Question

The ratio of the area of a circle with centre o and the area of rectangle ABCD inscribed in it is π:√3.E is a point on AB such that angle ADE=angle ODC. find AE:AD

Answer 1

Solution:

A circle with center O and , and a rectangle A B CD is inscribed in it.

It is given that , ratio of the area of a circle with centre o and the area of rectangle A B CD inscribed in it is π:√3 .

Also, it is given that, E is a point on AB such that ∠ A DE=∠ O DC.

In Δ ADE

Tan (∠ADE)= $\frac{AE}{AD}$

$\frac{\text{Area of circle}}{\text{Area of rectangle ABCD}} =\frac{\pi }{\sqrt{3}}\\\\ \frac{\pi r^2}{AB \times CB}=\frac{\pi }{\sqrt{3}}\\\\ AB \times CB=\sqrt{3}\times r^2$

In Δ ODC, Draw OM ⊥ DC

OD=r, CD=L

CM =MD=$\frac{L}{2}$→→→→Perpendicular from the center to the chord bisects the chord.

By , Pythagoras theorem

→OD²=OM² +D M²

→r²-DM²= OM²

→OM=

         $\sqrt{r^2-\frac{L^2}{4}}$

Draw, ON ⊥ AD

So, AN = ND→→→→Perpendicular from the center to the chord bisects the chord.

ND= OM= $\sqrt{r^2-\frac{L^2}{4}}$

Area of Rectangle ABCD= AB × BC

= 4 × ND×OM

= $4 \times\sqrt{r^2-\frac{L^2}{4}}\times \frac{L}{2}=L\sqrt{4r^2-L^2}$

Also, area of rectangle ABCD= √3 r²

$\sqrt{3}r^2= L\sqrt{4r^2-L^2}$

Squaring both sides

$3 r^4=L^2(4 r^2-L^2)\\\\ 3 r^4= 4 r^2 L^2-L^4\\\\ L^4-4 r^2L^2+3r^2=0\\\\ L^4-3 r^2L^2-r^2L^2+3r^2=0\\\\ L^2(L^2-3 r^2)-r^2(L^2-3 r^2)=0\\\\ (L^2-3 r^2)(L^2- r^2)=0\\\\ \frac{L^2}{r^2}=\frac{1}{3} \text{or} \frac{L^2}{r^2}=1$

------------------------------(1)

$\sqrt{4r^2-L^2}=\frac{\sqrt{3}r^2}{ L}$------(2)

Considering , Δ O MD and Δ D A E

∠AD E=∠OD M→→(Given)

∠DAE= ∠OMD=90°

So,  Δ O MD ~ Δ D A E →→[AA criterion]

As,when triangles are similar their sides are proportional.

$\frac{OM}{DA}=\frac{MD}{AE}\\\\ \frac{AE}{DA}=\frac{MD}{OM}\\\\  \frac{AE}{DA}=\frac{\frac{L}{2}}{\sqrt{r^2-\frac{L^2}{4}}}\\\\ \frac{AE}{DA}=\frac{L}{\sqrt{4r^2-L^2}}$

$\frac{AE}{DA}=\frac{L}{\sqrt{4r^2-L^2}}\\\\ \frac{AE}{DA}=\frac{L^2}{\sqrt{3}r^2}$

-------------using (2)

So, $\frac{AE}{DA}=\frac{1}{\sqrt{3}} \text{or} \frac{AE}{DA}=\frac{1}{3\sqrt{3}}$---Using (1)