# SSC > Quadrilateral and Parallelogram

Explore popular questions from Quadrilateral and Parallelogram for SSC. This collection covers Quadrilateral and Parallelogram previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

Correct2

Incorrect-0.5

The side of a rhombus whose diagonals are 16 cm and 12 cm respectively, is

A

14 cm

B

12 cm

10 cm

D

8 cm

##### Explanation

Given : {tex} d_{1}=16\ cm,\ d_{2}=12\ cm {/tex}

Side of rhombus = {tex} \frac{1}{2}\sqrt{d^{2}_{1}+d^{2}_{2}} {/tex}

= {tex} \frac{1}{2}\sqrt{16^{2}+12^{2}} {/tex}

= {tex} \frac{1}{2}\sqrt{400} {/tex}

= {tex} \frac{1}{2} \times 20 = 10\ cm {/tex}

Q 2.

Correct2

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If diagonals of a parallelogram are perpendicular, then it is a

rhombus

B

rectangle

C

D

none of these

Q 3.

Correct2

Incorrect-0.5

If a triangle and a rectangle have equal areas and equal altitude, then base of the triangle is equal to

A

base of the rectangle

twice the base of the rectangle

C

thrice the base of the rectangle

D

four times the base of the rectangle

##### Explanation

Let h be the common height of triangle and rectangle.

If a and b be respectively the bases of triangle and rectangle, then by hypothesis

{tex} \frac{1}{2}a \times h = b \times h{/tex}

=> {tex} \frac{1}{2}a = b {/tex}

=> {tex} a = 2b {/tex}

Q 4.

Correct2

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The number of sides of a regular polygon each of whose angles measures {tex} 156 ^{\circ} {/tex} is

A

8

B

10

C

12

15

##### Explanation

Sum of the interior angles of an n-sided regular polygon

= {tex} \left(2n-4\right) \times 90 ^{\circ} {/tex}

Therefore, {tex} \left(2n-4\right) \times 90 ^{\circ} = 156 \times n {/tex}

=> {tex} 180n - 360 = 156n {/tex}

=> {tex} 180n - 156n = 360 {/tex}

=> {tex} n = \frac{360}{24} = 15 {/tex}

Q 5.

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ABCD is a rhombus and O is the point of intersection of the diagonals AC and BD. The locus of a point which is equidistant from AB and AD is

AC

B

BD

C

CB

D

CD

Q 6.

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The area of a trapezium is {tex} 275 cm^{2} {/tex}. If its parallel sides are in the ratio 2 : 3 and perpendicular distance between them is 5 cm, then smaller of the parallel sides is

A

36 cm

B

40 cm

44 cm

D

48 cm

##### Explanation

Given : Area of trapezium = {tex} 275\ cm^{2} {/tex}

Let parallel sides be 2x and 3x.

Area of trapezium

= {tex} \frac{1}{2} \times sum\ of\ parallel\ side \times h {/tex}

Therefore, {tex} \frac{1}{2} \left(2x+3x\right) \times 5 = 275 {/tex}

{tex} x = \frac{2 \times 275}{25}\ =\ 22\ cm {/tex}

Therefore, Smaller of the parallel sides = 2x = 44 cm

Q 7.

Correct2

Incorrect-0.5

If each angle of a polygon is {tex} 165 ^{\circ} {/tex}, then number sides of the polygon is

24

B

30

C

35

D

40

##### Explanation

Exterior angle = {tex} 180 ^{\circ} - 165 ^{\circ} = 15 ^{\circ} {/tex}

Therefore, Number of sides = {tex} \frac{360 ^{\circ} }{Exterior\ angle} {/tex}

= {tex} \frac{360}{15} = 24 {/tex}

Q 8.

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ABCD is trapezium with AB parallel to DC. If AB=10 cm, AD = BC = 4 cm and {tex} \angle DAB = \angle CBA = 60 ^{\circ} {/tex}, then length of CD is equal to

A

3 cm

6 cm

C

7 cm

D

7.5 cm

##### Explanation

{tex} \frac{AP}{AD} = \cos 60 ^{\circ} {/tex}
=> {tex} \frac{x}{4} = \frac{1}{2} {/tex}
=> x = 2
Therefore, CD = AB - 2x = 10 - 4 = 6 cm

Q 9.

Correct2

Incorrect-0.5

The points A (-2, -1), B (1, 0), C (4, 3) and D (1, 2) are corners of a

A

Square

B

Trapezium

Parallelogram

D

Rectangle

##### Explanation

{tex} AB = \sqrt{3^{2}+1^{2}}=\sqrt{10} {/tex}
{tex} BC = \sqrt{3^{2}+3^{2}}=3\sqrt{2} {/tex}
{tex} CD = \sqrt{3^{2}+1^{2}}=\sqrt{10} {/tex}
{tex} DA = \sqrt{3^{2}+3^{2}}=3\sqrt{2} {/tex}
Therefore, Diagonal AC = {tex} \sqrt{6^{2}+4^{2}}=\sqrt{52} {/tex}
and diagonal BD = {tex} \sqrt{0+2^{2}}=2 {/tex}
Since opposite sides are equal and the diagonals are not equal, hence ABCD is a parallelogram.

Q 10.

Correct2

Incorrect-0.5

If two adjacent sides of a rectangle are 2 cm and 4 cm, then its perimeter is

A

6 cm

B

8 cm

12 cm

D

24 cm

##### Explanation

Perimeter = {tex} 2 \left(l+b\right)=2 \left(2+4\right) {/tex}

= 12 cm

Q 11.

Correct2

Incorrect-0.5

Sum of the interior angles of a regular polygon having 'n' sides is equal to

A

{tex} \left(n + 2\right) \pi {/tex}

B

{tex} \left(n + 1\right) \pi {/tex}

C

{tex} \left(n - 1\right) \pi {/tex}

{tex} \left(n - 2\right) \pi {/tex}

##### Explanation

Required sum of the interior angles of an n-side regular polygon

= {tex} n \pi - 2 \pi {/tex}

={tex} \left(n-2\right) \pi {/tex}

Q 12.

Correct2

Incorrect-0.5

In a regular polygon, if an interior angle is equal to four times the exterior angle, then number of sides in the polygon is

A

7

B

8

10

D

11

##### Explanation

Interior angle (I) + Exterior angle (E) = {tex} 180 ^{\circ} {/tex}

Therefore, I + E = {tex} 180 ^{\circ} {/tex}

Now I = AE (given)

Therefore, AE + E = {tex} 180 ^{\circ} {/tex}

E = {tex} 36 ^{\circ} {/tex}

Therefore, number of sides = {tex} \frac{360 ^{\circ} }{E} = \frac{360}{36} = 10 {/tex}

Q 13.

Correct2

Incorrect-0.5

If one side of a regular polygon with seven sides is produced, then exterior angle (in degrees) has the magnitude

A

60

{tex} 51\frac{3}{7} {/tex}

C

45

D

40

##### Explanation

Interior angle = {tex} \frac{ \left(n-2\right) \pi }{n}= \left(\frac{7-2}{7}\right) \pi = \frac{5 \pi }{7} {/tex}

Therefore, Exterior angle = {tex} \pi -\frac{5 \pi }{7} = \frac{2 \pi }{7} = \left(51\frac{3}{7}\right) ^{\circ} {/tex}

Q 14.

Correct2

Incorrect-0.5

The angle BDE in a regular hexagon ABCDEFFA is equal to

A

{tex} 120 ^{\circ} {/tex}

B

{tex} 105 ^{\circ} {/tex}

{tex} 90 ^{\circ} {/tex}

D

{tex} 60 ^{\circ} {/tex}