## 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

**Understanding Quadrilaterals Class 8 Worksheets With Answers Question 1.**

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

Solution:

AB = DC [Opposite sides of a parallelogram]

3x + 5 = 5x – 1

â‡’ 3x – 5x = -1 – 5

â‡’ -2x = -6

â‡’ x = 3

**Understanding Quadrilaterals Class 8 Worksheet Question 2.**

In the given figure find x + y + z.

Solution:

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

x + y + z = 360Â°

**Understanding Quadrilaterals Class 8 Extra Questions Question 3.**

In the given figure, find x.

Solution:

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

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

â‡’ x + 10 + 3x + 5 + 2x + 15 = 180

â‡’ 6x + 30 = 180

â‡’ 6x = 180 – 30

â‡’ 6x = 150

â‡’ x = 25

**Class 8 Maths Chapter 3 Extra Questions 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 = 360Â°

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

2x + 3x + 5x + 8x = 360Â°

â‡’ 18x = 360Â°

â‡’ x = 20Â°

Hence the angles are

2 Ã— 20 = 40Â°,

3 Ã— 20 = 60Â°,

5 Ã— 20 = 100Â°

and 8 Ã— 20 = 160Â°.

**Understanding Quadrilaterals Class 8 Questions With Solutions 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

**Understanding Quadrilaterals Class 8 Worksheet Pdf With Answers 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 Ã— 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.

**Class 8 Understanding Quadrilaterals Extra Questions Question 7.**

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

Solution:

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

x + 70Â° + 80Â° + 70Â° = 360Â°

â‡’ x + 220Â° = 360Â°

â‡’ x = 360Â° – 220Â° = 140Â°

**Understanding Quadrilaterals Extra Questions Question 8.**

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

Solution:

Given mâˆ Q = 110Â°

Then mâˆ S = 110Â° (Opposite angles are equal)

Since âˆ P and âˆ Q are supplementary.

Then mâˆ P + mâˆ Q = 180Â°

â‡’ mâˆ P + 110Â° = 180Â°

â‡’ mâˆ P = 180Â° – 110Â° = 70Â°

â‡’ mâˆ P = mâˆ R = 70Â° (Opposite angles)

Hence mâˆ P = 70, mâˆ R = 70Â°

and mâˆ S = 110Â°

**Class 8 Maths Ch 3 Extra Questions 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.

**Class 8 Quadrilaterals Extra Questions Question 10.**

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

Solution:

âˆ A + âˆ D = 180Â° (Adjacent angles)

â‡’ 125Â° + âˆ D = 180Â°

â‡’ âˆ D = 180Â° – 125Â°

x = 55Â°

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

â‡’ 125Â° = y + 56Â°

â‡’ y = 125Â° – 56Â°

â‡’ y = 69Â°

âˆ z + âˆ y = 180Â° (Adjacent angles)

â‡’ âˆ z + 69Â° = 180Â°

â‡’ âˆ z = 180Â° – 69Â° = 111Â°

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

**Class 8 Maths Chapter 3 Extra Questions With Solutions Question 11.**

Find x in the following figure. (NCERT Exemplar)

Solution:

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

âˆ 1 = 90Â°

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

x + 60Â° + 90Â° + 90Â° + 40Â° = 360Â°

â‡’ x + 280Â° = 360Â°

â‡’ x = 80Â°

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

**Quadrilaterals Class 8 Extra Questions Question 12.**

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

Solution:

âˆ A + âˆ B = 180Â°

3y + 2y – 5 = 180Â°

â‡’ 5y – 5 = 180Â°

â‡’ 5y = 180 + 5Â°

â‡’ 5y = 185Â°

â‡’ y = 37Â°

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

3y = 3x + 3

â‡’ 3 Ã— 37 = 3x + 3

â‡’ 111 = 3x + 3

â‡’ 111 – 3 = 3x

â‡’ 108 = 3x

â‡’ x = 36Â°

Hence, x = 36Â° and y – 37Â°.

**Questions On Quadrilaterals For Class 8 Question 13.**

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

Solution:

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

âˆ ADC = 126Â°

âˆ ODC = \(\frac { 1 }{ 2 }\) Ã— âˆ ADC (Diagonal of rhombus bisects the respective angles)

â‡’ âˆ ODC = \(\frac { 1 }{ 2 }\) Ã— 126Â° = 63Â°

â‡’ âˆ DOC = 90Â° (Diagonals of a rhombus bisect each other at 90Â°)

In Î”OCD,

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

â‡’ âˆ OCD + 63Â° + 90Â° = 180Â°

â‡’ âˆ OCD + 153Â° = 180Â°

â‡’ âˆ OCD = 180Â° – 153Â° = 27Â°

Hence âˆ OCD or âˆ ACD = 27Â°

**Extra Questions On Understanding Quadrilaterals Class 8 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

â‡’ x + x = 16 – 8

â‡’ 2x = 8

x = 4

Similarly, OB = OD

5y + 4 = 2y + 13

â‡’ 3y = 9

â‡’ 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 180Â°.

(e) A trapezium is a parallelogram.

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

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

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

(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 âˆ ADQ + âˆ CBP = âˆ A + âˆ C? Give reason.

(NCERT Exemplar)

Solution:

Join AC, then

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

Therefore,

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

= (âˆ BCA + âˆ ACD) + (âˆ BAC + âˆ DAC)

= âˆ C + âˆ 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}

â‡’ x^{2} + AB^{2} = 9x^{2}

â‡’ AB^{2} = 9x^{2} – x^{2}

â‡’ AB^{2} = 8x^{2}

â‡’ AB = âˆš8x = 2âˆš2x

Required ratio of AB : AD = 2âˆš2x : x = 2âˆš2 : 1

Question 18.

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

Solution:

In triangles AMD and CNB,

AD = BC (opposite sides of parallelogram)

âˆ AMB = âˆ CNB = 90Â°

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

So, âˆ†AMD = âˆ†CNB (AAS)