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Deepak

· started a discussion

· 1 Months ago

help (Question: 111072.0 )

koi easy solu explain kardo plz]

Question:
In the figure below, rectangle ABCD is inscribed in the circle with centre at O. The length of side AB is greater than the side BC. The ratio of the area of circle to the area of rectnalge ABCD is \(\pi:\sqrt[]{3}\) The line segment DE intersects AB at E such that \(\angle \)ODC = \(\angle \)ADE. What is the ratio of AE : AD?

Options:
A) \(1:\sqrt[]{3}\)
B) \(1:\sqrt[]{2}\)
C) \(1:2\sqrt[]{3}\)
D) 1 : 2
Solution:
Ans: (a)


We have, \(\cfrac{\pi R^2}{ab}=\cfrac{\pi}{\sqrt{3}}\)

 \(\therefore\)  \(\sqrt{3}R^2=ab\)

From \(\triangle\)DBC,

\(tan\theta=\cfrac{BC}{DC}=\cfrac{b}{a}\)

From \(\triangle\)DAE

\(tan\theta=\cfrac{AE}{AD}=\cfrac{AE}{b}\)

from (ii) and (iii) we get,

\(tan\theta=\cfrac{AE}{AD} =\cfrac{b}{a}\)

From triangle DBC,

4R2 = a2 +b2

\(4R^2=a^2+\cfrac{3R^4}{a^2}\Rightarrow\) a4 - 4R2a2 + 3R4 = 0

a4 - 3R2a2 - R2 a2 + 3R4 = 0 \(\Rightarrow\)  a2 (a2-3R2) - R2 (a2-3R2) = 0

(a2-R2) \(\Rightarrow\) (a2-3R2) = 0 \(\Rightarrow\) a2 = R2 and a2 = 3R2

a = R and a = \(\sqrt{3}R\)

and h = \(\sqrt{3}R\)

and when a= R

b= R and a= \(\sqrt{3}R\)

hence required ratio is \(1: \sqrt{3}\)

Vaibhav Yadav

· commented

· 1 Months ago

easiest .. ODC=30 =ADE use tan30 = ae/ad (DOC= 30 because x and x*3(1/2) are the sides)
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