Question

Can someone pls solve this?

In the following diagram, ABCD is a square while points P and Q lie on sides BC and CD
respectively such that APQ is an equilateral triangle. What is the ratio of the area of square
ABCD to that of equilateral triangle APO?

Solution: 4th option. 2 + root 3 / root 3​

Answer 1

Given:

ABCD is a square.

APQ is an equilateral triangle.

To find:

Ratio of area of square ABCD : Area of triangle APQ

Solution:

Please have a look at the labeling of the diagram as shown in the figure attached.

Let side of square = x

Let side of equilateral triangle = a

In $\triangle ABP$:

AB = a

BP = x and

Hypotenuse, AP = s

According to pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AP^{2} = AB^{2} + BP^{2}\\\Rightarrow s^2=a^{2} +x^{2} ........ (1)$

In $\triangle PQC$:

QC = PC = x - a and

Hypotenuse, PQ = s

According to pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow PQ^{2} = QC^{2} + PC^{2}\\\Rightarrow s^2=(a-x)^{2} +(a-x)^{2} ........ (2)$

Equating (1) and (2):

$a^{2}+ x^{2} = (a-x)^2+(a-x)^2\\\Rightarrow a^{2}+ x^{2} = 2a^2+2x^2-4ax\\\Rightarrow x^{2} -4ax+a^2=0$

It is a quadratic equation.

The solution of a quadratic equation $AX^2+BX+C=0$ is given as:

$X=\dfrac{-B\pm \sqrt{B^2-4AC}}{2A}$

Solving the above quadratic equation:

$x=\dfrac{4a\pm\sqrt{16a^2-4a^2}}{2}\\\Rightarrow x=\dfrac{4a\pm\sqrt{12a^2}}{2}\\\Rightarrow x=2a\pm\sqrt{3}a\\\\\Rightarrow x=a(2\pm\sqrt{3})$

Value $x=a(2-\sqrt3)$ will make $a-x$ as negative which is not possible so only one value $x=a(2+\sqrt3)$ is possible.

Putting in equation (1) to find $s^2$:

$s^2=a^2+(a(2-\sqrt3))^2\\\Rightarrow s^2=a^2+(a^2(4+3-4\sqrt3))\\\Rightarrow s^2=a^2(8-4\sqrt3)$

The required ratio:

$\dfrac{\text{Area of Square}}{\text{Area of Equilateral Triangle}}= \dfrac{a^2}{\dfrac{\sqrt3}{4}s^2}\\\Rightarrow \dfrac{4a^2}{\sqrt3 s^2}\\\Rightarrow \dfrac{4a^2}{4\sqrt3 (2-\sqrt3)a^2}\\\Rightarrow \dfrac{1}{\sqrt3 (2-\sqrt3)} \times \dfrac{2+\sqrt3}{2+\sqrt3}\\\Rightarrow \dfrac{2+\sqrt3}{\sqrt3 \times (4-3)}\\\Rightarrow \dfrac{2+\sqrt3}{\sqrt3}$

So, the required ratio is:

$\dfrac{2+\sqrt3}{\sqrt3}$