CLEAR
Toprankers-Info

+91 7676564400

Mobile App

Get it on Google Play
Discussions
Select Date
Tags:

Articles

Discussions
Submit Discussion
ASHWIN TANDEL

· started a discussion

· 1 Months ago

Still not clear (Question: 117234.0 )

though side of outermost circle is given as a, I still not clear what property we used to take EF = a/2. Please explain.

Question:
What is the area of the inner equilateral triangle if the side of the outermost square is ‘a’? (ABCD is a square).

Options:
A) \(\cfrac{3\sqrt[]{3}a^2}{32}\)
B) \(\cfrac{\sqrt[]{3}a^2}{16}\)
C) \(\cfrac{5\sqrt[]{3}a^2}{32}\)
D) \(\cfrac{5\sqrt[]{3}a^2}{64}\)
Solution:
Ans: (b) \(BD=a,EF=\cfrac{a}{2}\)

\(\therefore\) area of equilateral triangle EFG \(=\cfrac{\sqrt[]{3}}{4}\left ( \cfrac {a}{2} \right )^2=\cfrac{\sqrt[]{3}a^2}{16}\)

Knowledge Expert

· commented

· 1 Months ago

the side of the outermost square is given 'a'.......not of ABCD....

The line joining the midpoints of two sides of a triangle is parallel to third side and half of it.
-logo
We at aim to provide most comprehensive content & test for practice which is carefully divided into various chapters and topics to help you focus on your weak areas.Our Practice mock test gives you unmatched analytics to help you realize your strong areas,time management and improvement areas.Our Short & Crisp notes on each topic will help you to quickly revise the topic along with the tips & tricks of the exams.
More About
-Info
-Info

Mobile App

Get it on Google Play
About Us Campus Ambassador Careers Contact Us Refund Policy

All Rights Reserved Top Rankers

User Policy
Disclaimer
Terms & Conditions