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Saturday, July 18, 2015

UPSC Maths Problems Challenge : A Problem A Day! (Day 22)

Today's featured problem is from a basic problem from Mensuration,which can be confusing at the first glance but is nothing even similar to confusing,from any perspective.This is a very nice problem I have come across quite in a while (within solving Level 2 and Level 3 problems at Brilliant) apart from the problems included in the syllabus of Class XI WBCHSE.
As I said,unless you get to be a Class XI student,you know nothing about what a hard schedule might be like!
Read on to find out the solution to this problem.

Today's Problem(18/07/2015)

Question Credit :

(Prelim 2000) in the given figure, all line segments of the shaded portion are of same length and at right angles to each other. What is the area of the shaded portion?

Look at the given statement,it is already specified that all the line segments of the shaded portion are of same length and at right angles to each other.In other words,they are squares and if we stretch the boundary lines at these particular places,we will get squares.
So,if that is done,we get 3 such squares in each part of the un-shaded region.
Now,in each side measuring 10 cm,there are 5 such squares.So the length of each side of such small squares is (10/5)cm = 2 cm.
Therefore,the area of each one of the four distinct un-shaded regions is 3 * (2 * 2) cm^2 = 12 cm^2.
So,the area of all four of the distinct un-shaded regions is (4 * 12) cm^2 = 48 cm^2.
Now,the area of the entire figure given is (10 * 10) cm^2 = 100 cm^2.
Hence,Area of the shaded region = Area of Given Figure - Area of the region that is not shaded
i.e, Area of the shaded region = (100 - 48) cm^2 = 52 cm^2.

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