Mensuration Questions for SSC CGL Tier 2 PDF

Mensuration Questions for SSC CGL Tier 2 PDF

Download SSC CGL Tier 2 Mensuration Questions PDF. Top 10 SSC CGL Tier 2 Mensuration questions based on asked questions in previous exam papers very important for the SSC exam.

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Question 1: Length and breadth of a rectangle are increased by 40% and 70% respectively. What will be the percentage increase in the area of rectangle?

a) 118

b) 110

c) 138

d) 128

Question 2: The perimeter of a square is 40 cm, find its area?

a) 100 sq cm

b) 25 sq cm

c) 50 sq cm

d) 160 sq cm

Question 3: The circumference of a circle is 88 cm, find its area?

a) 616 sq cm

b) 308 sq cm

c) 154 sq cm

d) 77 sq cm

Question 4: The total surface area of a hemisphere is 166.32 sq cm, find its curved surface area?

a) 221.76 sq cm

b) 36.96 sq cm

c) 110.88 sq cm

d) 55.44 sq cm

Question 5: If G is the centroid of triangle ABC and area of triangle ABC = 48cm², then the area of triangle BGC is

a) 8 cm²

b) 16 cm²

c) 24 cm²

d) 32 cm²

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Question 6: In $\triangle$ PQR, the line drawn from the vertex P intersects QR at a point S. If QR = 4.5 cm and SR = 1.5 cm then the ratios of the area of triangle PQS and triangle PSR is

a) 4 : 1

b) 3 : 1

c) 3 : 2

d) 2 : 1

Question 7: If G is the centroid and AD, BE, CF are three medians of triangle ABC with area 72 sq cm , then the area of triangle BDG is :

a) 12 sq cm

b) 16 sq cm

c) 24 sq cm

d) 8 sq cm

Question 8: The three medians AD, BE and CF of triangle ABC intersect at point G. If the area of triangle ABC is 60 sq.cm. then the area of the quadrilateral BDGF is :

a) 10 sq.cm

b) 15 sq.cm

c) 20 sq.cm

d) 30 sq.cm

Question 9: In triangle ABC, AD, BE and CF are the medians intersecting at point G and area of triangle ABC is 156 cm2. What is the area $(in cm^2)$ of triangle FGE?

a) 13

b) 26

c) 39

d) 52

Question 10: The tangents drawn at points A and B of a circle with centre O, meet at P. If ∠AOB = 120° and AP = 6 cm, then what is the area of triangle (in cm$^2)$ APB?

a) $6\sqrt{3}$

b) $8\sqrt{3}$

c) $9$

d) $9\sqrt{3}$

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Answers & Solutions:

1) Answer (C)

Let length and breadth of rectangle initially be $10$ units

=> Area, $A=10\times10=100$ sq.units

Length is increased by 40%, => New length = $10+(\frac{40}{100}\times10)=14$ units

Similarly, new breadth = $10+(\frac{70}{100}\times10)=17$ units

=> New area, $A’=14\times17=238$ sq.units

$\therefore$ % increase in area = $\frac{(238-100)}{100}\times100=138\%$

=> Ans – (C)

2) Answer (A)

Let side of square = $s$ cm

Perimeter of square = $4 \times s = 40$

=> $s = \frac{40}{4} = 10$ cm

$\therefore$ Area of square = $(s)^2$

= $(10)^2 = 100 cm^2$

=> Ans – (A)

3) Answer (A)

Let the radius of circle = $r$ cm

Circumference of circle = $2 \pi r = 88$

=> $2 \times \frac{22}{7} \times r = 88$

=> $r = \frac{88 \times 7}{44} = 14$ cm

$\therefore$ Area of circle = $\pi r^2$

= $\frac{22}{7} \times (14)^2$

= $22 \times 2 \times 14 = 616 cm^2$

=> Ans – (A)

4) Answer (C)

Let radius of hemisphere = $r$ cm

Total surface area of hemisphere = $3 \pi r^2 = 166.32$ ————(i)

