence, Option 1 is the correct option.

Since triangles are similar hence the ratio of their corresponding sides are equal.

$$\begin{aligned}\therefore \frac{AC}{QR} = \frac{BC}{PR} \\[1em] \Rightarrow \frac{6}{3} = \frac{4.5}{x} \\[1em] \Rightarrow 2 = \frac{4.5}{x} \\[1em] \Rightarrow x = \frac{4.5}{2} \\[1em] \Rightarrow x = 2.25\end{aligned}$$

Hence, Option 1 is the correct option.

Considering △APB and △CPD,

$\frac{AP}{PD} = \frac{6}{5} \text{ and } \frac{BP}{CP} = \frac{3}{2.5} = \frac{6}{5}.$ and ∠APB = ∠CPD (Vertically opposite angles are equal)

∴ △APB ~ △CPD

Hence, ∠PAB = ∠PDC = 30°

∠PBA = 180° - (∠PAB + ∠APB) = 180° - (30° + 50°) = 180° - 80° = 100°.

Hence, Option 4 is the correct option.

Given,

∠B = ∠E, ∠F = ∠C and AB = 3DE.

∴ Two angles of the one triangles are equal to corresponding two angles of the other, but sides are not equal.

∴ Triangles are similar but not congruent.

Hence, Option 2 is the correct option.

Given AB || DE.

Considering △ABC and △DEC,

∠ ACB = ∠ DCE (Vertically opposite angles are equal)
∠ ABC = ∠ CDE (Alternate angles are equal)

Hence, by AA axiom △ABC ~ △DEC.

Since, both the triangles are similar, hence ratio of their corresponding sides are equal,

$$\begin{aligned}\therefore \frac{AB}{DE} = \frac{BC}{CD} \\[1em] \Rightarrow \frac{5}{3} = \frac{4.5}{CD} \\[1em] \Rightarrow CD = \frac{3 \times 4.5}{5} \\[1em] \Rightarrow CD = 2.7\end{aligned}$$

Hence, Option 2 is the correct option.

Let perimeter of △PQR be x cm.

Since triangles are similar,

$$\begin{aligned}\therefore \frac{\text{PQ}}{\text{AB}} = \frac{\text{Perimeter of △PQR}}{\text{Perimeter of △ABC}} \\[1em] \Rightarrow \frac{6}{8} = \frac{x}{36} \\[1em] \Rightarrow x = \frac{6 \times 36}{8} \\[1em] \Rightarrow x = 27.\end{aligned}$$

∴ Perimeter of △PQR = 27 cm.

Hence, Option 2 is the correct option.

Given DE || BC.

Considering △ABC and △ADE,

∠ A = ∠ A (Common angles)
∠ ABC = ∠ ADE (Corresponding angles are equal)

Hence, by AA axiom △ABC ~ △ADE.

Let the length of AE be x cm.

Since, both the triangles are similar,

$$\begin{aligned}\therefore \frac{AD}{AB} = \frac{AE}{AC} \\[1em] \Rightarrow \frac{AD}{AD + DB} = \frac{AE}{AE + EC} \\[1em] \Rightarrow \frac{3}{3 + 4} = \frac{x}{x + 3} \\[1em] \Rightarrow \frac{3}{7} = \frac{x}{x + 3} \\[1em] \Rightarrow 3(x + 3) = 7x \\[1em] \Rightarrow 3x + 9 = 7x \\[1em] \Rightarrow 4x = 9 \\[1em] \Rightarrow x = 2.25\end{aligned}$$

∴ Length of AE = 2.25 cm.

Hence, Option 2 is the correct option.

Given PQ || CA.

Considering △ABC and △PBQ,

∠ B = ∠ B (Common angles)
∠ CAB = ∠ QPB (Corresponding angles are equal)

Hence, by AA axiom △ABC ~ △PBQ.

Let length of QC be x cm.

