Quadrilaterals

Let the measure of the fourth angle be $x^{\circ}$.

We know that the sum of the angles of a quadrilateral is 360°. Then,

$$\begin{aligned}\Rightarrow 68^{\circ} + 74^{\circ} + 108^{\circ} + x^{\circ} = 360^{\circ} \\[1em] \Rightarrow 250^{\circ} + x^{\circ} = 360^{\circ} \\[1em] \Rightarrow x^{\circ} = 360^{\circ} - 250^{\circ} \\[1em] \Rightarrow x^{\circ} = 110^{\circ}\end{aligned}$$

Hence, the measure of the fourth angle is 110°.

Let the measures of the angles of the given quadrilateral be $(4x)^{\circ}, (3x)^{\circ}, (2x)^{\circ}$ and $(6x)^{\circ}$ respectively.

We know that the sum of the angles of a quadrilateral is 360°. Then,

$$\begin{aligned}\Rightarrow (4x)^{\circ} + (3x)^{\circ} + (2x)^{\circ} + (6x)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (15x)^{\circ} = 360^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{360^{\circ}}{15} \\[1em] \Rightarrow x^{\circ} = 24^{\circ}\end{aligned}$$

∴ First angle = $(4x)^{\circ}$ = (4 × 24)° = 96°

Second angle = $(3x)^{\circ}$ = (3 × 24)° = 72°

Third angle = $(2x)^{\circ}$ = (2 × 24)° = 48°

Fourth angle = $(6x)^{\circ}$ = (6 × 24)° = 144°

Hence, the angles of the quadrilateral are 96°, 72°, 48° and 144°.

Let the measures of the three angles of the given quadrilateral be $(2x)^{\circ}, (5x)^{\circ}$ and $(6x)^{\circ}$ respectively, and the fourth angle measures 100°.

We know that the sum of the angles of a quadrilateral is 360°. Then,

$$\begin{aligned}\Rightarrow (2x)^{\circ} + (5x)^{\circ} + (6x)^{\circ} + 100^{\circ} = 360^{\circ} \\[1em] \Rightarrow (13x)^{\circ} + 100^{\circ} = 360^{\circ} \\[1em] \Rightarrow (13x)^{\circ} = 360^{\circ} - 100^{\circ} \\[1em] \Rightarrow (13x)^{\circ} = 260^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{260^{\circ}}{13} \\[1em] \Rightarrow x^{\circ} = 20^{\circ}\end{aligned}$$

∴ First angle = $(2x)^{\circ}$ = (2 × 20)° = 40°

Second angle = $(5x)^{\circ}$ = (5 × 20)° = 100°

Third angle = $(6x)^{\circ}$ = (6 × 20)° = 120°

Hence, the other three angles of the quadrilateral are 40°, 100° and 120°.

Let the measures of the remaining two angles of the given quadrilateral be $(3x)^{\circ}$ and $(5x)^{\circ}$ respectively.

We know that the sum of the angles of a quadrilateral is 360°. Then,

$$\begin{aligned}\Rightarrow 72^{\circ} + 56^{\circ} + (3x)^{\circ} + (5x)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (8x)^{\circ} + 128^{\circ} = 360^{\circ} \\[1em] \Rightarrow (8x)^{\circ} = 360^{\circ} - 128^{\circ} \\[1em] \Rightarrow (8x)^{\circ} = 232^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{232^{\circ}}{8} \\[1em] \Rightarrow x^{\circ} = 29^{\circ}\end{aligned}$$

∴ Third angle = $(3x)^{\circ}$ = (3 × 29)° = 87°

Fourth angle = $(5x)^{\circ}$ = (5 × 29)° = 145°

Hence, the angles of the quadrilateral are 72°, 56°, 87° and 145°.

Let the measure of each of the equal angles be $x^{\circ}$.

We know that the sum of the angles of a quadrilateral is 360°. Then,

$$\begin{aligned}\Rightarrow 116^{\circ} + 84^{\circ} + x^{\circ} + x^{\circ} = 360^{\circ} \\[1em] \Rightarrow (2x)^{\circ} + 200^{\circ} = 360^{\circ} \\[1em] \Rightarrow (2x)^{\circ} = 360^{\circ} - 200^{\circ} \\[1em] \Rightarrow (2x)^{\circ} = 160^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{160^{\circ}}{2} \\[1em] \Rightarrow x^{\circ} = 80^{\circ}\end{aligned}$$

Hence, the measure of each of the equal angles is 80°.

Let the measure of each of the three equal angles be $x^{\circ}$.

We know that the sum of the angles of a quadrilateral is 360°. Then,

$$\begin{aligned}\Rightarrow x^{\circ} + x^{\circ} + x^{\circ} + 75^{\circ} = 360^{\circ} \\[1em] \Rightarrow (3x)^{\circ} + 75^{\circ} = 360^{\circ} \\[1em] \Rightarrow (3x)^{\circ} = 360^{\circ} - 75^{\circ} \\[1em] \Rightarrow (3x)^{\circ} = 285^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{285^{\circ}}{3} \\[1em] \Rightarrow x^{\circ} = 95^{\circ}\end{aligned}$$

Hence, the measure of each of the equal angles is 95°.

(i) We know that the sum of the angles of a quadrilateral is 360°.

$$\begin{aligned}\Rightarrow (5x)^{\circ} + (3x + 10)^{\circ} + (6x - 20)^{\circ} + (x + 25)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (5x + 3x + 10 + 6x - 20 + x + 25)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (15x + 15)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (15x)^{\circ} = 360^{\circ} - 15^{\circ} \\[1em] \Rightarrow (15x)^{\circ} = 345^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{345^{\circ}}{15} \\[1em] \Rightarrow x^{\circ} = 23^{\circ}\end{aligned}$$

Hence, $\bm{x = 23}$.

(ii) First angle = $(5x)^{\circ}$ = (5 × 23)° = 115°

Second angle = $(3x + 10)^{\circ}$ = (3 × 23 + 10)° = (69 + 10)° = 79°

Third angle = $(6x - 20)^{\circ}$ = (6 × 23 − 20)° = (138 − 20)° = 118°

Fourth angle = $(x + 25)^{\circ}$ = (23 + 25)° = 48°

Hence, the angles of the quadrilateral are 115°, 79°, 118° and 48°.

Given,

In quadrilateral ABCD, AB ∥ DC, ∠A : ∠D = 2 : 3 and ∠B : ∠C = 4 : 5.

Let ∠A = $(2x)^{\circ}$ and ∠D = $(3x)^{\circ}$.

As AB ∥ DC and AD is a transversal, ∠A and ∠D are co-interior angles.

