# Understanding Quadrilaterals Class 8 Extra Questions Maths Chapter 3

**Extra Questions for Class 8 Maths Chapter 3 Understanding Quadrilaterals**

### Understanding Quadrilaterals Class 8 Extra Questions Very Short Answer Type

Question 1.

In the given figure, ABCD is a parallelogram. Find x.

Solution:

AB = DC [Opposite sides of a parallelogram]

3x + 5 = 5x – 1

⇒ 3x – 5x = -1 – 5

⇒ -2x = -6

⇒ x = 3

Question 2.

In the given figure find x + y + z.

Solution:

We know that the sum of all the exterior angles of a polygon = 360°

x + y + z = 360°

Question 3.

In the given figure, find x.

Solution:

∠A + ∠B + ∠C = 180° [Angle sum property]

(x + 10)° + (3x + 5)° + (2x + 15)° = 180°

⇒ x + 10 + 3x + 5 + 2x + 15 = 180

⇒ 6x + 30 = 180

⇒ 6x = 180 – 30

⇒ 6x = 150

⇒ x = 25

Question 4.

The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle.

Solution:

Sum of all interior angles of a quadrilateral = 360°

Let the angles of the quadrilateral be 2x°, 3x°, 5x° and 8x°.

2x + 3x + 5x + 8x = 360°

⇒ 18x = 360°

⇒ x = 20°

Hence the angles are

2 × 20 = 40°,

3 × 20 = 60°,

5 × 20 = 100°

and 8 × 20 = 160°.

Question 5.

Find the measure of an interior angle of a regular polygon of 9 sides.

Solution:

Measure of an interior angle of a regular polygon

Question 6.

Length and breadth of a rectangular wire are 9 cm and 7 cm respectively. If the wire is bent into a square, find the length of its side.

Solution:

Perimeter of the rectangle = 2 [length + breadth]

= 2[9 + 7] = 2 × 16 = 32 cm.

Now perimeter of the square = Perimeter of rectangle = 32 cm.

Side of the square = \(\frac { 32 }{ 4 }\) = 8 cm.

Hence, the length of the side of square = 8 cm.

Question 7.

In the given figure ABCD, find the value of x.

Solution:

Sum of all the exterior angles of a polygon = 360°

x + 70° + 80° + 70° = 360°

⇒ x + 220° = 360°

⇒ x = 360° – 220° = 140°

Question 8.

In the parallelogram given alongside if m∠Q = 110°, find all the other angles.

Solution:

Given m∠Q = 110°

Then m∠S = 110° (Opposite angles are equal)

Since ∠P and ∠Q are supplementary.

Then m∠P + m∠Q = 180°

⇒ m∠P + 110° = 180°

⇒ m∠P = 180° – 110° = 70°

⇒ m∠P = m∠R = 70° (Opposite angles)

Hence m∠P = 70, m∠R = 70°

and m∠S = 110°

Question 9.

In the given figure, ABCD is a rhombus. Find the values of x, y and z.

Solution:

AB = BC (Sides of a rhombus)

x = 13 cm.

Since the diagonals of a rhombus bisect each other

z = 5 and y = 12

Hence, x = 13 cm, y = 12 cm and z = 5 cm.

Question 10.

In the given figure, ABCD is a parallelogram. Find x, y and z.

Solution:

∠A + ∠D = 180° (Adjacent angles)

⇒ 125° + ∠D = 180°

⇒ ∠D = 180° – 125°

x = 55°

∠A = ∠C [Opposite angles of a parallelogram]

⇒ 125° = y + 56°

⇒ y = 125° – 56°

⇒ y = 69°

∠z + ∠y = 180° (Adjacent angles)

⇒ ∠z + 69° = 180°

⇒ ∠z = 180° – 69° = 111°

Hence the angles x = 55°, y = 69° and z = 111°

Question 11.

Find x in the following figure. (NCERT Exemplar)

Solution:

In the given figure ∠1 + 90° = 180° (linear pair)

∠1 = 90°

Now, sum of exterior angles of a polygon is 360°, therefore,

x + 60° + 90° + 90° + 40° = 360°

⇒ x + 280° = 360°

⇒ x = 80°

### Understanding Quadrilaterals Class 8 Extra Questions Short Answer Type

Question 12.

