Pythagoras Theorem

By Pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

In △ ABC,

⇒ AB² = AC² + BC²

Given, BC = a, AC = b and AB = c

⇒ c² = b² + a²

(i) Given,

a = 8 cm and b = 6 cm

⇒ c² = 6² + 8²

⇒ c² = 36 + 64

⇒ c² = 100

⇒ c = $\sqrt{100}$

⇒ c = 10 cm

Hence, c = 10 cm.

(ii) Given,

c = 25 cm and b = 7 cm

⇒ 25² = 7² + a²

⇒ 625 = 49 + a²

⇒ a² = 625 - 49

⇒ a² = 576

⇒ a = $\sqrt{576}$

⇒ a = 24 cm

Hence, a = 24 cm.

(iii) Given,

c = 13 cm and a = 5 cm

⇒ 13² = b² + 5²

⇒ 169 = b² + 25

⇒ b² = 169 - 25

⇒ b² = 144

⇒ b = $\sqrt{144}$

⇒ b = 12 cm

Hence, b = 12 cm.

Choose the greatest length. Check whether the square of greatest length is equal to the sum of squares of other two lengths.

(i) Here greatest length is 5 cm and other lengths are 3 cm, 4 cm.

Note that 5² = 25 and 3² + 4² = 9 + 16 = 25.

Thus, 5² = 3² + 4²

Hence, the triangle with given lengths of sides is a right triangle.

(ii) Here greatest length is 15 cm and other lengths are 10 cm, 13 cm.

Note that 15² = 225 and 10² + 13² = 100 + 169 = 269.

Thus, 225 ≠ 269

Hence, the triangle with given lengths of sides is not a right triangle.

(iii) Here greatest length is 100 cm and other lengths are 80 cm, 60 cm.

Note that 100² = 10000 and 80² + 60² = 6400 + 3600 = 10000

Thus, 10000 = 10000

Hence, the triangle with given lengths of sides is a right triangle.

Let, diagonal of the rectangle be BD.

BC = 30 m and CD = 40 m

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In right-angled triangle BCD,

⇒ BD² = BC² + CD²

⇒ BD² = 30² + 40²

⇒ BD² = 900 + 1600

⇒ BD² = 2500

⇒ BD = $\sqrt{2500}$

⇒ BD = 50 m

Hence, the length of its diagonal is 50 m.

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AC² = BC² + AB²

⇒ AC² = 8² + 15²

⇒ AC² = 64 + 225

⇒ AC² = 289

⇒ AC = $\sqrt{289}$

⇒ AC = 17 m.

Hence, the man is 17 m far from the starting point.

Let AB be the ladder, BC be the building and B be the window.

Then, AB = 17 m, BC = 15 m

AC be the distance of the foot of the ladder from the building.

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AB² = BC² + AC²

⇒ 17² = 15² + AC²

⇒ 289 = 225 + AC²

⇒ AC² = 289 - 225

⇒ AC² = 64

⇒ AC = $\sqrt{64}$

⇒ AC = 8 m.

Hence, the distance of the foot of the ladder from the building is 8 m.

Let AB be the ladder, BC be the vertical wall and AC be the distance of the foot of the ladder from the wall.

Then, AB = 13 m, AC = 5 m

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AB² = BC² + AC²

⇒ 13² = BC² + 5²

⇒ 169 = BC² + 25

⇒ BC² = 169 - 25

⇒ BC² = 144

⇒ BC = $\sqrt{144}$

⇒ BC = 12 m.

Hence, the distance of the other end of the ladder from the ground is 12 m.

From figure,

Let width of the street be AB = AC + BC

Let CD and CE be the ladder at different positions.

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ADC,

⇒ CD² = AD² + AC²

⇒ 15² = 9² + AC²

⇒ 225 = 81 + AC²

⇒ AC² = 225 - 81

⇒ AC² = 144

⇒ AC = $\sqrt{144}$

⇒ AC = 12 m

In triangle BCE,

⇒ CE² = BE² + BC²

⇒ 15² = 12² + BC²

⇒ 225 = 144 + BC²

⇒ BC² = 225 - 144

⇒ BC² = 81

⇒ BC = $\sqrt{81}$

⇒ BC = 9 m

AB = AC + BC = 12 + 9 = 21 m.