⇒ ∠A + ∠D = 180°

$$\begin{aligned}\Rightarrow (2x)^{\circ} + (3x)^{\circ} = 180^{\circ} \\[1em] \Rightarrow (5x)^{\circ} = 180^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{180^{\circ}}{5} \\[1em] \Rightarrow x^{\circ} = 36^{\circ}\end{aligned}$$

∴ ∠A = (2 × 36)° = 72° and ∠D = (3 × 36)° = 108°

Again, let ∠B = $(4y)^{\circ}$ and ∠C = $(5y)^{\circ}$.

As AB ∥ DC and BC is a transversal, ∠B and ∠C are co-interior angles.

⇒ ∠B + ∠C = 180°

$$\begin{aligned}\Rightarrow (4y)^{\circ} + (5y)^{\circ} = 180^{\circ} \\[1em] \Rightarrow (9y)^{\circ} = 180^{\circ} \\[1em] \Rightarrow y^{\circ} = \frac{180^{\circ}}{9} \\[1em] \Rightarrow y^{\circ} = 20^{\circ}\end{aligned}$$

∴ ∠B = (4 × 20)° = 80° and ∠C = (5 × 20)° = 100°

Hence, ∠A = 72°, ∠B = 80°, ∠C = 100° and ∠D = 108°.

Given,

In quadrilateral ABCD, the side BA is produced to E and the side DC is produced to F.

As BA is produced to E, ∠DAE and ∠DAB form a linear pair.

⇒ $b^{\circ}$ + ∠A = 180° .....(1)

As DC is produced to F, ∠BCF and ∠BCD form a linear pair.

⇒ $a^{\circ}$ + ∠C = 180° .....(2)

Adding (1) and (2), we get :

⇒ $a^{\circ} + b^{\circ}$ + ∠A + ∠C = 360° .....(3)

Also, in quadrilateral ABCD,

⇒ $x^{\circ} + y^{\circ}$ + ∠A + ∠C = 360° [Sum of the angles of a quadrilateral is 360°] .....(4)

From (3) and (4), we get :

⇒ $a^{\circ} + b^{\circ}$ + ∠A + ∠C = $x^{\circ} + y^{\circ}$ + ∠A + ∠C

⇒ $a^{\circ} + b^{\circ} = x^{\circ} + y^{\circ}$

Hence, $\bm{a + b = x + y}$.

Given,

O is a point within quadrilateral ABCD and AC, BD are its diagonals.

We know that the sum of any two sides of a triangle is greater than the third side.

In ΔOAC, we have :

⇒ OA + OC > AC .....(1)

In ΔOBD, we have :

⇒ OB + OD > BD .....(2)

Adding (1) and (2), we get :

⇒ OA + OC + OB + OD > AC + BD

Hence, OA + OB + OC + OD > AC + BD.

Given,

ABCD is a parallelogram in which ∠A = 50°.

As AD ∥ BC and AB cuts them, ∠A and ∠B are co-interior angles.

⇒ ∠A + ∠B = 180° [Co-interior angles]

⇒ 50° + ∠B = 180°

⇒ ∠B = (180° − 50°) = 130°

Since the opposite angles of a parallelogram are equal, we have :

⇒ ∠C = ∠A = 50° and ∠D = ∠B = 130°

Hence, ∠B = 130°, ∠C = 50° and ∠D = 130°.

Given,

In parallelogram ABCD, ∠A = $(2x + 35)^{\circ}$ and ∠C = $(3x - 5)^{\circ}$.

(i) Since the opposite angles of a parallelogram are equal, we have :

⇒ ∠A = ∠C

$$\begin{aligned}\Rightarrow (2x + 35)^{\circ} = (3x - 5)^{\circ} \\[1em] \Rightarrow (3x)^{\circ} - (2x)^{\circ} = 35^{\circ} + 5^{\circ} \\[1em] \Rightarrow x^{\circ} = 40^{\circ}\end{aligned}$$

Hence, $\bm{x = 40}$.

(ii) ∠A = (2 × 40 + 35)° = (80 + 35)° = 115° and ∠C = (3 × 40 − 5)° = (120 − 5)° = 115°

We know that ∠B = ∠D [Opposite angles of a parallelogram]

Now, ∠A + ∠B + ∠C + ∠D = 360° [Sum of the angles of a quadrilateral is 360°]

⇒ 115° + ∠B + 115° + ∠B = 360°

⇒ 230° + 2∠B = 360°

⇒ 2∠B = (360° − 230°) = 130°

⇒ ∠B = $\frac{130^{\circ}}{2}$ = 65°

∴ ∠D = ∠B = 65°

Hence, ∠A = 115°, ∠B = 65°, ∠C = 115°, ∠D = 65°.

Given,

ABCD is a parallelogram in which the side AB has been produced to E and ∠CBE = 75°.

As AB is produced to E, ∠ABC and ∠CBE form a linear pair.

⇒ ∠ABC + ∠CBE = 180°

⇒ ∠ABC + 75° = 180°

⇒ ∠ABC = (180° − 75°) = 105°

Since the opposite angles of a parallelogram are equal, we have :

⇒ ∠ADC = ∠ABC = 105°

As AB ∥ DC and BC cuts them, ∠ABC and ∠BCD are co-interior angles.

⇒ ∠ABC + ∠BCD = 180° [Co-interior angles]

⇒ 105° + ∠BCD = 180°

⇒ ∠BCD = (180° − 105°) = 75°

∴ ∠BAD = ∠BCD = 75° [Opposite angles of a parallelogram]

Hence, ∠ABC = 105°, ∠ADC = 105°, ∠BCD = 75° and ∠BAD = 75°.

Given,

ABCD is a parallelogram in which AB = 8.5 cm and its perimeter is 24 cm.

Since the opposite sides of a parallelogram are equal, we have :

⇒ CD = AB and AD = BC

Now, perimeter of parallelogram ABCD = AB + BC + CD + AD

⇒ 24 = AB + BC + AB + BC

⇒ 24 = 2(AB + BC)

⇒ AB + BC = 12

⇒ 8.5 + BC = 12

⇒ BC = (12 − 8.5) = 3.5 cm

∴ CD = AB = 8.5 cm and AD = BC = 3.5 cm

Hence, BC = 3.5 cm, CD = 8.5 cm and AD = 3.5 cm.

Given,

ABCD is a rhombus in which ∠BAC = 40°.

(i) Since all the sides of a rhombus are equal, in ΔABC we have AB = BC.

⇒ ∠ACB = ∠BAC = 40° [Angles opposite to equal sides are equal]

Hence, ∠ACB = 40°.