Since triangles are similar,

$$\begin{aligned}\therefore \frac{BQ}{BC} = \frac{BP}{BA} \\[1em] \Rightarrow \frac{BQ}{BQ + QC} = \frac{BP}{BP + PA} \\[1em] \Rightarrow \frac{5}{5 + x} = \frac{4}{4 + 2.4} \\[1em] \Rightarrow \frac{5 \times 6.4}{4} = x + 5 \\[1em] \Rightarrow 8 = x + 5 \\[1em] \Rightarrow x = 3.\end{aligned}$$

BC = BQ + QC = 5 + x = 5 + 3 = 8 cm.

Hence, Option 3 is the correct option.

Given MN || QR.

Considering △PMN and △PQR,

∠P = ∠P (Common angles)
∠PMN = ∠PQR (Corresponding angles are equal)

Hence, by AA axiom △PMN ~ △PQR.

Let length of PM be x cm.

Since triangles are similar by basic proportionality theorem,

$$\begin{aligned}\therefore \frac{PM}{MQ} = \frac{PN}{NR} \\[1em] \Rightarrow \frac{PM}{PQ - PM} = \frac{PN}{NR} \\[1em] \Rightarrow \frac{x}{5 - x} = \frac{3.6}{2.4} \\[1em] \Rightarrow \frac{x}{5 - x} = \frac{36}{24} \\[1em] \Rightarrow 24x = 36(5 - x) \\[1em] \Rightarrow 24x = 180 - 36x \\[1em] \Rightarrow 24x + 36x = 180 \\[1em] \Rightarrow 60x = 180 \\[1em] \Rightarrow x = 3.\end{aligned}$$

∴ Length of PM = 3 cm.

Hence, Option 4 is the correct option.

Given $\frac{BC}{QR} = \frac{1}{3}$

So, $\frac{QR}{BC} = \frac{3}{1}$.

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △PQR}}{\text{Area of △ABC}} = \frac{QR^2}{BC^2} \\[1em] = \frac{3^2}{1^2} \\[1em] = \frac{9}{1} \\[1em] = 9.\end{aligned}$$

Hence, Option 1 is the correct option.

Given, ratio of the areas of the two similar triangles = 4 : 9

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\text{Ratio of corresponding sides } = \sqrt{\text{Ratio of the areas of similar triangles}} = \sqrt{4} : \sqrt{9}$ = 2 : 3.

Hence, Option 3 is the correct option.

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △PQR}} = \frac{BC^2}{QR^2} \\[1em] = \frac{8^2}{6^2} \\[1em] = \frac{64}{36} \\[1em] = \frac{16}{9} \\[1em] = 16 : 9.\end{aligned}$$

Hence, Option 4 is the correct option.

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \frac{\text{Area of △ABC}}{\text{Area of △PQR}} = \frac{BC^2}{PR^2}$

Given, $\frac{\text{area of △ABC}}{\text{area of △PQR}} = \frac{9}{4}$.

So,

$$\begin{aligned}\Rightarrow \frac{BC^2}{PR^2} = \frac{9}{4} \\[1em] \Rightarrow \frac{15^2}{PR^2} = \frac{9}{4} \\[1em] \Rightarrow PR^2 = \frac{225 \times 4}{9} \\[1em] \Rightarrow PR^2 = 100 \\[1em] \Rightarrow PR = \sqrt{100} \\[1em] \Rightarrow PR = 10.\end{aligned}$$

∴ PR = 10 cm.

Hence, Option 1 is the correct option.

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △PQR}} = \frac{BC^2}{QR^2} \\[1em] \Rightarrow \frac{BC^2}{6^2} = \frac{81}{144} \\[1em] \Rightarrow BC^2 = \frac{81 \times 36}{144} \\[1em] \Rightarrow BC^2 = \frac{2916}{144} \\[1em] \Rightarrow BC^2 = 20.25 \\[1em] \Rightarrow BC = \sqrt{20.25} \\[1em] \Rightarrow BC = 4.5\end{aligned}$$

∴ BC = 4.5 cm.

Hence, Option 2 is the correct option.

Given DE || CA.

Considering △BDE and △BCA,

∠B = ∠B (Common angles)
∠BDE = ∠BCA (Corresponding angles are equal)

Hence, by AA axiom △BDE ~ △BCA.