Multiplying equation (i) by $\frac{2}{3}$

=> $\frac{2}{3} \times 3 \pi r^2 = \frac{2}{3} \times 166.32$

=> Curved Surface area of hemisphere = $2 \pi r^2 = 2 \times 55.44 = 110.88 cm^2$

=> Ans – (C)

5) Answer (B)

As we know area of triangle, with centroid as one of the vertices and remaining 2 triangle vertices, is $\frac{1}{3}$rd of the area of whole triangle.
Hence area will be $\frac{48}{3}$ = 16

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6) Answer (D)

NOTE :- The ratio of area of two triangles on same base is equal to the ratio of two corresponding sides of the two triangles.

---

Given : QR = 4.5 cm and SR = 1.5 cm

=> QS = QR – SR = 4.5-1.5 = 3 cm

Since, the two triangles PQS and PSR have same base PS

=> $\frac{area(\triangle PQS)}{area(\triangle PSR)} = \frac{QS}{SR}$

= $\frac{3}{1.5}$ = 2 : 1

7) Answer (A)

The area of triangle formed by any two vertices and centroid is (1/3) times the area of ABC.
Also the median divides the triangle into two equal areas.
So, area of BDG = (1/6) times of ABC
= (1/6)*72
=12

8) Answer (C)

Given ∆ABC, G is the centroid and AD,BE, CF are three medians and the area of ∆AGE = 10
As we know the median divides the triangle into 6 triangles of equal area
Hence area of the quadrilateral BDGF = 2*∆AGE = 2*10
area of the quadrilateral BDGF = 20

9) Answer (A)

Medians of a triangle divides the triangle into 6 parts of equal areas.

Also, ar($\triangle$ ABC) = 156

=> ar($\triangle$ AFG) = ar($\triangle$ FBG) = ar($\triangle$ BGD) = ar($\triangle$ DGC) = ar($\triangle$ CGE) = ar($\triangle$ EGA) = $\frac{156}{6}=26$ $cm^2$

=> ar(AFGE) = ar($\triangle$ AFG) + ar($\triangle$ EAG)

= $26+26=52$ $cm^2$

$\therefore$ ar($\triangle$ FGE) = $\frac{1}{4}\times$ ar(AFGE)

= $\frac{52}{4}=13$ $cm^2$

=> Ans – (A)

10) Answer (D)

Given : ∠AOB = 120° and AP = 6 cm and $\angle$ OAP = 90°

To find : ar($\triangle$ APB) = ?

Solution : $\angle$ AOP = $\frac{1}{2} \angle$ AOB = $\frac{120}{2}=60^\circ$

In $\triangle$ AOP,

=> $tan(\angle AOP)=\frac{AP}{OA}$

=> $tan(60^\circ)=\frac{6}{OA}$

=> $\sqrt3=\frac{6}{OA}$

=> $OA=\frac{6}{\sqrt3}=2\sqrt3$ cm

Thus, area of $\triangle$ AOP = $\frac{1}{2}\times(OA)\times(AP)$

$\frac{1}{2}\times(2\sqrt3)\times(6)=6\sqrt3$ $cm^2$ ————-(i)

Now, in $\triangle$ AOM

=> $sin(\angle AOM)=\frac{AM}{OA}$

=> $sin(60^\circ)=\frac{AM}{2\sqrt3}$

=> $\frac{\sqrt3}{2}=\frac{AM}{2\sqrt3}$

=> $AM=3$ cm

Similarly, $OM = \sqrt3$ cm

Thus, area of $\triangle$ AOM = $\frac{1}{2}\times(OM)\times(AM)$

$\frac{1}{2}\times(\sqrt3)\times(3)=1.5\sqrt3$ $cm^2$ ————-(ii)

=> $ar(\triangle AMP)=ar(\triangle AOP)-ar(\triangle AOM)$

= $6\sqrt3-1.5\sqrt3=4.5\sqrt3$ $cm^2$

$\therefore$ $ar(\triangle APB)=2ar(\triangle AMP)$

= $2\times4.5\sqrt3=9\sqrt3$ $cm^2$

=> Ans – (D)

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