(ii) In ΔABC, we have :

⇒ ∠BAC + ∠ACB + ∠ABC = 180° [Sum of the angles of a triangle is 180°]

⇒ 40° + 40° + ∠ABC = 180°

⇒ 80° + ∠ABC = 180°

⇒ ∠ABC = (180° − 80°) = 100°

Hence, ∠ABC = 100°.

(iii) ∠ADC = ∠ABC = 100° [Opposite angles of a parallelogram]

Hence, ∠ADC = 100°.

(iv) As AB ∥ DC and BC cuts them, ∠ABC and ∠BCD are co-interior angles.

⇒ ∠ABC + ∠BCD = 180° [Co-interior angles]

⇒ 100° + ∠BCD = 180°

⇒ ∠BCD = (180° − 100°) = 80°

The diagonals of a rhombus bisect its angles, so AC bisects ∠BCD.

⇒ ∠ACD = $\frac{1}{2}$ ∠BCD

⇒ ∠ACD = $\frac{1}{2}$ × 80° = 40°

Hence, ∠ACD = 40°.

(v) ∠BAD = ∠BCD = 80° [Opposite angles of a parallelogram]

The diagonal AC bisects ∠BAD.

⇒ ∠CAD = $\frac{1}{2}$ ∠BAD

⇒ ∠CAD = $\frac{1}{2}$ × 80° = 40°

Hence, ∠CAD = 40°.

Given,

ABCD is a rectangle whose diagonals AC and BD intersect at O and ∠AOB = 110°.

(i) As ∠AOD and ∠AOB form a linear pair, we have :

⇒ ∠AOD + ∠AOB = 180°

⇒ ∠AOD + 110° = 180°

⇒ ∠AOD = (180° − 110°) = 70°

We know that the diagonals of a rectangle are equal and bisect each other. Then, OA = $\frac{1}{2}$ AC and OD = $\frac{1}{2}$ BD.

We have,

⇒ AC = BD

⇒ $\frac{1}{2}$ AC = $\frac{1}{2}$ BD

⇒ OA = OD

In ΔOAD, OA = OD, so ∠OAD = ∠ODA [Angles opposite to equal sides are equal]

Now, ∠OAD + ∠ODA + ∠AOD = 180° [Sum of the angles of a triangle is 180°]

⇒ ∠ODA + ∠ODA + 70° = 180° [∵ ∠OAD = ∠ODA]

⇒ 2∠ODA + 70° = 180°

⇒ 2∠ODA = (180° − 70°) = 110°

⇒ ∠ODA = $\frac{110^{\circ}}{2}$ = 55°

Hence, ∠ODA = 55°.

(ii) ∠COD = ∠AOB = 110° [Vertically opposite angles]

We know that the diagonals of a rectangle are equal and bisect each other. Then, OC = $\frac{1}{2}$ AC and OD = $\frac{1}{2}$ BD.

We have,

⇒ AC = BD

⇒ $\frac{1}{2}$ AC = $\frac{1}{2}$ BD

⇒ OC = OD

In ΔOCD, OC = OD, so ∠OCD = ∠ODC [Angles opposite to equal sides are equal]

Now, ∠OCD + ∠ODC + ∠COD = 180° [Sum of the angles of a triangle is 180°]

⇒ ∠OCD + ∠OCD + 110° = 180° [∵ ∠ODC = ∠OCD]

⇒ 2∠OCD + 110° = 180°

⇒ 2∠OCD = (180° − 110°) = 70°

⇒ ∠OCD = $\frac{70^{\circ}}{2}$ = 35°

Hence, ∠OCD = 35°.

Given,

The diagonals AC and BD of parallelogram ABCD intersect at O, ∠OBC = 36°, ∠ODC = 28° and ∠AOD = 64°.

(i) As AD ∥ BC and BD cuts them, we have :

⇒ ∠ADB = ∠DBC = 36° [Alternate angles]

i.e., ∠ODA = 36°

In ΔAOD, we have :

⇒ ∠OAD + ∠ODA + ∠AOD = 180° [Sum of the angles of a triangle is 180°]

⇒ ∠OAD + 36° + 64° = 180°

⇒ ∠OAD + 100° = 180°

⇒ ∠OAD = (180° − 100°) = 80°

Hence, ∠OAD = 80°.

(ii) As ∠DOC and ∠AOD form a linear pair, we have :

⇒ ∠DOC + ∠AOD = 180°

⇒ ∠DOC + 64° = 180°

⇒ ∠DOC = (180° − 64°) = 116°

In ΔDOC, we have :

⇒ ∠OCD + ∠ODC + ∠DOC = 180° [Sum of the angles of a triangle is 180°]

⇒ ∠OCD + 28° + 116° = 180°

⇒ ∠OCD + 144° = 180°

⇒ ∠OCD = (180° − 144°) = 36°

Hence, ∠OCD = 36°.

Given,

In square ABCD, AB = $(3x - 8)$ cm and BC = $(x + 6)$ cm.

Since all the sides of a square are equal, we have :

$$\begin{aligned}\Rightarrow AB = BC \\[1em] \Rightarrow 3x - 8 = x + 6 \\[1em] \Rightarrow 3x - x = 6 + 8 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \frac{14}{2} \\[1em] \Rightarrow x = 7\end{aligned}$$

∴ Each side of the square = AB = (3 × 7 − 8) cm = (21 − 8) cm = 13 cm

Hence, $\bm{x = 7}$ and the length of each side of the square is 13 cm.

Given,

ABCD is a parallelogram in which the diagonals AC and BD are equal, i.e., AC = BD.

In ΔABC and ΔBAD, we have :

AC = BD [Given]

BC = AD [Opposite sides of a parallelogram]

AB = BA [Common]

∴ ΔABC ≅ ΔBAD [S.S.S. axiom]

⇒ ∠ABC = ∠BAD [c.p.c.t.]

i.e., ∠B = ∠A

But, AD ∥ BC and AB cuts them, so we have :

⇒ ∠A + ∠B = 180° [Co-interior angles]

⇒ ∠A + ∠A = 180°

⇒ 2∠A = 180°

⇒ ∠A = $\frac{180^{\circ}}{2}$ = 90°

∴ ∠B = ∠A = 90°

Also, ∠C = ∠A = 90° and ∠D = ∠B = 90° [Opposite angles of a parallelogram]

Thus, ABCD is a parallelogram each of whose angles measures 90°.

Hence, the parallelogram is a rectangle.

Given,

ABCD is a square and CDE is an equilateral triangle.