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \frac{\text{Area of △ABC}}{\text{Area of △BDE}} = \frac{BC^2}{BD^2}$ .....(i)

Given,

$$\begin{aligned}\Rightarrow \frac{BD}{DC} = \frac{2}{1} \\[1em] \Rightarrow \frac{BD}{BC - BD} = \frac{2}{1} \\[1em] \Rightarrow BD = 2(BC - BD) \\[1em] \Rightarrow BD = 2BC - 2BD \\[1em] \Rightarrow BD + 2BD = 2BC \\[1em] \Rightarrow 3BD = 2BC \\[1em] \Rightarrow \frac{BC}{BD} = \frac{3}{2}.\end{aligned}$$

Putting this value in (i) we get,

$$\begin{aligned}\Rightarrow \frac{\text{Area of △ABC}}{\text{Area of △BDE}} = \frac{BC^2}{BD^2} \\[1em] = \frac{3^2}{2^2} \\[1em] = \frac{9}{4} \\[1em] = 9 : 4.\end{aligned}$$

Hence, Option 3 is the correct option.

Since triangles ABC and BDE are equilateral triangles so, each angle will be equal to 60°.

Since all angles are equal to 60°.

Hence, by AAA axiom △ABC ~ △BDE.

Since D is the midpoint of BC so,

$$\begin{aligned}\Rightarrow BC = 2BD \\[1em] \Rightarrow \frac{BC}{BD} = \frac{2}{1}.\end{aligned}$$

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △BDE}} = \frac{BC^2}{BD^2} \\[1em] = \frac{2^2}{1^2} \\[1em] = \frac{4}{1} \\[1em] = 4 : 1.\end{aligned}$$

Hence, Option 4 is the correct option.

Let the altitude of bigger triangle be x cm.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding altitudes.

$$\begin{aligned}\therefore \frac{\text{Area of bigger triangle}}{\text{Area of smaller triangle}} = \frac{(\text{Bigger triangle altitude})^2}{(\text{Smaller triangle altitude})^2} \\[1em] \Rightarrow \frac{81}{49} = \frac{x^2}{(3.5)^2} \\[1em] \Rightarrow x^2 = \frac{81 \times (3.5)^2}{49} \\[1em] \Rightarrow x^2 = \frac{992.25}{49} \\[1em] \Rightarrow x^2 = 20.25 \\[1em] \Rightarrow x = \sqrt{20.25} \\[1em] \Rightarrow x = 4.5\end{aligned}$$

Hence, altitude of the bigger triangle is 4.5 cm.

Hence, Option 4 is the correct option.

Given, △ABC ~ △PQR.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding medians.

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △PQR}} = \frac{AD^2}{PM^2} \\[1em] \Rightarrow \frac{54}{24} = \frac{x^2}{10^2} \\[1em] \Rightarrow x^2 = \frac{54 \times 100}{24} \\[1em] \Rightarrow x^2 = \frac{5400}{24} \\[1em] \Rightarrow x^2 = 225 \\[1em] \Rightarrow x = 15.\end{aligned}$$

Hence, length of AD = 15 cm.

Hence, Option 3 is the correct option.

Given,

△ ABC ~ △ PQR

⇒ ∠B = ∠Q (Corresponding angles of similar triangle are equal)

In △ ABD and △ PQS,

⇒ ∠B = ∠Q (Proved above)

⇒ ∠D = ∠S (Both equal to 90°)

∴ △ ABD ~ △ PQS (By A.A. axiom)

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \frac{AB}{PQ} = \frac{AD}{PS} = \frac{3}{8}$.

∴ AB : PQ = 3 : 8.

Hence, Option 3 is the correct option.

Given,

∆ABC ∼ ∆PQR

∴ ∠A = ∠P

⇒ $\frac{∠A}{2} = \frac{∠P}{2}$

⇒ ∠BAD = ∠QPS

∠B = ∠Q [∵ ∆ABC ∼ ∆PQR]

In ∆ABD ∼ ∆PQS,

⇒ ∠BAD = ∠QPS

⇒ ∠B = ∠Q

∴ ∆ABD ∼ ∆PQS (By A.A. axiom)

Hence, Option 2 is the correct option.