Hence, the width of the street is 21 m.

In right angled triangle BCD,

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle BCD,

⇒ BD² = CD² + BC²

⇒ BD² = 12² + 3²

⇒ BD² = 144 + 9

⇒ BD² = 153

⇒ BD = $\sqrt{153}$ cm.

In triangle ABD,

⇒ AD² = AB² + BD²

⇒ 13² = AB² + ($\sqrt{153}$)²

⇒ 169 = AB² + 153

⇒ AB² = 169 - 153

⇒ AB² = 16

⇒ AB = $\sqrt{16}$

⇒ AB = 4 cm.

Hence, length of AB = 4 cm.

In right angled triangle PQS,

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ PQ² = PS² + QS²

⇒ 10² = PS² + 6²

⇒ 100 = PS² + 36

⇒ PS² = 100 - 36

⇒ PS² = 64

⇒ PS = $\sqrt{64}$

⇒ PS = 8 cm.

From figure,

RS = RQ + QS = 9 + 6 = 15 cm

In right angled △ PRS,

Hypotenuse² = Perpendicular² + Base²

⇒ PR² = RS² + PS²

⇒ PR² = 15² + 8²

⇒ PR² = 225 + 64

⇒ PR² = 289

⇒ PR = $\sqrt{289}$

⇒ PR = 17 cm.

Hence, the length of PR = 17 cm.

We know that,

The diagonals of a rhombus bisect each other at right angles.

From figure,

OP = OR = 8 cm

PQ = 17 cm

In right angled △ POQ,

Hypotenuse² = Perpendicular² + Base²

⇒ PQ² = OP² + OQ²

⇒ 17² = 8² + OQ²

⇒ 289 = 64 + OQ²

⇒ OQ² = 289 - 64

⇒ OQ² = 225

⇒ OQ = $\sqrt{225}$

⇒ OQ = 15 cm.

Since, OQ = OS = 15 cm (the diagonals of a rhombus bisect each other)

QS = OQ + OS = 15 + 15 = 30 cm

∴ Diagonals are PR = 16 cm and QS = 30 cm.

As we know,

Area of rhombus = $\frac{1}{2}$ × Product of diagonals

= $\frac{1}{2} \times (30 × 16) = \frac{1}{2} \times 480$ = 240 cm².

Hence, the area of rhombus = 240 cm².

From figure,

AB || DC

Distance between parallel sides, AE = BC = 4 cm

In right angled △ AED,

Hypotenuse² = Perpendicular² + Base²

⇒ AD² = AE² + ED²

⇒ 5² = 4² + ED²

⇒ 25 = 16 + ED²

⇒ ED² = 25 - 16

⇒ ED² = 9

⇒ ED = $\sqrt{9}$

⇒ ED = 3 cm.

∴ DC = EC - ED = 5 - 3 = 2 cm

As we know,

Area of trapezium = $\frac{1}{2} \times (\text{sum of parallel sides} \times \text{distance between them})$

$$\begin{aligned}= \frac{1}{2} \times (\text{AB + DC}) \times \text{BC} \\[1em] = \frac{1}{2} \times (5 + 2) \times 4 \\[1em] = \frac{1}{2} \times 7 \times 4 \\[1em] = 7 \times 2 \\[1em] = 14 \text{ cm}^2.\end{aligned}$$

Hence, the area of trapezium ABCD is 14 cm².

Consider △ABC as a right angled triangle.

AB = 5x cm and BC = (3x - 1) cm

We know that,

Area of △ABC = ${1}{2}$ × base × height = $\frac{1}{2}$ × BC × AB

Substituting the values we get,

⇒ 60 = $\frac{1}{2}$ × (3x - 1) × 5x

⇒ 120 = 5x(3x - 1)

⇒ 120 = 15x² - 5x

⇒ 15x² - 5x - 120 = 0

⇒ 5(3x² - x - 24) = 0

⇒ 3x² - x - 24 = 0

⇒ 3x² - 9x + 8x - 24 = 0

⇒ 3x(x - 3) + 8(x - 3) = 0

⇒ (3x + 8)(x - 3) = 0

⇒ 3x + 8 = 0 or x - 3 = 0

⇒ 3x = -8 or x = 3

⇒ x = $-\frac{8}{3}$ or x = 3.