(i) AD = DC [Sides of a square] and DC = DE [Sides of an equilateral triangle]

⇒ AD = DE

Also, ∠ADC = 90° [Each angle of a square = 90°] and ∠EDC = 60° [Each angle of an equilateral triangle = 60°]

⇒ ∠ADE = ∠ADC − ∠EDC = (90° − 60°) = 30°

In ΔADE, AD = DE, so ∠AED = ∠DAE [Angles opposite to equal sides are equal]

Now, ∠ADE + ∠AED + ∠DAE = 180° [Sum of the angles of a triangle is 180°]

⇒ 30° + ∠AED + ∠AED = 180° [∵ ∠DAE = ∠AED]

⇒ 30° + 2∠AED = 180°

⇒ 2∠AED = (180° − 30°) = 150°

⇒ ∠AED = $\frac{150^{\circ}}{2}$ = 75°

Hence, ∠AED = 75°.

(ii) ∠DAE = ∠AED = 75°

Also, ∠DAB = 90° [Each angle of a square = 90°]

⇒ ∠EAB = ∠DAB − ∠DAE = (90° − 75°) = 15°

Hence, ∠EAB = 15°.

(iii) ∠AEC = ∠AED + ∠DEC = (75° + 60°) = 135°

⇒ reflex ∠AEC = (360° − 135°) = 225°

Hence, reflex ∠AEC = 225°.

Given,

ABCD is a parallelogram in which X and Y are points on the diagonal BD such that AX ∥ CY.

(i) In ΔABX and ΔCDY, we have :

∠ABX = ∠CDY [Alternate angles, as AB ∥ DC and BD is a transversal]

∠AXB = ∠CYD [Alternate angles, as AX ∥ CY and BD is a transversal]

AB = CD [Opposite sides of a parallelogram]

∴ ΔABX ≅ ΔCDY [A.A.S. axiom]

⇒ AX = CY [c.p.c.t.]

Hence, AX = CY.

(ii) AX ∥ CY [Given]

and AX = CY [Proved above]

Thus, in quadrilateral AXCY, one pair of opposite sides is equal and parallel.

Hence, AXCY is a parallelogram.

Given,

ABCD is a parallelogram in which ∠A = 120° and the bisectors of ∠A and ∠B meet at P.

As AD ∥ BC and AB cuts them, we have :

⇒ ∠A + ∠B = 180° [Co-interior angles]

⇒ 120° + ∠B = 180°

⇒ ∠B = (180° − 120°) = 60°

As AP bisects ∠A, we have :

⇒ ∠PAB = $\frac{1}{2}$ × 120° = 60°

As BP bisects ∠B, we have :

⇒ ∠ABP = $\frac{1}{2}$ × 60° = 30°

In ΔAPB, we have :

⇒ ∠PAB + ∠ABP + ∠APB = 180° [Sum of the angles of a triangle is 180°]

⇒ 60° + 30° + ∠APB = 180°

⇒ 90° + ∠APB = 180°

⇒ ∠APB = (180° − 90°) = 90°

Hence, ∠APB is a right angle.

Given,

ABCD is a parallelogram and EF is a line segment through D such that EF ∥ AC, ∠ADE = 32° and ∠CDF = 26°.

As EF ∥ AC and AD is a transversal, we have :

⇒ ∠DAC = ∠ADE = 32° [Alternate angles are equal]

As EF ∥ AC and DC is a transversal, we have :

⇒ ∠DCA = ∠CDF = 26° [Alternate angles are equal]

In ΔADC, we have :

⇒ ∠DAC + ∠DCA + ∠ADC = 180° [Sum of the angles of a triangle is 180°]

⇒ 32° + 26° + ∠ADC = 180°

⇒ 58° + ∠ADC = 180°

⇒ ∠ADC = (180° − 58°) = 122°

∴ ∠ABC = ∠ADC = 122° [Opposite angles of a parallelogram]

Hence, ∠ABC = 122°.

Given,

AB = AC, CP ∥ BA and AP is the bisector of ∠CAD.

(i) In ΔABC, AB = AC.

⇒ ∠ABC = ∠ACB [Angles opposite to equal sides are equal]

As ∠CAD is the exterior angle of ΔABC at A, we have :

⇒ ∠CAD = ∠ABC + ∠ACB [Exterior angle property of a triangle]

⇒ ∠CAD = ∠ACB + ∠ACB = 2∠ACB

As AP bisects ∠CAD, we have :

⇒ ∠PAC = $\frac{1}{2}$ ∠CAD = $\frac{1}{2}$ × 2∠ACB = ∠ACB = ∠BCA

Hence, ∠PAC = ∠BCA.

(ii) ∠PAC and ∠BCA are the alternate angles formed when AP and BC are cut by the transversal AC.

⇒ AP ∥ BC

Also, CP ∥ BA [Given]

Thus, in quadrilateral ABCP, both the pairs of opposite sides are parallel.

Hence, ABCP is a parallelogram.

Given,

The diagonals AC and BD of a rhombus ABCD intersect each other at O.

We know that the diagonals of a rhombus bisect each other at right angles.

⇒ OA = OC, OB = OD and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

In right angled ΔAOB, by Pythagoras theorem, we have :

⇒ AB² = OA² + OB²

Similarly, in right angled ΔBOC, ΔCOD and ΔDOA, we have :

⇒ BC² = OB² + OC² = OB² + OA² [∵ OC = OA]

⇒ CD² = OC² + OD² = OA² + OB² [∵ OC = OA and OD = OB]

⇒ DA² = OD² + OA² = OB² + OA² [∵ OD = OB]

Adding all of them, we get :

⇒ AB² + BC² + CD² + DA² = (OA² + OB²) + (OB² + OA²) + (OA² + OB²) + (OB² + OA²)

⇒ AB² + BC² + CD² + DA² = 4OA² + 4OB²

⇒ AB² + BC² + CD² + DA² = 4(OA² + OB²)

Hence, AB² + BC² + CD² + DA² = 4(OA² + OB²).

Let ABCD be an isosceles trapezium in which AB ∥ DC and AD = BC.

We know that the angles on the same base of an isosceles trapezium are equal.

⇒ ∠A = ∠B and ∠C = ∠D

As AB ∥ DC and AD cuts them, we have :

⇒ ∠A + ∠D = 180° [Co-interior angles]

As AB ∥ DC and BC cuts them, we have :

⇒ ∠B + ∠C = 180° [Co-interior angles]

Now, ∠A + ∠C = ∠B + ∠C [∵ ∠A = ∠B]

⇒ ∠A + ∠C = 180° [∵ ∠B + ∠C = 180°]

Also, ∠B + ∠D = ∠A + ∠D [∵ ∠B = ∠A]

⇒ ∠B + ∠D = 180° [∵ ∠A + ∠D = 180°]

Here, (∠A, ∠C) and (∠B, ∠D) are the two pairs of opposite angles of the trapezium.

Hence, in an isosceles trapezium the opposite angles are supplementary.