By angle bisector theorem,

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

In a ΔABC, if AD is the internal angle bisector of ∠A, then it divides the opposite side BC in the ratio:

$\frac{BD}{DC} = \frac{AB}{AC}$

So, reason (R) is true.

In a ΔABC, if D is a point on the side BC such that AD divides BC in ratio AB : AC, then AD is the bisector of ∠A.

This is the converse of the Angle Bisector Theorem.

So, assertion (A) is true.

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given Δ ABC ∼ Δ PQR

If area of Δ ABC : area of Δ PQR = 16 : 25

We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of Δ ABC}}{\text{Area of Δ PQR}} = \frac{AB^2}{PQ^2} \\[1em] \Rightarrow \frac{AB^2}{PQ^2} = \frac{16}{25} \\[1em] \Rightarrow \frac{AB}{PQ} = \sqrt{\frac{16}{25}} \\[1em] \Rightarrow \frac{AB}{PQ} = \frac{\sqrt{16}}{\sqrt{25}} \\[1em] \Rightarrow \frac{AB}{PQ} = \frac{4}{5}.\end{aligned}$$

Since, corresponding sides of similar triangle are proportional.

$\therefore \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$

We know that,

For any two or more equal ratios, each ratio is equal to the ratio between sum of their antecedents and sum of their consequents.

$$\begin{aligned}\Rightarrow \frac{AB}{PQ} = \frac{AB + BC + AC}{PQ + QR + PR}\\[1em] \Rightarrow \frac{AB}{PQ} = \frac{\text{Perimeter of Δ ABC}}{\text{Perimeter of Δ PQR}} \\[1em] \Rightarrow \frac{4}{5} = \frac{\text{Perimeter of Δ ABC}}{\text{Perimeter of Δ PQR}} \\[1em]\end{aligned}$$

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given Δ PQR ∼ Δ DEF and PX is median of triangle PQR, DY is median of triangle DEF.

Since Δ PQR ∼ Δ DEF, corresponding sides of similar triangle are proportional.

$$\begin{aligned}\therefore\frac{PQ}{DE} = \frac{QR}{EF}\\[1em] \Rightarrow \frac{PQ}{DE} = \frac{\frac{1}{2}QR}{\frac{1}{2}EF}\\[1em] \Rightarrow \frac{PQ}{DE} = \frac{QX}{EY}\end{aligned}$$

And we also know that corresponding angles of similar triangles are equal.

∴ ∠Q = ∠E

Now, in Δ PQX and Δ DEY,

$\Rightarrow \frac{PQ}{DE} = \frac{QX}{EY}$

⇒ ∠Q = ∠E

Using SAS similarity,

⇒ Δ PQX ∼ Δ DEY

Since, corresponding sides of similar triangle are proportional,

$\Rightarrow \frac{PQ}{DE} = \frac{QX}{EY} = \frac{PX}{DY}$

If two triangles are similar, then the ratio of their areas equals the square of the ratio of their corresponding sides.

$\therefore\frac{\text{area of ΔPQR}}{\text{area of ΔDEF}} = \frac{PQ^2}{DE^2} = \frac{PX^2}{DY^2}$

So, for the similar triangles, the ratio of their corresponding sides is equal to the ratio of their corresponding medians.

So, reason (R) is true.

Given,

area of Δ PQR : area of Δ DEF = 9 : 49

$$\begin{aligned}\Rightarrow \frac{\text{area of ΔPQR}}{\text{area of ΔDEF}} = \frac{9}{49} \\[1em] \Rightarrow \frac{PX^2}{DY^2} = \frac{9}{49}\\[1em] \Rightarrow \frac{PX}{DY} = \frac{\sqrt{9}}{\sqrt{49}}\\[1em] \Rightarrow \frac{PX}{DY} = \frac{3}{7}\end{aligned}$$

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given Δ ABC ∼ Δ DEF.

As we know that the ratio of area of two similar triangle is equal to the ratio of square of their corresponding sides.