Since, x cannot be negative as length of a side cannot be negative. So, x = 3.

AB = 5 × 3 = 15 cm

BC = (3 × 3 - 1) = 9 - 1 = 8 cm

In right angled △ABC,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC = $\sqrt{289}$

⇒ AC = 17 cm.

Hence, the length of the sides of the triangle is 17 cm, 15 cm and 8 cm.

In △ ABC,

Draw altitude AD perpendicular to BC.

In an equilateral triangle, the altitude also acts as the median (bisecting the base).

∴ BD = $\frac{1}{2}$ × BC = $\frac{1}{2} \times 5\sqrt{3} = \frac{5\sqrt{3}}{2}$

In right angled △ABD,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AD² + BD²

$$\begin{aligned}\Rightarrow (5\sqrt{3})^2 = \text{AD}^2 + \Big(\frac{5\sqrt{3}}{2}\Big)^2 \\[1em] \Rightarrow 25 \times 3 = \text{AD}^2 + \frac{25 \times 3}{4} \\[1em] \Rightarrow 75 = \text{AD}^2 + \frac{75}{4} \\[1em] \Rightarrow \text{AD}^2 = 75 - \frac{75}{4} \\[1em] \Rightarrow \text{AD}^2 = \frac{300 - 75}{4} \\[1em] \Rightarrow \text{AD}^2 = \frac{225}{4} \\[1em] \Rightarrow \text{AD} = \sqrt{\frac{225}{4}} \\[1em] \Rightarrow \text{AD} = \frac{15}{2} \\[1em] \Rightarrow \text{AD} = 7.5 \text{ cm}.\end{aligned}$$

Hence, the altitude is 7.5 cm.

In rhombus, AB = BC = CD = AD (All sides of a rhombus are equal)

Diagonal AC and BD intersect at point O.

Diagonals bisect each other at right angles.

So, AO = OC = $\frac{1}{2}$ × AC and BO = OD = $\frac{1}{2}$ × BD

In right angled △AOB,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AO² + BO²

$$\begin{aligned}\Rightarrow \text{AB}^2 = \Big(\frac{\text{AC}}{2}\Big)^2 + \Big(\frac{\text{BD}}{2}\Big)^2 \\[1em] \Rightarrow \text{AB}^2 = \frac{\text{AC}^2}{4} + \frac{\text{BD}^2}{4} \\[1em] \Rightarrow \text{AB}^2 = \frac{\text{AC}^2 + \text{BD}^2}{4}\end{aligned}$$

∴ 4AB² = AC² + BD²

Hence, proved that AC² + BD² = 4AB².

(i) Given,

ABC is a triangle such that ∠ABC = 90°. D is the mid-point of BC.

In △ ABD, using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AD² = AB² + BD² ......(1)

Similarly, in △ ABC,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (BD + DC)²

⇒ AC² = AB² + BD² + DC² + 2 × BD × DC

As D is the midpoint of BC, BD = DC.

⇒ AC² = AB² + BD² + CD² + 2 × CD × CD

⇒ AC² = AB² + BD² + CD² + 2CD²

⇒ AC² = AB² + BD² + 3CD²

Using equation (1), we get

⇒ AC² = AD² + 3CD²

Hence, proved that AC² = AD² + 3 CD².

(ii) In △ ABD,

⇒ AD² = AB² + BD²

BD = $\frac{\text{BC}}{2}$

$$\begin{aligned}\Rightarrow \text{AD}^2 = \text{AB}^2 + \Big(\frac{\text{BC}}{2}\Big)^2 \\[1em] \Rightarrow \text{AD}^2 = \text{AB}^2 + \frac{\text{BC}^2}{4} \\[1em] \Rightarrow \text{AD}^2 - \text{AB}^2 = \frac{\text{BC}^2}{4} \\[1em] \Rightarrow \text{BC}^2 = 4 (\text{AD}^2 - \text{AB}^2)\end{aligned}$$

Hence, proved that BC² = 4 (AD² - AB²).