Given,

ABCD is a parallelogram, E is the mid-point of AB and CE bisects ∠BCD.

(i) As AB ∥ DC and CE cuts them, we have :

⇒ ∠BEC = ∠ECD [Alternate angles are equal]

As CE bisects ∠BCD, we have :

⇒ ∠ECD = ∠ECB

⇒ ∠BEC = ∠ECB

In ΔBEC, ∠BEC = ∠ECB.

⇒ EB = BC [Sides opposite to equal angles are equal]

As E is the mid-point of AB, we have AE = EB.

Also, AD = BC [Opposite sides of a parallelogram]

⇒ AE = EB = BC = AD

Hence, AE = AD.

(ii) In ΔADE, AE = AD.

⇒ ∠ADE = ∠AED [Angles opposite to equal sides are equal]

As AB ∥ DC and DE cuts them, we have :

⇒ ∠AED = ∠EDC [Alternate angles are equal]

⇒ ∠ADE = ∠EDC

Hence, DE bisects ∠ADC.

(iii) As AD ∥ BC and DC cuts them, we have :

⇒ ∠ADC + ∠BCD = 180° [Co-interior angles]

⇒ $\frac{1}{2}$ (∠ADC) + $\frac{1}{2}$ (∠BCD) = 90°

⇒ ∠EDC + ∠DCE = 90°

In ΔDEC, we have :

⇒ ∠EDC + ∠DCE + ∠DEC = 180° [Sum of the angles of a triangle is 180°]

⇒ 90° + ∠DEC = 180°

⇒ ∠DEC = 180° - 90° = 90°

Hence, ∠DEC = 90°.

Given,

P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD.

In the given figure, let the diagonals AC and BD intersect at O.

We know that the diagonals of a parallelogram bisect each other.

⇒ OA = OC and OB = OD .....(1)

As P and Q are the points of trisection of BD, we have :

⇒ BP = PQ = QD = $\frac{1}{3}$ (BD)

⇒ BP = DQ .....(2)

Now, OP = OB − BP and OQ = OD − DQ

From (1) and (2), OB = OD and BP = DQ.

⇒ OB − BP = OD − DQ

⇒ OP = OQ .....(3)

In quadrilateral APCQ, the diagonals AC and PQ intersect at O.

From (1), OA = OC, so O is the mid-point of AC.

From (3), OP = OQ, so O is the mid-point of PQ.

Thus, the diagonals AC and PQ of quadrilateral APCQ bisect each other.

⇒ APCQ is a parallelogram [A quadrilateral whose diagonals bisect each other is a parallelogram]

⇒ CQ ∥ AP [Opposite sides of a parallelogram are parallel]

Hence, CQ ∥ AP.

Given,

ABCD is a parallelogram whose diagonals intersect at O and EOF is a line segment meeting AB at E and DC at F.

In ΔOAE and ΔOCF, we have :

∠OAE = ∠OCF [Alternate angles, as AB ∥ DC and AC is a transversal]

OA = OC [Diagonals of a parallelogram bisect each other]

∠AOE = ∠COF [Vertically opposite angles]

∴ ΔOAE ≅ ΔOCF [A.S.A. axiom]

⇒ OE = OF [c.p.c.t.]

Hence, OE = OF.

Let ABCD be a rhombus whose diagonals AC and BD intersect at O.

Then, AB = 10 cm and BD = 16 cm.

We know that the diagonals of a rhombus bisect each other at right angles.

⇒ OB = OD = $\frac{16}{2}$ = 8 cm and ∠AOB = 90°

In right angled ΔAOB, by Pythagoras theorem, we have :

⇒ AB² = OA² + OB²

⇒ (10)² = OA² + (8)²

⇒ 100 = OA² + 64

⇒ OA² = (100 − 64) = 36

⇒ OA = $\sqrt{36}$ = 6 cm

∴ AC = 2 × OA = (2 × 6) cm = 12 cm

Hence, the length of the other diagonal is 12 cm.

(i) Every square is a rectangle.

A square is a parallelogram each of whose angles measures 90°, which is exactly the definition of a rectangle.

⇒ The statement is true.

(ii) Every square is a rhombus.

A square is a parallelogram having all its sides equal, which is exactly the definition of a rhombus.

⇒ The statement is true.

(iii) Every rectangle is a parallelogram.

In a rectangle, each angle measures 90°, so both the pairs of opposite angles are equal.

A quadrilateral in which both the pairs of opposite angles are equal is a parallelogram. So, every rectangle is a parallelogram.

⇒ The statement is true.

(iv) Every parallelogram is a rhombus.

In a parallelogram, the opposite sides are equal, but the adjacent sides need not be equal, whereas in a rhombus all the four sides are equal.

So, a parallelogram is a rhombus only when its adjacent sides are equal. Thus, every parallelogram need not be a rhombus.

⇒ The statement is false.

(v) Every rhombus is a square.

In a rhombus the angles need not measure 90°, whereas in a square each angle measures 90°.

⇒ The statement is false.

(vi) The diagonals of a parallelogram are equal.

The diagonals of a parallelogram bisect each other but they are not equal, except when the parallelogram is a rectangle or a square.

⇒ The statement is false.

(vii) The diagonals of a parallelogram bisect each other at right angles.

The diagonals of a parallelogram bisect each other, but not at right angles, except when the parallelogram is a rhombus or a square.

⇒ The statement is false.

Hence, the statements (i), (ii) and (iii) are true.

The diagonals of a kite intersect each other at right angles.

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

The diagonals of a square are equal and bisect each other at right angles.

The diagonals of a rectangle are equal and bisect each other, but they need not intersect at right angles.

Hence, option 2 is the correct option.

A rectangle, a square and a rhombus are all parallelograms and the diagonals of a parallelogram bisect each other.

In a kite, only one of the diagonals is bisected by the other, so its diagonals need not bisect each other.

Hence, option 4 is the correct option.

The diagonals of an isosceles trapezium are equal in length.

The diagonals of a rectangle are equal and the diagonals of a square are also equal.

The diagonals of a rhombus bisect each other at right angles, but they need not be equal.

Hence, option 4 is the correct option.

Let ABCD be a parallelogram in which the adjacent angles ∠A and ∠B are equal.

As AD ∥ BC and AB cuts them, we have :

⇒ ∠A + ∠B = 180° [Co-interior angles]

⇒ ∠A + ∠A = 180° [∵ ∠A = ∠B]

⇒ 2∠A = 180°

⇒ ∠A = $\frac{180^{\circ}}{2}$ = 90°

∴ ∠B = ∠A = 90°, ∠C = ∠A = 90° and ∠D = ∠B = 90°

Thus, the parallelogram has each of its angles equal to 90°, so it is a rectangle.

