wrong ans (Question: 107692.0 )

Online Courses

Test Series

Resources

Offline Courses

Community

News

+91 7676564400

support@toprankers.com

Mobile App

Mock Tests

Practice

Assignment

Study Notes

SSC Articles

Discussions

Online Coaching

GK- Current Affairs

Bookmarked Questions

Descriptive Papers

Discussions

Select Date

Tags:

Submit Discussion

venkatesh

· started a discussion

· 1 Months ago

wrong ans (Question: 107692.0 )

option b is correct

This topic is not avialable

satish

· started a discussion

· 1 Months ago

wrong ans. (Question: 107692.0 )

option b is correct

Question:

ABCD is a square of side a cm. AB, BC, CD and AD all are the chords of circles with equal radii each. If the chords subtends an angle of 120° at their respective centres, find the total area of the given figure, where arcs are the parts of the circles:

Options:

A) \( \left [ a^2 +4\left ( \cfrac {\pi a^2}{9 }-\cfrac{a^2}{3\sqrt[]{2}} \right ) \right ]\)

B) \( \left [ a^2 +4\left ( \cfrac {\pi a^2}{9 }-\cfrac{a^2}{4\sqrt[]{3}} \right ) \right ]\)

C) \(\big[ 9a^2 -4\pi + 3\sqrt[]{3a^2}\big]\)

D) None of these

Solution:

Ans: (d)

In \(\Delta\)OAB, from 30 - 60 - 90 theorem

\(OC=\cfrac{a}{2\sqrt[]{3}}\)

\(OA=\cfrac{a}{\sqrt[]{3}}\)

Area of sector OAB = \(\cfrac{120}{360}\times \pi \left ( \cfrac{a}{\sqrt[]{3}} \right )^2\) = \(\cfrac{\pi a^2}{9}\)

Area of section AB = \(\left ( \cfrac{\pi a^2}{9}-\cfrac{a^2}{2\sqrt[]{3}} \right )\)

Hence area of fig. = 4 \(\left ( \cfrac{\pi a^2}{9}-\cfrac{a^2}{2\sqrt[]{3}} \right )+a^2\)

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