Mensuration Practice Set-1 for SSC CGL, CHSL, IBPS PO and Clerk Exams

Mensuration Practice Set-1 for SSC CGL, CHSL, IBPS PO and Clerk Exams. This is Mensuration Practice Set-1 for SSC CGL, CHSL, IBPS PO and Clerk Exams. These Multiple Choice Questions has been collected by our team from different previous years exams. We hope this will helpful  in enhancing your knowledge.

1. The base of a right prism is a triangle whose perimeter is 28 cm and the in radius of the triangle is 4 cm. If the volume of the prism is 366 cc, then its height is [SSC 10+2-2013]

(a) 4 cm

(b) 8 cm

(c) 6 cm

(d) 6.5 cm

Explanation/Hint :

Given, Perimeter = 28 cm or s = 28/2 = 14 cm

Also, Radius of triangle (r) = 4 cm.

Volume of the prism = 366 cc.

We know, Area of base of prism = $\frac{1}{2}\times r\times \left(a+b+c \right)$

= r x s = 4 x 14 =56 cm2

Volume of prism = Area of base of prism x height of prism

or 366 cc = 56 x h

or h = 6.53 cm

2. The sum of the interior angles of a polygon is 1444°. The number of sides of the polygon is [SSC 10+2-2014]

(a) 6

(b) 9

(c) 10

(d) 12

Explanation/Hint :

Given, Sum of interior angles of polygon = 10

Sum of interior angles of polygon = (n - 2) x 180°

or n = 10

3. The sides of triangle is given in the ratio of $\frac{1}{2}$ : $\frac{1}{4}$ :$\frac{1}{5}$ and its perimeter is 94 cm. Find the length of largest side of the triangle :

1. 48 cm
2. 42.5 cm
3. 40 cm
4. 47 cm

Explanation/Hint :

Let the sides of triangle be x,
Given sides of triangle $\frac{1}{2}$ : $\frac{1}{4}$ : $\frac{1}{5}$
= $\frac{60}{3}$ : $\frac{60}{4}$ :$\frac{60}{5}$L.C.M of 3, 4 ,5)
= 20 : 15: 12 or 20x, 15x and 12x

Now,
20x + 15x + 12x = 94 ( a+b+c = perimeter of triangle )
x = 2, The length of largest side = 20 x 2 = 40 cm

4. The perimeters of two similar triangles ΔABC and ΔPQR are 36 cm and 24 cm respectively. If PQ = 10 cm, the AB is [SSC 10+2-2014]

(a) 15 cm

(b) 12 cm

(c) 14 cm

(d) 26 cm

Explanation/Hint :

Given, ΔABC ≈ ΔPQR

Thus, $\frac{AB}{PQ}$ = $\frac{BC}{QR}$ = $\frac{AC}{PR}$ = $\frac{AB+BC+AC}{PQ+QR+PR}$

or $\frac{AB+BC+AC}{PQ+QR+PR}$ = $\frac{AB}{PQ}$

or $\frac{Perimeter of ABC}{Perimeter of PQR}$ = $\frac{AB}{PQ}$

or $\frac{36}{24}$ = $\frac{AB}{10}$

or AB =15 cm

5. Two circles intersect each other at the points A and B. A straight line parallel to AB intersects the circles at C, D, E and F. If CD = 4.5 cm, then the measure of EF is [SSC 10+2-2014]

(a) 1.50 cm

(b) 2.25 cm

(c) 4.50 cm

(d) 9.00

6. If the isosceles right angle triangle has area of 4000 cm 2 the length of its hypotenuse (approx) is :

1. 25 cm
2. 25√20 cm
3. 40 cm
4. 47 cm

Explanation/Hint :

Area of right angle triangle = $\frac{1}{2}$ x base x perpendicular
⇒ 400 = $\frac{1}{2}$ x a2
⇒ a2 = 800 = a = 17.88
Now  h2 = p2 + b2
h2 =  (8√5)2 + (8√5)2
⇒ h = 25.2 (approx) cm

7. The area of right angled triangle is 60 cm 2. If its base is equal to 48 cm . Find its height.

1. 3$\frac{3}{6}$ cm
2. 4$\frac{3}{6}$√20 cm
3. 2$\frac{3}{6}$ cm
4. 6$\frac{3}{6}$cm

Answer : c. 2 $\frac{3}{6}$ cm

Explanation/Hint :

Area of right angle triangle = $\frac{1}{2}$ x base x height
⇒  60 = $\frac{1}{2}$ x 48 x h = 2$\frac{3}{6}$ cm

8. A triangle with three equal sides has its area equal to 8/3 cm2. Find its perimeter.

1. 12√2 cm
2. 10√2 cm
3. 12√3 cm
4. 9√3  cm

Explanation/Hint :

Area of Equilateral Triangle = $\frac{\sqrt3}4$ a2 cm

⇒ 8√3 = $\frac{\sqrt3}4$ a2

⇒  a2  = 32

⇒  a = 4√2

Perimeter =  3a = 3 x 4 x √2  = 12√2 cm

9. Find the area of triangle whose sides measures 10 cm , 12 cm, and 14 cm.

1. 22√6 cm2
2. 28√3 cm2
3. 24√6 cm2
4. 18√3 cm2

Explanation/Hint :