Hence, option 1 is the correct option.

Let ABCD be a parallelogram in which the adjacent sides AB and BC are equal.

Since the opposite sides of a parallelogram are equal, we have :

⇒ CD = AB and AD = BC

⇒ AB = BC = CD = AD [∵ AB = BC]

Thus, the parallelogram has all its sides equal, so it is a rhombus.

Hence, option 3 is the correct option.

(i) The diagonals of a parallelogram bisect each other, but they need not be equal. So, the statement is false.

(ii) The diagonals of a square are equal and bisect each other at right angles, so they are perpendicular to each other. So, the statement is true.

(iii) The diagonals of a kite also intersect at right angles, but a kite need not be a rhombus. So, the statement is true.

(iv) A quadrilateral may have no pair of parallel sides and no pair of equal adjacent sides, so it need not be a trapezium or a parallelogram or a kite. So, the statement is false.

Thus, only the statements (ii) and (iii) are correct.

Hence, option 3 is the correct option.

Let the measures of the angles of the quadrilateral be $(2x)^{\circ}, (3x)^{\circ}, (4x)^{\circ}$ and $(9x)^{\circ}$ respectively.

We know that the sum of the angles of a quadrilateral is 360°. Then,

$$\begin{aligned}\Rightarrow (2x)^{\circ} + (3x)^{\circ} + (4x)^{\circ} + (9x)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (18x)^{\circ} = 360^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{360^{\circ}}{18} \\[1em] \Rightarrow x^{\circ} = 20^{\circ}\end{aligned}$$

∴ Smallest angle = $(2x)^{\circ}$ = (2 × 20)° = 40°

Hence, option 4 is the correct option.

Let the measures of the angles ∠A, ∠B, ∠C and ∠D be $x^{\circ}, (3x)^{\circ}, (7x)^{\circ}$ and $(9x)^{\circ}$ respectively.

We know that the sum of the angles of a quadrilateral is 360°. Then,

$$\begin{aligned}\Rightarrow x^{\circ} + (3x)^{\circ} + (7x)^{\circ} + (9x)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (20x)^{\circ} = 360^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{360^{\circ}}{20} \\[1em] \Rightarrow x^{\circ} = 18^{\circ}\end{aligned}$$

∴ ∠A = 18°, ∠B = (3 × 18)° = 54°, ∠C = (7 × 18)° = 126° and ∠D = (9 × 18)° = 162°

Now, ∠A + ∠D = (18° + 162°) = 180°

As ∠A and ∠D are co-interior angles formed by the transversal AD, we get AB ∥ DC.

Also, ∠A + ∠B = (18° + 54°) = 72° ≠ 180°, so BC is not parallel to AD.

Thus, ABCD has exactly one pair of parallel sides, so it is a trapezium with AB ∥ CD.

Hence, option 3 is the correct option.

The adjacent angles of a parallelogram are supplementary.

$$\begin{aligned}\Rightarrow (5x - 10)^{\circ} + (x + 10)^{\circ} = 180^{\circ} \\[1em] \Rightarrow (6x)^{\circ} = 180^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{180^{\circ}}{6} \\[1em] \Rightarrow x^{\circ} = 30^{\circ}\end{aligned}$$

∴ First angle = (5 × 30 − 10)° = (150 − 10)° = 140° and second angle = (30 + 10)° = 40°

∴ Ratio of these angles = 140 : 40 = 7 : 2

Hence, option 4 is the correct option.

Given,

ABCD is a parallelogram in which ∠A = 100° and the side CB is produced beyond B.

Since the opposite angles of a parallelogram are equal, we have :

⇒ $x$ = ∠BCD = ∠A

⇒ $x$ = 100°

As AD ∥ BC and AB cuts them, we have :

⇒ ∠A + ∠ABC = 180° [Co-interior angles]

⇒ 100° + $z$ = 180° [∵ ∠ABC = z]

⇒ $z$ = (180° − 100°) = 80°

As CB is produced beyond B, the angle $y$ and ∠ABC form a linear pair.

$$\begin{aligned}\Rightarrow y + ∠ABC = 180^{\circ} \\[1em] \Rightarrow y + 80^{\circ} = 180^{\circ} \\[1em] \Rightarrow y = (180^{\circ} − 80^{\circ}) \\[1em] \Rightarrow y = 100^{\circ}\end{aligned}$$

∴ $x = 100^{\circ}, y = 100^{\circ}$ and $z = 80^{\circ}$

Hence, option 2 is the correct option.

Given,

$(x + 16)^{\circ}, (2x - 6)^{\circ}, (3x + 15)^{\circ}$ and $(4x - 5)^{\circ}$ are the exterior angles of quadrilateral ABCD at D, C, A and B respectively.

We know that the sum of the exterior angles of a polygon taken in order is 360°.

$$\begin{aligned}\Rightarrow (x + 16)^{\circ} + (2x - 6)^{\circ} + (3x + 15)^{\circ} + (4x - 5)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (10x + 20)^{\circ} = 360^{\circ} \\[1em] \Rightarrow (10x)^{\circ} = 360^{\circ} - 20^{\circ} \\[1em] \Rightarrow (10x)^{\circ} = 340^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{340^{\circ}}{10} \\[1em] \Rightarrow x^{\circ} = 34^{\circ}\end{aligned}$$

∴ Exterior angles = (34 + 16)° = 50°, (2 × 34 − 6)° = (68 − 6)° = 62°, (3 × 34 + 15)° = (102 + 15)° = 117° and (4 × 34 − 5)° = (136 − 5)° = 131°

As each interior angle and the corresponding exterior angle form a linear pair, we have :

⇒ ∠D = (180° − 50°) = 130°

⇒ ∠C = (180° − 62°) = 118°

⇒ ∠A = (180° − 117°) = 63°

⇒ ∠B = (180° − 131°) = 49°

∴ The smallest angle of quadrilateral ABCD = 49°.

Hence, option 1 is the correct option.

Given,

In parallelogram ABCD the bisectors of ∠A and ∠B meet at E.

As AD ∥ BC and AB cuts them, we have :

⇒ ∠A + ∠B = 180° [Co-interior angles]

⇒ $\frac{1}{2}$ ∠A + $\frac{1}{2}$ ∠B = 90°

⇒ ∠EAB + ∠EBA = 90°

In ΔAEB, we have :

⇒ ∠EAB + ∠EBA + ∠AEB = 180° [Sum of the angles of a triangle is 180°]

⇒ 90° + ∠AEB = 180°

⇒ ∠AEB = 90°

Hence, option 3 is the correct option.

Given,

ABCD is a rectangle whose diagonals intersect at O, OD = $5x + 1$ and OA = $4x + 19$.