So, reason (R) is true.

$$\begin{aligned}\Rightarrow \frac{\text{area of ΔABC}}{\text{area of ΔDEF}} = \frac{BC^2}{EF^2} \\[1em] \Rightarrow \frac{64}{49} = \frac{4^2}{EF^2} \\[1em] \Rightarrow \frac{64}{49} = \frac{16}{EF^2} \\[1em] \Rightarrow EF^2 = \frac{49 \times 16}{64} \\[1em] \Rightarrow EF^2 = \frac{49}{4} \\[1em] \Rightarrow EF = \sqrt{\frac{49}{4}} \\[1em] \Rightarrow EF = \frac{\sqrt{49}}{\sqrt{4}} \\[1em] \Rightarrow EF = \frac{7}{2}.\end{aligned}$$

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given, ∠1 = ∠2

Adding ∠QPT to both the sides,

∠1 + ∠QPT = ∠2 + ∠QPT

∴ ∠SPT = ∠QPR

∠PST = ∠PQR (As ∠3 = ∠4)

Hence, by AA axiom △PQR ~ △PST.

Since, triangles are similar so ratio of their corresponding sides will be equal.

$$\begin{aligned}\Rightarrow \frac{PT}{PR} = \frac{ST}{QR} \\[1em] \Rightarrow PT \times QR = PR \times ST.\end{aligned}$$

Hence, proved that PT × QR = PR × ST.

Considering △PNC and △PMB,

∠PNC = ∠PMB (Both are equal to 90°)

∠NCP = ∠PBM (As AB = AC)

Hence by AA axiom △PNC ~ △PMB.

Since, triangles are similar so ratio of their corresponding sides will be equal.

$$\begin{aligned}\Rightarrow \frac{PC}{PB} = \frac{PN}{PM} \\[1em] \Rightarrow PC \times PM = PN \times PB.\end{aligned}$$

Hence, proved that PC × PM = PN × PB.

Considering △ABC and △ADE,

∠AED = ∠ABC (Given)

∠A = ∠A (Common angles)

Hence by AA axiom △ABC ~ △ADE.

Since, triangles are similar so ratio of their corresponding sides will be equal.

$$\begin{aligned}\Rightarrow \frac{AD}{AC} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{AD}{AE + EC} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{3}{4 + 2} = \frac{y}{10} \\[1em] \Rightarrow \frac{3}{6} = \frac{y}{10} \\[1em] \Rightarrow y = \frac{30}{6} \\[1em] \Rightarrow y = 5.\end{aligned}$$

Similarly,

$$\begin{aligned}\Rightarrow \frac{AB}{AE} = \frac{BC}{DE} \\[1em] \Rightarrow \frac{AD + DB}{AE} = \frac{BC}{DE} \\[1em] \Rightarrow \frac{3 + x}{4} = \frac{10}{y} \\[1em] \Rightarrow \frac{3 + x}{4} = \frac{10}{5} \\[1em] \Rightarrow 3 + x = \frac{40}{5} \\[1em] \Rightarrow 3 + x = 8 \\[1em] \Rightarrow x = 5.\end{aligned}$$

Hence, the value of x = 5 and y = 5.

(i) Considering △FGE and △CGB,

∠FGE = ∠BGC (Vertically opposite angles are equal)

∠GFE = ∠GCB (Alternate angles are equal)

Hence by AA axiom △FGE ~ △CGB.

(ii) Considering △AFE and △ABC,

∠A = ∠A (Common angles)

∠AFE = ∠ABC (Corresponding angles are equal)

Hence by AA axiom △AFE ~ △ABC.

Given BE is the median of AC, so

AE = EC

AC = AE + EC = AE + AE = 2AE.

$\therefore \frac{AE}{AC} = \frac{1}{2}.$

Since, △AFE ~ △ABC, so the ratio of their corresponding sides are equal,

$$\begin{aligned}\therefore \frac{FE}{BC} = \frac{AE}{AC} \\[1em] \Rightarrow \frac{FE}{BC} = \frac{1}{2}.\end{aligned}$$

Since, △FGE ~ △CGB, so the ratio of their corresponding sides are equal,

$$\begin{aligned}\therefore \frac{FE}{BC} = \frac{GE}{BG} \\[1em] \Rightarrow \frac{GE}{BG} = \frac{1}{2} \\[1em] \Rightarrow BG = 2GE.\end{aligned}$$

Hence, proved that BG = 2GE.