Area of Isosceles triangle = $\sqrt{s(s-a)(s-b)(s-c)}$ , where s = $\frac{a+b+c}{2}$

Here a = 10, b = 12, c = 14

s = $\frac{10+12+14}{2}$ = 18 cm

Area =$\sqrt{18(18-10)(18-12)(18-14)}$ = $\sqrt{18\times 8\times 6\times 4}$$\frac{24}{6}$ cm2

10. The base of the triangle is 5 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is

1. 9 cm
2. 18 cm
3. 20 cm
4. 12.5 cm

Explanation/Hint :

$\frac{1}{2}$ x 15 x 12 = $\frac{1}{2}$  x 20 x h

⇒ h = 18 cm

11. If the total surface area of a cube is 96 cm2, its volume is  [SSC 10+2-2013]

(a) 36 cm3

(b) 56 cm3

(c) 16 cm3

(d) 64 cm3

Explanation/Hint :

Total surface area of cube = 6a2 ..................1

Also, Total surface area of cube = 96 cm2 .............2

From eqn 1 and 2,

6a2 = 96 cm2

or a = 16 cm2

Volume of cube = a3 = 64 cm3

12. 360 sq. cm and 250 sq. cm are the areas of two similar triangles. If the length of one of the sides of the first triangle be 8 cm, then the length of the corresponding side of the second triangle is [SSC 10+2-2013]

(a) 6 cm

(b) 6$\frac{4}{3}$ cm

(c) 6$\frac{1}{3}$ cm

(d) 6$\frac{2}{3}$ cm

Answer : (d) 6$\frac{2}{3}$ cm

Explanation/Hint :

$\left(\frac{360}{250}\right)$ = $\left(\frac{8}{x}\right)^{2}$

or x = 6$\frac{2}{3}$ cm

13. The length and breadth of a rectangle are doubled. Percentage increase in area is [SSC 10+2-2013]

(a) 400%

(b) 150%

(c) 200%

(d) 300%

Explanation/Hint :

A = lb

A' = 2L x 2b = 4 lb = 4A

%  Change = $\frac{4A-A}{A}\times 100$ =360 %

14. The difference between the circumference and the radius of a circle is 37 cm. The area of circle is (Take π = $\frac{22}{7}$[SSC 10+2-2013]

(a) 154 cm2

(b) 259 cm2

(c) 148 cm2

(d) 111 cm2

Explanation/Hint :

Given, circumference - radius of a circle = 37 cm

or 2πr - r = 37 or r = 7

Now, Area of circle = πr2 = 154 cm2

15. If the sides of a right angled triangle are three consecutive integers, then the length of the smallest side is [SSC 10+2-2014]

(a) 3 units

(b) 2 units

(c) 4 units

(d) 5 units

Explanation/Hint :

Given, three consecutive integers = 3, 4 and 5

then the length of the smallest side = 3

16. If the three medians of a triangle are same, then the triangle is [SSC 10+2-2014]

(a) equilateral

(b) isosceles

(c) right-angled

(d) obtuse-angled

Explanation/Hint :

If the three medians of a triangle are same, then the triangle is equilateral.

17.Area of a regular hexagon with side 'a' is [SSC 10+2-2014]

(a) $\frac{3\sqrt{3}}{4}{a}^{2}$ sq. unit

(b) $\frac{12}{2\sqrt{3}}{a}^{2}$ sq. unit

(c) $\frac{9}{2\sqrt{3}}{a}^{2}$ sq. unit

(d) $\frac{6}{\sqrt{2}}{a}^{2}$ sq. unit

Answer : (c) $\frac{9}{2\sqrt{3}}{a}^{2}$ sq. unit

18. If the sum of the dimensions of a rectangular parallelepiped is 24 cm and the length of the diagonal is 15 cm, then the total surface area of it is [SSC 10+2-2014]

(a) 420 cm2

(b) 275 cm2

(c) 351 cm2

(d) 378 cm2

Explanation/Hint :

l+b+h = 24

or  $\sqrt{{l}^{2}+\quad{b}^{2\quad}+\quad{h}^{2}}$ =15

or  l2 + b2 + h2= 225

we know, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc

or a2 + b2 + c2 = 2ab + 2ac + 2bc - (a + b + c)2

thus, l2 + b2 + h= (l+b+h)2- 2(lb+bh+lh) = 225

or (l+b+h)2- 2(lb+bh+lh) = 225

or (24)2- 225 = 2(lb+bh+lh)     [ (l+b+h) = 24 given ]

or 2(lb+bh+lh) =Total surface area = 351 cm2

19. The sides of a triangle are 16 cm, 12 cm and 20 cm. Find the area [SSC 10+2-2013]

(a) 81 cm2

(b) 64 cm2

(c) 112 cm2

(d) 96 cm2

Area of Isosceles triangle = $\sqrt{s(s-a)(s-b)(s-c)}$ , where s = $\frac{a+b+c}{2}$