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jai ho

· started a discussion

· 1 Months ago

wrong question (Question: 111072.0 )

tanx = 1 as DO = OC (diag of rectangle are equal) ..
ans should be 1:1

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

answer is correct ...doc=30 so ade = 30 use tan30 = ae/ad

Ambarish Gupta

· commented

· 1 Months ago

you might be right bro..i have also deduced the same thing without using 45 degree angle and getting the same ratio 1:1....used sin theta/side ratio ..
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