(i) Given, AP : PB = 4 : 3.

Since, PQ || AC. By using Basic Proportionality Theorem,

$$\begin{aligned}\Rightarrow \frac{AP}{PB} = \frac{CQ}{QB} \\[1em] \Rightarrow \frac{CQ}{QB} = \frac{4}{3} \\[1em] \Rightarrow \frac{BC - BQ}{BQ} = \frac{4}{3} \\[1em] \Rightarrow 3(BC - BQ) = 4BQ \\[1em] \Rightarrow 3BC - 3BQ = 4BQ \\[1em] \Rightarrow 3BC = 7BQ \\[1em] \Rightarrow \frac{BQ}{BC} = \frac{3}{7} \qquad \text{[....Eq 1]}\end{aligned}$$

Considering △PBQ and △ABC,

∠QPB = ∠CAB (Corresponding angles are equal)

∠PQB = ∠ACB (Corresponding angles are equal)

Hence by AA axiom △PBQ ~ △ABC. Since triangles are similar so the ratio of the corresponding sides are equal,

$$\begin{aligned}\Rightarrow \frac{PQ}{AC} = \frac{BQ}{BC} \\[1em] \Rightarrow \frac{PQ}{AC} = \frac{3}{7} \qquad \text{[From Eq 1]}.\end{aligned}$$

Hence, PQ : AC = 3 : 7.

(ii) Considering △ARC and △QSP,

∠ARC = ∠QSP (Both are equal to 90°)

∠ACR = ∠SPQ (Alternate angles are equal)

Hence by AA axiom △ARC ~ △QSP. Since triangles are similar so the ratio of the corresponding sides are equal,

$$\begin{aligned}\Rightarrow \frac{AR}{QS} = \frac{AC}{PQ} \\[1em] \Rightarrow AR = \frac{AC}{PQ} \times QS \\[1em]\end{aligned}$$

We calculated PQ : AC = 3 : 7 above.

$\therefore \frac{AC}{PQ} = \frac{7}{3}$

Putting this value of $\frac{AC}{PQ}$ we get,

$$\begin{aligned}\Rightarrow AR = \frac{7}{3} \times 6 \\[1em] \Rightarrow AR = 7 \times 2 \\[1em] \Rightarrow AR = 14.\end{aligned}$$

Hence, length of AR = 14 cm.

The below figure shows △ABC and the points D and E on the sides AB and AC respectively:

Considering △ABC and △ADE,

∠A = ∠A (Common angles)

∠ADE = ∠ABC (Corresponding angles are equal)

Hence by AA axiom △ABC ~ △ADE. Since triangles are similar so the ratio of the corresponding sides are equal,

$\therefore \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$

Consider $\frac{AD}{AB} = \frac{DE}{BC}$

$$\begin{aligned}\Rightarrow \frac{2.4}{AB} = \frac{2}{5} \\[1em] \Rightarrow AB = \frac{2.4 \times 5}{2} \\[1em] \Rightarrow AB = \frac{12}{2} \\[1em] \Rightarrow AB = 6.\end{aligned}$$

Now consider $\frac{AE}{AC} = \frac{DE}{BC}$

$$\begin{aligned}\Rightarrow \frac{3.2}{AC} = \frac{2}{5} \\[1em] \Rightarrow AC = \frac{3.2 \times 5}{2} \\[1em] \Rightarrow AC = \frac{16}{2} \\[1em] \Rightarrow AC = 8.\end{aligned}$$

From figure we see that,

⇒ BD = AB - AD = 6 - 2.4 = 3.6 cm.

⇒ CE = AC - AE = 8 - 3.2 = 4.8 cm.

Hence, the length of BD = 3.6 cm and CE = 4.8 cm.

