# Important Questions of Mensuration: Quantitative Aptitude Mensuration is one the toughest topic of quantitative aptitude section. The only thing is it

Tazmin Horton 2021-01-04 Answered
Important Questions of Mensuration: Quantitative Aptitude Mensuration is one the toughest topic of quantitative aptitude section. The only thing is it takes time to analyze the question. Rest is just clarification and formula learning ability of candidate. This chapter is a part of quantitative aptitude section of SSC CGL and SBI PO. Today I will discuss some questions related to basic terms of mensuration.
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Answered 2021-04-21 Author has 162 answers

### Examples with Solution

Example1: What will be the area and perimeter of triangular plot whose sides are 17 m, 8m and 15m long?
Solution: Firstly, we will check which kind of triangle it is. Since, ${8}^{2}+{15}^{2}={17}^{2}.......\left({H}^{2}={P}^{2}+{B}^{2}\right)64+225=289=289$

Therefore, given triangle is a right angle triangle with: Hypotenuse $={17}^{2}$ Base or Perpendicular $=8$ or 15

Next thing we need is: Formula of Area and perimeter Area of triangle $=1/2$ Basetimes perpendicular $⇒1/28×15⇒60{m}^{2}$
Also, Perimeter $=8+15+17=40m$
Example2: What will be the area of equilateral triangle whose side is 4 cm.
Solution: Very simple question. You just need formula of area of equilateral triangle ( Equilateral triangle whose sides are equal)

Area of equilateral triangle $=\sqrt{3}/4×\left(side{\right)}^{2}⇒\sqrt{3}/4×{4}^{2}⇒4\sqrt{3}c{m}^{2}$
Example3: The radius of a circle is 6cm. What is radius of another circle whose area is 36 times that of first?
Solution: We know that, Area is directly proportional to square of radius $⇒\left(\text{Area of 2nd}\right)/\left(\text{Area of 1st}\right)=\left(\text{Radius of 2nd}{\right)}^{2}/\left(\text{Radius of 1st}{\right)}^{2}⇒36/1=\left(\text{Radius of 2nd}{\right)}^{2}/36⇒\text{Radius}=36m$
Example4: One brick measures $30cm×20cm×15cm$ How many bricks will be required for a wall 30m times 2m times 1.5m?
Solution: Obviously, Number of bricks = (Total  volume  of  wall) / (volume  of  1  brick)
Note: Units should be same
Number of bricks $=\left(30×2×1.5×100×100×100\right)\left(30×20×15\right)⇒10000$ bricks.
Example5: Surface area of cube is $150c{m}^{2}$. Find its volume and diagonal.
Solution: Surface area of cube $=6{a}^{2}=150⇒a=5cm$ Now, Volume of cube $={a}^{3}={5}^{3}=125c{m}^{3}$
Diagonal $=a\sqrt{3}=5×\sqrt{3}⇒8.66cm$

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