We know that the diagonals of a rectangle are equal and bisect each other. Then, OD = $\frac{1}{2}$ BD and OA = $\frac{1}{2}$ AC.

We have,

⇒ BD = AC

⇒ $\frac{1}{2}$ BD = $\frac{1}{2}$ AC

⇒ OD = OA

$$\begin{aligned}\Rightarrow 5x + 1 = 4x + 19 \\[1em] \Rightarrow 5x - 4x = 19 - 1 \\[1em] \Rightarrow x = 18\end{aligned}$$

Hence, option 4 is the correct option.

The diagonals of the given quadrilateral are divided into the parts $(3b - 2), 2a$ and $(5b - 9), (3b + 3)$ at their point of intersection.

The given quadrilateral will be a parallelogram only when its diagonals bisect each other.

$$\begin{aligned}\Rightarrow 5b - 9 = 3b + 3 \\[1em] \Rightarrow 5b - 3b = 3 + 9 \\[1em] \Rightarrow 2b = 12 \\[1em] \Rightarrow b = \frac{12}{2} \\[1em] \Rightarrow b = 6\end{aligned}$$

Also, $3b - 2 = 2a$

$$\begin{aligned}\Rightarrow 3 \times 6 - 2 = 2a \\[1em] \Rightarrow 2a = 16 \\[1em] \Rightarrow a = \frac{16}{2} \\[1em] \Rightarrow a = 8\end{aligned}$$

Hence, option 4 is the correct option.

Given,

ABCD is a rhombus in which ∠BAC = $x^{\circ}$.

We know that the diagonals of a rhombus bisect its angles, so AC bisects ∠A.

⇒ ∠A = 2∠BAC = $(2x)^{\circ}$

As AD ∥ BC and AB cuts them, we have :

⇒ ∠A + ∠B = 180° [Co-interior angles]

⇒ $(2x)^{\circ}$ + ∠B = 180°

⇒ ∠B = $(180 - 2x)^{\circ}$

Hence, option 4 is the correct option.

Given,

DEFG is a rhombus in which ∠FDE = 50°.

We know that the diagonals of a rhombus bisect its angles, so DF bisects ∠GDE.

⇒ ∠GDE = 2∠FDE = (2 × 50°) = 100°

As GF ∥ DE and GD cuts them, we have :

⇒ ∠DGF + ∠GDE = 180° [Co-interior angles]

⇒ ∠DGF + 100° = 180°

⇒ ∠DGF = (180° − 100°) = 80°

The diagonal GE bisects ∠DGF.

⇒ $p$ = ∠FGE = $\frac{1}{2}$ × 80° = 40°

Also, the diagonals of a rhombus bisect each other at right angles.

⇒ $q$ = 90°

∴ $q - p$ = (90° − 40°) = 50°

Hence, option 4 is the correct option.

In ∥gm PQRS, ∠P and ∠Q are adjacent angles, so they are supplementary.

⇒ ∠P + ∠Q = 180°

⇒ $a^{\circ}$ = 180°, i.e., $a = 180$

Also, ∠P and ∠R are opposite angles of the parallelogram, so they are equal.

⇒ ∠P − ∠R = 0°

⇒ $b^{\circ}$ = 0°, i.e., $b = 0$

∴ $a$ and $b$ are 180° and 0° respectively.

Hence, option 3 is the correct option.

(i) A rectangle whose adjacent sides are equal becomes a square.

(ii) The adjacent angles of a parallelogram are supplementary.

(iii) As all the sides of a square are equal, we have AB = BC.

$$\begin{aligned}\Rightarrow 2x + 3 = 3x - 5 \\[1em] \Rightarrow 3x - 2x = 3 + 5 \\[1em] \Rightarrow x = 8\end{aligned}$$

Hence, in a square ABCD, AB = $(2x + 3)$ cm and BC = $(3x - 5)$ cm. Then the value of $x$ is 8.

(iv) A rectangle is a parallelogram each of whose angles measures 90°.

A rhombus is a parallelogram having all its sides equal.

So, a parallelogram which is both a rectangle and a rhombus has each of its angles equal to 90° and all its sides equal, i.e., it is a square.

Hence, if a parallelogram is both, a rectangle and a rhombus, it is a square.

(v) In a rhombus opposite angles are equal.

(vi) A quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measures is a kite.

(i) Every square is a rhombus.

A square is a parallelogram having all its sides equal, so it is a rhombus.

⇒ The statement is true.

(ii) Diagonals of a rectangle are perpendicular to each other.

The diagonals of a rectangle are equal and bisect each other, but they need not be perpendicular to each other.

⇒ The statement is false.

(iii) Diagonals of a rhombus are equal and perpendicular to each other.

The diagonals of a rhombus bisect each other at right angles, but they need not be equal.

⇒ The statement is false.

(iv) Every rhombus is a square.

The angles of a rhombus need not measure 90°, whereas each angle of a square measures 90°.

⇒ The statement is false.

(v) If the diagonals of a quadrilateral bisect each other, it must be a parallelogram.

A quadrilateral whose diagonals bisect each other is a parallelogram.

⇒ The statement is true.

(vi) If one pair of opposite angles of a quadrilateral are equal, it must be a parallelogram.

In a kite, one pair of opposite angles is equal, but a kite need not be a parallelogram.

⇒ The statement is false.

(vii) If the diagonals of a quadrilateral are perpendicular to each other, it is definitely a rhombus.

The diagonals of a kite are also perpendicular to each other, but a kite need not be a rhombus.

⇒ The statement is false.

A trapezium is a quadrilateral in which one pair of opposite sides is parallel.

In a kite, two pairs of adjacent sides are equal, but no pair of opposite sides is parallel. So, a kite is not a trapezium.

⇒ Assertion (A) is false.

By definition, a trapezium is a quadrilateral in which a pair of opposite sides is parallel.

⇒ Reason (R) is true.

Hence, option 4 is the correct option.

The sum of the interior angles of a quadrilateral is 360° and the sum of its exterior angles taken in order is also 360°.

So, for a quadrilateral, the sum of the exterior angles is equal to the sum of the interior angles.

⇒ Assertion (A) is true.

The sum of the exterior angles of a polygon taken in order is 360°.

⇒ Reason (R) is true.

The interior angles of a quadrilateral add up to 360°, and Reason (R) states that its exterior angles also add up to 360°. This is exactly why the two sums are equal.

So, Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

⇒ Assertion (A) is true.

The diagonals of a parallelogram bisect each other, but they need not be equal. They are equal only when the parallelogram is a rectangle or a square.

⇒ Reason (R) is false.

Hence, option 3 is the correct option.

Given,

ABCD is a rhombus, DE ⊥ AB and E is the mid-point of AB.