EC = AC - AE = 8.8 - 3.3 = 5.5 cm.

$$\begin{aligned}\frac{AD}{DB} = \frac{5.7}{9.5} \\[1em] = \frac{57}{95} \\[1em] = \frac{3}{5}.\end{aligned}$$

Calculating $\frac{AE}{EC}$,

$$\begin{aligned}\frac{AE}{EC} = \frac{3.3}{5.5} \\[1em] = \frac{33}{55} \\[1em] = \frac{3}{5}.\end{aligned}$$

So, $\frac{AD}{DB} = \frac{AE}{EC}$.

Hence, by basic proportionality theorem DE || BC.

Let the corresponding side of the second triangle be x cm.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of } △_1}{\text{Area of } △_2} = \frac{(\text{Side of } △_1)^2}{(\text{Side of } △_2)^2} \\[1em] \Rightarrow \frac{360}{250} = \frac{8^2}{x^2} \\[1em] \Rightarrow x^2 = \frac{64 \times 250}{360} \\[1em] \Rightarrow x^2 = \frac{16000}{360} \\[1em] \Rightarrow x^2 = \frac{1600}{36} \\[1em] \Rightarrow x = \sqrt{\frac{1600}{36}} \\[1em] \Rightarrow x = \frac{40}{6} \\[1em] \Rightarrow x = \frac{20}{3} = 6\frac{2}{3}.\end{aligned}$$

Hence, the length of corresponding side of second triangle is $6\frac{2}{3}$ cm.

(i) Considering △ABC and △ACD,

∠C = ∠C (Common angles)

∠ABC = ∠CAD (Given)

Hence by AA axiom △ABC ~ △ACD. Since triangles are similar hence the ratio of the corresponding sides will be equal

$$\begin{aligned}\therefore \frac{AB}{AD} = \frac{BC}{AC} \\[1em] \Rightarrow \frac{5}{4} = \frac{BC}{3} \\[1em] \Rightarrow BC = \frac{15}{4} \\[1em] \Rightarrow BC = 3.75\end{aligned}$$

Hence, the length of BC = 3.75 cm.

(ii) Since triangles △ABC and △ACD are similar hence the ratio of the corresponding sides will be equal.

$$\begin{aligned}\therefore \frac{AB}{AD} = \frac{AC}{DC} \\[1em] \Rightarrow \frac{5}{4} = \frac{3}{DC} \\[1em] \Rightarrow DC = \frac{12}{5} \\[1em] \Rightarrow DC = 2.4\end{aligned}$$

Hence, the length of DC = 2.4 cm.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ACD}}{\text{Area of △BCA}} = \frac{AD^2}{AB^2} \\[1em] = \frac{4^2}{5^2} \\[1em] = \frac{16}{25}\end{aligned}$$

Hence, the ratio of area of △ACD : area of △BCA is 16 : 25.

In the figure,

Diagonals of parallelogram ABCD are AC and BD which intersect each other at O. OE is drawn parallel to CB to meet AB in E.

In the figure four triangles have equal area.

So, Area of △OAB = $\frac{1}{4}$ Area of parallelogram ABCD

Then, O is the midpoint of AC as diagonals of parallelogram intersect at centre.
OE || CB. We know that, ABCD is a parallelogram and opposite sides are parallel in parallelogram. Thus OE || AD also,

∴ E is the midpoint of AB.

∴ OE is the median of △AOB.

$$\begin{aligned}\text{Area of △AOE} = \frac{1}{2} \text{Area of △AOB} \\[1em] = \frac{1}{2} \times \frac{1}{4} \text{Area of parallelogram ABCD} \\[1em] = \frac{1}{8} \text{Area of parallelogram ABCD} \\[1em] \therefore \text{Area of △AOE} = \frac{1}{8} \text{Area of parallelogram ABCD} \\[1em] \therefore \frac{\text{Area of △AOE}}{\text{Area of parallelogram ABCD}} = \frac{1}{8}.\end{aligned}$$

Hence, the ratio of area of △AOE : area of ||gm ABCD is 1 : 8.

Given, 2AB = 3DC

$\Rightarrow \frac{AB}{DC} = \frac{3}{2}$.