Let AB be the smaller pole and CD the bigger pole.

From figure,

⇒ CE = AB = 9 m and AE = BC = 12 m

⇒ CD = CE + ED

⇒ 14 = 9 + ED

⇒ ED = 14 - 9

⇒ ED = 5 m

From figure,

△ADE is right angled triangle.

By pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AD² = AE² + ED²

⇒ AD² = (12)² + (5)²

⇒ AD² = 144 + 25

⇒ AD² = 169

⇒ AD = $\sqrt{169}$

⇒ AD = 13 m.

Hence, the distance between their tops is 13 m.

ABC is a triangle in which AB = AC and D is any point on BC.

In △ ABE and △ ACE,

⇒ AB = AC (Given)

⇒ AE = AE (Common)

⇒ ∠AEB = ∠AEC (both are 90°)

Using RHS congruency criterion,

△ ABE ≅ △ ACE

⇒ BE = CE (by C.P.C.T.)

In △ ABE, using Pythagorean theorem,

Hypotenuse² = Base² + Height²

⇒ AB² = AE² + BE² ......(1)

In △ ADE, using Pythagorean theorem,

⇒ AD² = AE² + DE² ......(2)

Subtracting equation (2) from (1), we get:

⇒ AB² - AD² = (AE² + BE²) - (AE² + DE²)

⇒ AB² - AD² = AE² + BE² - AE² - DE²

⇒ AB² - AD² = BE² - DE²

⇒ AB² - AD² = (BE - DE)(BE + DE)

⇒ AB² - AD² = (BE - DE)(CE + DE) [∴ BE = CE]

⇒ AB² - AD² = BD × CD

Hence, proved that AB² - AD² = BD × CD.

Choose the greatest length. Check whether the square of greatest length is equal to the sum of squares of other two lengths.

10, 24, 26

Here greatest length is 26 cm and other lengths are 24 cm, 10 cm.

Note that 26² = 676 and 24² + 10² = 576 + 100 = 676.

Thus, 26² = 24² + 10².

Thus, triangle with sides 10, 24, 26 form a right-angled triangle.

Hence, option 2 is the correct option.

In △ ADB,

Using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AB² = BD² + AD²

⇒ BD² = AB² - AD² .....(1)

In △ BDC,

Using Pythagoras theorem,

⇒ BC² = BD² + CD²

⇒ BD² = BC² - CD² ......(2)

Equating eq.(1) and (2), we get:

⇒ AB² - AD² = BC² - CD²

⇒ AB² - BC² = AD² - CD².

Hence, option 2 is the correct option.

From figure,

⇒ BE = AD = 20 cm and AB = DE = 40 cm

⇒ CE = BC - BE

⇒ CE = 29 - 20

⇒ CE = 9 cm

From figure,

△ CDE is right angled triangle.

By pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ CD² = DE² + CE²

⇒ CD² = 40² + 9²

⇒ CD² = 1600 + 81

⇒ CD² = 1681

⇒ CD = $\sqrt{1681}$

⇒ CD = 41 cm.

Hence, option 3 is the correct option.

In △ BCD,

By pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ CD² = BD² + BC²

⇒ 10² = BD² + 8²

⇒ 100 = BD² + 64

⇒ BD² = 100 - 64

⇒ BD² = 36

⇒ BD = $\sqrt{36}$

⇒ BD = 6 cm

From figure,

AB = AD + BD = 9 + 6 = 15 cm

In △ ABC,

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC = $\sqrt{289}$

⇒ AC = 17 cm

Hence, option 1 is the correct option.

Let AC = 5 cm and BD = 12 cm.

We know that,

Diagonals of rhombus are perpendicular and bisect each other,

OB = $\frac{1}{2}$ BD = 6 cm and AO = $\frac{1}{2}$ AC = 2.5 cm.

In right triangle AOB,

By pythagoras theorem we get,

Hypotenuse² = Base² + Height²

⇒ AB² = AO² + OB²

⇒ AB² = (2.5)² + 6²

⇒ AB² = 6.25 + 36

⇒ AB² = 42.25

⇒ AB = $\sqrt{42.25}$ = 6.5 cm.

