Question

A square has 2 of its vertices, one each on a longer side of a rectangle and one vertex on a shorter side. If the fourth vertex is collinear with midpoints on longer sides. Then the ratio of the area of the square to that of the rectangle is

(Given that one side of the square makes angle $\theta$ with the rectangle where tan $\theta$ = 3)

• A. 7 : 19
• B. 5 : 13
• C. 5 : 16
• D. 7 : 16

Answer 1

The correct option is C 5 : 16

$tan\theta =3\Rightarrow ta{n}^2\theta =9\Rightarrow se{c}^2\theta -1=9$

$sec\theta =\sqrt{9+1}=\sqrt{10}\Rightarrow cos\theta =\frac{1}{\sqrt{10}}$

$sin\theta =\frac{3}{\sqrt{10}}$

Let side of a square formed be a

Length of rectangle $=2(asin\theta +acos\theta )$

$=2a(sin\theta +cos\theta )$

$=2a(\frac{3}{\sqrt{10}}+\frac{1}{\sqrt{10}})$

$=\frac{8a}{\sqrt{10}}$

Breadth of rectangle $=a(sin\theta +cos\theta )$

$=a(\frac{3}{\sqrt{10}}+\frac{1}{\sqrt{10}})$

$=\frac{4a}{\sqrt{10}}$

Area of rectangle $=\frac{3a}{\sqrt{10}}\ast \frac{4a}{\sqrt{10}}=\frac{32a^2}{10}=\frac{16a^2}{5}$

Required ratio $=\frac{a^2}{\frac{16a^2}{5}}=\frac{5}{16}=5:16$