Considering △AOB and △COD,

∠AOB = ∠COD (Vertically opposite angles are equal)

∠OAB = ∠OCD (Alternate angles are equal)

Hence by AA axiom △AOB ~ △COD.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △AOB}}{\text{Area of △COD}} = \frac{AB^2}{DC^2} \\[1em] = \frac{3^2}{2^2} \\[1em] = \frac{9}{4}\end{aligned}$$

Hence, the ratio of area of △AOB : area of △COD is 9 : 4.

(i) Given E is the mid-point of BC,

∴ 2EC = BC

Since ABCD is a parallelogram so, BC = AD or, 2EC = AD.

Considering △AOD and △EDC,

∠AOD = ∠EOC (Vertically opposite angles are equal)

∠OAD = ∠OCB (Alternate angles are equal)

Hence by AA axiom △AOD ~ △EOC. Since triangles are similar so the ratio of their corresponding sides are equal.

$$\begin{aligned}\therefore \frac{DO}{OE} = \frac{AD}{EC} = \frac{2EC}{EC} \\[1em] = \frac{2}{1} \\[1em]\end{aligned}$$

Hence, proved that DO : OE = 2 : 1.

(ii) From (i) we have proved that △AOD ~ △EOC.

$$\begin{aligned}\therefore \frac{\text{Area of △OEC}}{\text{Area of △AOD}} = \frac{OE^2}{DO^2} \\[1em] = \frac{1^2}{2^2} \\[1em] = \frac{1}{4} \\[1em] = 1 : 4.\end{aligned}$$

Hence, proved that area of △OEC : area of △OAD = 1 : 4.

(a) In ∆APD and ∆BPC,

⇒ ∠APD = ∠BPC (Vertically opposite angles are equal)

⇒ ∠ADP = ∠BCP (Both equal to 90°)

Hence, proved that ∆APD ∼ ∆BPC.

(b) In ∆ADB,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ 10² = 6² + BD²

⇒ BD² = 100 - 36

⇒ BD² = 64

⇒ BD = $\sqrt{64}$ = 8 cm.

⇒ PB = BD - PD = 8 - 4.5 = 3.5 cm

Hence, BD = 8 cm and PB = 3.5 cm.

(c) In ∆APD,

By pythagoras theorem,

⇒ AP² = AD² + DP²

⇒ AP² = 6² + (4.5)²

⇒ AP² = 36 + 20.25

⇒ AP² = 56.25

⇒ AP = $\sqrt{56.25}$ = 7.5 cm

Hence, length of AP = 7.5 cm.

(d) We know that,

Ratio of area of similar triangles is equal to the square of the corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APD}}{\text{Area of △BPC}} = \frac{AD^2}{BC^2} \\[1em] = \frac{6^2}{(2.4)^2} \\[1em] = \frac{6 \times 6}{2.4 \times 2.4} \\[1em] = \frac{1 \times 1}{0.4 \times 0.4} \\[1em] = \frac{10 \times 10}{4 \times 4} \\[1em] = \frac{100}{16} \\[1em] = \frac{25}{4} \\[1em] = 25 : 4.\end{aligned}$$

Hence, area ∆APD : area ∆BPC = 25 : 4.

(i) Since, the model of the ship is made to the scale of 1 : 250.

∴ K (Scale factor) = 250.

Actual length of the ship = k × (the length of model) = 250 × 1.6 = 400 m.

Hence, the length of the ship is 400 m.

(ii) Area of the deck of the ship = k² × (Area of the deck of the model)
= (250)² x 2.4
= 250 x 250 x 2.4
= 1,50,000 m²

Hence, the area of the deck of the ship is 1,50,000 m².

(iii) Volume of the ship = k³ × (the volume of the model)

Given, volume of ship = 1 km³ = (1000)³ m³

Let the volume of the model be x m³.

$$\begin{aligned}\Rightarrow 1000^3 = 250^3 \times x \\[1em] \Rightarrow x = \frac{1000 \times 1000 \times 1000}{250 \times 250 \times 250} \\[1em] \Rightarrow x = 4 \times 4 \times 4 \\[1em] \Rightarrow x = 64.\end{aligned}$$

Hence, the volume of the model of the ship is 64 m³.