∴ The length of each side of the rhombus is 6.5 cm.

Hence, option 2 is the correct option.

Let AC = 6 cm and BD = 8 cm.

We know that,

Diagonals of rhombus are perpendicular and bisect each other,

OB = $\frac{1}{2}$ BD = 4 cm and AO = $\frac{1}{2}$ AC = 3 cm.

In right triangle AOB,

By pythagoras theorem we get,

Hypotenuse² = Base² + Height²

⇒ AB² = AO² + OB²

⇒ AB² = 3² + 4²

⇒ AB² = 9 + 16

⇒ AB² = 25

⇒ AB = $\sqrt{25}$ = 5 cm.

∴ The length of each side of the rhombus is 5 cm.

Hence, option 2 is the correct option.

In right triangle ABC,

By pythagoras theorem we get,

Hypotenuse² = Base² + Height²

⇒ AC² = BC² + AB²

⇒ 3² = (x - 2)² + $(2\sqrt{2}\text{x})^2$

⇒ 9 = x² + 4 - 4x + 4 × 2 x²

⇒ 9 = x² + 4 - 4x + 8x²

⇒ 9x² - 4x + 4 - 9 = 0

⇒ 9x² - 4x - 5 = 0

⇒ 9x² - 9x + 5x - 5 = 0

⇒ 9x(x - 1) + 5(x - 1) = 0

⇒ (9x + 5) = 0 or (x - 1) = 0

⇒ x = - $\frac{5}{9}$ or x = 1

Since, length cannot be negative.

⇒ x = 1 cm.

Hence, option 1 is the correct option.

In △ ABC,

AB = BC = AC = a units

Draw altitude AD perpendicular to BC.

In an equilateral triangle, the altitude also acts as the median (bisecting the base).

∴ BD = $\frac{1}{2} \times BC = \frac{1}{2} \times a = \frac{\text{a}}{2}$

In right angled △ABD,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AD² + BD²

$$\begin{aligned}\Rightarrow \text{a}^2 = \text{AD}^2 + \Big(\frac{\text{a}}{2}\Big)^2 \\[1em] \Rightarrow \text{AD}^2 = \text{a}^2 - \frac{\text{a}^2}{4} \\[1em] \Rightarrow \text{AD}^2 = \frac{4\text{a}^2 - \text{a}^2}{4} \\[1em] \Rightarrow \text{AD}^2 = \frac{3\text{a}^2}{4} \\[1em] \Rightarrow \text{AD} = \sqrt{\frac{3\text{a}^2}{4}} \\[1em] \Rightarrow \text{AD} = \frac{\text{a} \sqrt{3}}{2} \text{ units.}\end{aligned}$$

Hence, option 1 is the correct option.

Let CD = x cm

BD = BC + CD = (6 + x) cm

In △ ABD, using Pythagorean theorem,

Hypotenuse² = Base² + Height²

⇒ AB² = BD² + AD²

⇒ 8² = (6 + x)² + AD²

⇒ 64 - (6 + x)² = AD²

⇒ AD² = 64 - (6 + x)²

⇒ AD² = 64 - (36 + x² + 12x)

⇒ AD² = 64 - 36 - x² - 12x

⇒ AD² = 28 - x² - 12x .....(1)

In △ ADC, using Pythagorean theorem,

⇒ AC² = CD² + AD²

⇒ 3² = x² + AD²

⇒ 9 = x² + AD²

⇒ AD² = 9 - x² .....(2)

From eq.(1) and (2), we have :

⇒ 9 - x² = 28 - x² - 12x

⇒ 9 = 28 - 12x

⇒ 12x = 28 - 9

⇒ 12x = 19

⇒ x = $\frac{19}{12}$

⇒ x = $1\frac{7}{12}$ cm.

Hence, option 1 is the correct option.

Let width of the street be AB = AC + BC

Let CD and CE be the ladder at different positions.