In the given parallelogram ABCD, find the value of x andy.

Solution:

∠A + ∠B = 180°

3y + 2y – 5 = 180°

⇒ 5y – 5 = 180°

⇒ 5y = 180 + 5°

⇒ 5y = 185°

⇒ y = 37°

Now ∠A = ∠C [Opposite angles of a parallelogram]

3y = 3x + 3

⇒ 3 × 37 = 3x + 3

⇒ 111 = 3x + 3

⇒ 111 – 3 = 3x

⇒ 108 = 3x

⇒ x = 36°

Hence, x = 36° and y – 37°.

Question 13.

ABCD is a rhombus with ∠ABC = 126°, find the measure of ∠ACD.

Solution:

∠ABC = ∠ADC (Opposite angles of a rhombus)

∠ADC = 126°

∠ODC = \(\frac { 1 }{ 2 }\) × ∠ADC (Diagonal of rhombus bisects the respective angles)

⇒ ∠ODC = \(\frac { 1 }{ 2 }\) × 126° = 63°

⇒ ∠DOC = 90° (Diagonals of a rhombus bisect each other at 90°)

In ΔOCD,

∠OCD + ∠ODC + ∠DOC = 180° (Angle sum property)

⇒ ∠OCD + 63° + 90° = 180°

⇒ ∠OCD + 153° = 180°

⇒ ∠OCD = 180° – 153° = 27°

Hence ∠OCD or ∠ACD = 27°

Question 14.

Find the values of x and y in the following parallelogram.

Solution:

Since, the diagonals of a parallelogram bisect each other.

OA = OC

x + 8 = 16 – x

⇒ x + x = 16 – 8

⇒ 2x = 8

x = 4

Similarly, OB = OD

5y + 4 = 2y + 13

⇒ 3y = 9

⇒ y = 3

Hence, x = 4 and y = 3

Question 15.

Write true and false against each of the given statements.

(a) Diagonals of a rhombus are equal.

(b) Diagonals of rectangles are equal.

(c) Kite is a parallelogram.

(d) Sum of the interior angles of a triangle is 180°.

(e) A trapezium is a parallelogram.

(f) Sum of all the exterior angles of a polygon is 360°.

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

(h) Triangle is possible with angles 60°, 80° and 100°.

(i) In a parallelogram, the opposite sides are equal.

Solution:

(a) False

(b) True

(c) False

(d) True

(e) False

(f) True

(g) False

(h) False

(i) True

Question 16.

The sides AB and CD of a quadrilateral ABCD are extended to points P and Q respectively. Is ∠ADQ + ∠CBP = ∠A + ∠C? Give reason.

(NCERT Exemplar)

Solution:

Join AC, then

∠CBP = ∠BCA + ∠BAC and ∠ADQ = ∠ACD + ∠DAC (Exterior angles of triangles)

Therefore,

∠CBP + ∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC

= (∠BCA + ∠ACD) + (∠BAC + ∠DAC)

= ∠C + ∠A

### Understanding Quadrilaterals Class 8 Extra Questions Higher Order Thinking Skills (HOTS)

Question 17.

The diagonal of a rectangle is thrice its smaller side. Find the ratio of its sides.

Solution:

Let AD = x cm

diagonal BD = 3x cm

In right-angled triangle DAB,

AD^{2} + AB^{2} = BD^{2} (Using Pythagoras Theorem)

x^{2} + AB^{2} = (3x)^{2}

⇒ x^{2} + AB^{2} = 9x^{2}

⇒ AB^{2} = 9x^{2} – x^{2}

⇒ AB^{2} = 8x^{2}

⇒ AB = √8x = 2√2x

Required ratio of AB : AD = 2√2x : x = 2√2 : 1

Question 18.

If AM and CN are perpendiculars on the diagonal BD of a parallelogram ABCD, Is ∆AMD = ∆CNB? Give reason. (NCERT Exemplar)

Solution:

In triangles AMD and CNB,

AD = BC (opposite sides of parallelogram)

∠AMB = ∠CNB = 90°

∠ADM = ∠NBC (AD || BC and BD is transversal.)

So, ∆AMD = ∆CNB (AAS)