Arun Sharma Solutions for Exercise 3: Extra Practice Exercise on Geometry and Mensuration

Author:Arun Sharma

Arun Sharma Quantitative Aptitude Solutions for Exercise - Arun Sharma Solutions for Exercise 3: Extra Practice Exercise on Geometry and Mensuration

Attempt the practice questions from Exercise 3: Extra Practice Exercise on Geometry and Mensuration with hints and solutions to strengthen your understanding. How to prepare for Quantitative Aptitude solutions are prepared by Experienced Embibe Experts.

Questions from Arun Sharma Solutions for Exercise 3: Extra Practice Exercise on Geometry and Mensuration with Hints & Solutions

EASY
IPMAT: Rohtak
IMPORTANT

If the sides of a triangle measures 13, 14, 15 cm respectively, what is the height of the triangle for the base is 14.

EASY
IPMAT: Rohtak
IMPORTANT

A right angled triangle is drawn on a plane such that sides adjacent to right angle are 3 cm and 4 cm. Now three semi-circles are drawn taking all three sides of the triangle as diameters respectively (as shown in the figure). What is the area of the shaded regions A1+A2

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EASY
IPMAT: Rohtak
IMPORTANT

In the figure given below, AB=16, CD=12 and OM=6. Calculate ON.

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EASY
IPMAT: Rohtak
IMPORTANT

In the figure, M is the centre of the circle. 1QS=102 , 1PR=1RS and PR is parallel to QS. Find the area of the shaded region.

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EASY
IPMAT: Rohtak
IMPORTANT

In the given figure, PBC and PKH are straight lines. If AH=AKb=70o, c=40o, the value of d is

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EASY
IPMAT: Rohtak
IMPORTANT

Six solid hemispherical balls have to be arranged one upon the other vertically. Find the minimum total surface area of the cylinder in which the hemispherical balls can be arranged, if the radii of each hemispherical ball is 7 cm.

EASY
IPMAT: Rohtak
IMPORTANT

On a semicircle with diameter AD, Chord BC is parallel to the diameter. Further each of the chords AB and CD has Length 2 cm while AD has length 8 cm. Find the length of BC.

EASY
IPMAT: Rohtak
IMPORTANT

In the given figure, B and C are points on the diameter AD of the circle such that AB = BC = CD. Then find the ratio of area of the shaded portion to that of the whole circle.

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