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ADC,

⇒ CD² = AD² + AC²

⇒ 50² = 14² + AC²

⇒ 2500 = 196 + AC²

⇒ AC² = 2500 - 196

⇒ AC² = 2304

⇒ AC = $\sqrt{2304}$

⇒ AC = 48 ft

In triangle BCE,

⇒ CE² = BE² + BC²

⇒ 50² = 48² + BC²

⇒ 2500 = 2304 + BC²

⇒ BC² = 2500 - 2304

⇒ BC² = 196

⇒ BC = $\sqrt{196}$

⇒ BC = 14 ft

AB = AC + BC = 48 + 14 = 62 ft.

Hence, option 2 is the correct option.

1. Given, Speed of aeroplane P = 400 km/h

As we know,

Distance traveled = Speed × Time taken

= 400 × 1.5

= 600 km.

Hence, Option (a) is the correct option.

2. Given, Speed of aeroplane Q = 300 km/h

As we know,

Distance traveled = Speed × Time taken

= 300 × 1.5

= 450 km.

Hence, Option (d) is the correct option.

3. From figure,

Distance between the two aeroplanes after a certain period of time is represented by the line segment PQ.

Hence, Option (c) is the correct option.

4. From above,

Distance traveled by aeroplane P in 1.5 hours = 600 km

Distance traveled by aeroplane Q in 1.5 hours = 450 km

600 - 450 = 150 km

⇒ Aeroplane P traveled longer distance by 150 km.

Hence, Option (a) is the correct option.

5. Distance traveled by aeroplane P in 1.5 hours = 600 km

⇒ AP = 600 km

Distance traveled by aeroplane Q in 1.5 hours = 450 km

⇒ AQ = 450 km

From figure,

Let ∠A = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle APQ,

⇒ PQ² = AP² + AQ²

⇒ PQ² = 600² + 450²

⇒ PQ² = 360000 + 202500

⇒ PQ² = 562500

⇒ PQ = $\sqrt{562500}$

⇒ PQ = 750 km.

Hence, Option (d) is the correct option.

We know that,

In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares of the other two sides.

∴ Reason (R) is false.

From figure,

Choose the greatest length. Check whether the square of greatest length is equal to the sum of squares of other two lengths.

Here greatest length is 10 cm and other lengths are 8 cm, 6 cm.

Note that 10² = 100 and 8² + 6² = 64 + 36 = 100.

Thus, 10² = 8² + 6²

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ AC is the hypotenuse and triangle ABC is right angled at B.

∴ Assertion (A) is true.

Hence, Option 1 is the correct option.

In a right angled triangle, the longest side is called hypotenuse and the other two sides are called its legs. Also, the sum of any two sides is always greater than the third side.

∴ Assertion (A) is false.

In a right angled triangle, the side opposite to the right angle is called its hypotenuse.

∴ Reason (R) is true.

Hence, Option 2 is the correct option.

Let one leg = x cm and the other = (x + 5) cm

From figure,

AB = (x + 5) cm, BC = x cm and AC = 25 cm, ∠B = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AC² = AB² + BC²

⇒ 25² = (x + 5)² + x²

⇒ 625 = x² + 25 + 10x + x²

⇒ 2x² + 10x + 25 - 625 = 0

⇒ 2x² + 10x - 600 = 0

⇒ 2(x² + 5x - 300) = 0

⇒ x² + 5x - 300 = 0

⇒ x² - 15x + 20x - 300 = 0

⇒ x(x - 15) + 20(x - 15) = 0

⇒ (x + 20)(x - 15) = 0

⇒ x = -20 or x = 15

Since, length cannot be negative.

x = 15 cm.

Sum of the lengths = x + (x + 5) = 15 + (15 + 5) = 15 + 20 = 35 cm.

Hence, Option 2 is the correct option.

Let aeroplane flying in north direction at 300 km/h be at point P after 1.5 hours and aeroplane flying in west direction at 400 km/h be at point Q after 1.5 hours.

Given, 90 minutes = 1.5 hours

Speed of aeroplane P = 300 km/h

As we know,

Distance traveled = Speed × Time taken

AP = 300 × 1.5 = 450 km.

Speed of aeroplane Q = 400 km/h

As we know,

Distance traveled = Speed × Time taken

AQ = 400 × 1.5 = 600 km.

