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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.

Q 1.

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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.

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

rhombus

B

rectangle

C

quadrilateral

D

none of these

Q 3.

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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.

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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.

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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} \triangle ADP, {/tex}
{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.

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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

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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.

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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.

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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.

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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.

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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}