Join DB.

In ΔAED and ΔBED, we have :

AE = BE [∵ E is the mid-point of AB]

∠AED = ∠BED [Each equal to 90°]

DE = DE [Common]

∴ ΔAED ≅ ΔBED [S.A.S. axiom]

⇒ AD = BD [c.p.c.t.]

Also, AD = AB [Sides of a rhombus]

⇒ AD = AB = BD

So, ΔABD is an equilateral triangle.

⇒ ∠A = 60°

Hence, option 2 is the correct option.

Given,

CP ∥ DB, ∠DAB = 100°, ∠ABD = 40°, ∠ADC = 85° and ∠BCD = 70°.

In quadrilateral ABCD, we have :

⇒ ∠DAB + ∠ABC + ∠BCD + ∠ADC = 360° [Sum of the angles of a quadrilateral is 360°]

⇒ 100° + ∠ABC + 70° + 85° = 360°

⇒ 255° + ∠ABC = 360°

⇒ ∠ABC = (360° − 255°) = 105°

Now, ∠DBC = ∠ABC − ∠ABD = (105° − 40°) = 65°

As CP ∥ DB and BC is a transversal, we have :

⇒ ∠PCB = ∠DBC = 65° [Alternate angles]

Hence, option 1 is the correct option.

Given,

ABCD is a parallelogram in which BM ⊥ DC and BN ⊥ AD, where M lies on DC and N lies on AD, and ∠MBC = 40°.

Here, $z$ = ∠BCM, $y$ = ∠NBM and $x$ = ∠DAB.

In right angled ΔBMC, we have :

⇒ ∠BMC + ∠MBC + ∠BCM = 180° [Sum of the angles of a triangle is 180°]

⇒ 90° + 40° + $z$ = 180°

⇒ 130° + $z$ = 180°

⇒ $z$ = (180° − 130°) = 50°

Since the opposite angles of a parallelogram are equal, we have :

⇒ $x$ = ∠DAB = ∠BCD = $z$ = 50°

In right angled ΔANB, we have :

⇒ ∠ANB + ∠NAB + ∠ABN = 180° [Sum of the angles of a triangle is 180°]

⇒ 90° + 50° + ∠ABN = 180°

⇒ 140° + ∠ABN = 180°

⇒ ∠ABN = (180° − 140°) = 40°

As AD ∥ BC and AB cuts them, we have :

⇒ ∠DAB + ∠ABC = 180° [Co-interior angles]

⇒ 50° + ∠ABC = 180°

⇒ ∠ABC = (180° − 50°) = 130°

Now, ∠ABC = ∠ABN + $y$ + ∠MBC

⇒ 130° = 40° + $y$ + 40°

⇒ 130° = 80° + $y$

⇒ $y$ = (130° − 80°) = 50°

∴ $x + y + z$ = (50° + 50° + 50°) = 150°

Hence, option 3 is the correct option.

Given,

ABCD is a rhombus in which ∠BAC = 40°.

We know that the diagonals of a rhombus bisect its angles, so AC bisects ∠BAD.

⇒ ∠BAD = 2∠BAC = (2 × 40°) = 80°

As AD ∥ BC and AB cuts them, we have :

⇒ ∠BAD + ∠ABC = 180° [Co-interior angles]

⇒ 80° + ∠ABC = 180°

⇒ ∠ABC = (180° − 80°) = 100°

∴ The two adjacent angles of the rhombus are 100° and 80°.

Hence, option 1 is the correct option.

Given,

ABCD is a rhombus in which ∠ACB = 30° and the diagonal AC is extended to F so that CF is a diagonal of parallelogram CEFG.

We know that the diagonals of a rhombus bisect its angles, so CA bisects ∠BCD.

⇒ ∠ACD = ∠ACB = 30°

As B, C, E are collinear and A, C, F are collinear, we have :

⇒ ∠ECF = ∠ACB = 30° [Vertically opposite angles]

As D, C, G are collinear and A, C, F are collinear, we have :

⇒ ∠GCF = ∠ACD = 30° [Vertically opposite angles]

⇒ ∠ECF = ∠GCF = 30°

As EF ∥ CG and CF is a transversal, we have :

⇒ ∠EFC = ∠GCF = 30° [Alternate angles]

In ΔCEF, ∠ECF = ∠EFC = 30°.

⇒ CE = EF [Sides opposite to equal angles are equal]

Also, CE = FG and EF = CG [Opposite sides of a parallelogram]

⇒ CE = EF = FG = CG

So, CEFG is a parallelogram having all its sides equal, i.e., it is a rhombus.

Also, ∠ECG = ∠ECF + ∠GCF = (30° + 30°) = 60° ≠ 90°, so CEFG is not a square.

Hence, option 3 is the correct option.

In a kite, two pairs of adjacent sides are equal, one of the diagonals is bisected by the other and the diagonals are perpendicular to each other.

Divya describes a quadrilateral whose opposite sides are equal and parallel, whose diagonals bisect each other and are perpendicular to each other. This is the description of a rhombus, not of a kite.

Kavya describes a quadrilateral having two pairs of equal consecutive sides, in which one of the diagonals bisects the other and the diagonals are perpendicular to each other. This is the correct description of a kite.

Hence, option 2 is the correct option.

The diagonals of a square are equal in length and the diagonals of a rectangle are also equal in length.

The diagonals of a parallelogram bisect each other, but they need not be equal in length.

So, a quadrilateral windowframe having one of its diagonals longer than the other can only be a parallelogram.

Hence, option 3 is the correct option.

Given,

Four similar right triangles with legs 3 cm and 2 cm are joined at their right angled vertices to form quadrilateral PQRS.

Then, the two diagonals of PQRS meet at right angles at the common vertex and we have :

⇒ SQ = (3 + 3) cm = 6 cm and RP = (2 + 2) cm = 4 cm

So, the diagonals of PQRS bisect each other at right angles, but they are not equal.

∴ PQRS is a rhombus, but it is not a square.

Hence, option 3 is the correct option.

Given,

ABCD is a square which is cut along XY, where X lies on DC and Y lies on AB with XY ⊥ AB.

In quadrilateral ADXY, we have DX ∥ AY and ∠A = ∠D = ∠AYX = ∠DXY = 90°.

So, ADXY is a rectangle.

(i) The diagonals of a rectangle are equal, so the lengths of the diagonals of ADXY are equal. This statement is correct.

(ii) The diagonals of a rectangle are perpendicular to each other only when the rectangle is a square. Since Y lies between A and B, AY < AB = AD, i.e., AD ≠ AY. So, ADXY is not a square and its diagonals are not perpendicular to each other. This statement is not correct.

Hence, option 1 is the correct option.