From figure,

Let ∠A = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle APQ,

⇒ PQ² = AP² + AQ²

⇒ PQ² = 450² + 600²

⇒ PQ² = 202500 + 360000

⇒ PQ² = 562500

⇒ PQ = $\sqrt{562500}$

⇒ PQ = 750 km.

Hence, Option 4 is the correct option.

From figure,

Distance between two cities A and B = AB = 26 km

Since, AC ⟂ CB

∠A = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AB² = AC² + CB²

⇒ 26² = (2x)² + [2(x + 7)]²

⇒ 676 = 4x² + 4(x + 7)²

⇒ 676 = 4x² + 4(x² + 49 + 14x)

⇒ 676 = 4x² + 4x² + 196 + 56x

⇒ 8x² + 196 + 56x - 676 = 0

⇒ 8x² + 56x - 480 = 0

⇒ 8(x² + 7x - 60) = 0

⇒ x² + 7x - 60 = 0

⇒ x² + 12x - 5x - 60 = 0

⇒ x(x + 12) - 5(x + 12) = 0

⇒ (x - 5)(x + 12) = 0

⇒ x = 5 or x = -12.

Since, length cannot be negative.

x = 5

Distance traveled via city C = 2x + 2(x + 7) = 2x + 2x + 14 = 4x + 14 = 4 × 5 + 14 = 20 + 14 = 34 km.

Difference in distance traveled via city C and after construction of highway = 34 - 26 = 8 km.

Hence, 8 km distance will be saved in reaching city B from city A after the construction of the highway.

First ladder reaches at A, thus AB = 4 m and AC = 5 m.

Given,

Foot of the ladder is moved 1.6 m towards the wall

The distance by which the top of the ladder would slide upwards on the wall = AE

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AC² = AB² + BC²

⇒ 5² = 4² + BC²

⇒ 25 = 16 + BC²

⇒ BC² = 25 - 16

⇒ BC² = 9

⇒ BC = $\sqrt{9}$

⇒ BC = 3 m

BD = BC - CD = 3 - 1.6 = 1.4 m

In triangle EBD,

Ladder ED = 5 m

⇒ ED² = EB² + BD²

⇒ 5² = EB² + (1.4)²

⇒ 25 = EB² + 1.96

⇒ EB² = 25 - 1.96

⇒ EB² = 23.04

⇒ EB = $\sqrt{23.04}$

⇒ EB = 4.8 m

From figure,

AE = EB - AB = 4.8 - 4 = 0.8 m

Hence, the distance by which the top of the ladder would slide upwards on the wall is 0.8 m.

Given, squares are formed joining mid-points of square ABCD.

AE = AH = $\frac{1}{2}$ × AB = $\frac{1}{2}$ × 20 = 10 cm.

In square ABCD,

∠A = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle AEH,

⇒ EH² = AE² + AH²

⇒ EH² = 10² + 10²

⇒ EH² = 100 + 100

⇒ EH² = 200

⇒ EH = $\sqrt{200} = \sqrt{2 \times 100}$

⇒ EH = $10\sqrt{2}$ cm.

In square EFGH,

∠E = 90°

EM = EN = $\frac{1}{2}$ × EH = $\frac{1}{2} \times 10 \sqrt{2} = 5\sqrt{2} \text{ cm.}$

By Pythagoras theorem,

In triangle EMN,

⇒ MN² = EN² + EM²

⇒ MN² = $(5\sqrt{2})^2 + (5\sqrt{2})^2$

⇒ MN² = 50 + 50

⇒ MN² = 100

⇒ MN = $\sqrt{100}$

⇒ MN = 10 cm.

In square MNOP,

∠M = 90°

IM = ML = $\frac{1}{2} × MN = \frac{1}{2} \times 10 = 5 \text{ cm.}$

By Pythagoras theorem,

In triangle IML,

⇒ IL² = IM² + ML²

⇒ IL² = 5² + 5²

⇒ IL² = 25 + 25

⇒ IL² = 50

⇒ IL = $\sqrt{25 \times 2}$

⇒ IL = $5\sqrt{2}$ cm.

Hence, the length of a side of the smallest square is $5\sqrt{2}$ cm.