Can someone pls solve this?In the following diagram, ABCD is a square

1.	Can someone pls solve this?In the following diagram, ABCD is a square while points P and Q lie on sides BC and CDrespectively such that APQ is an equilateral triangle. What is the ratio of the area of squareABCD to that of equilateral triangle APO?Solution: 4th option. 2 + root 3 / root 3
Answer» 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: $\dfrac{2+\sqrt3}{\sqrt3}$

Can someone pls solve this?In the following diagram, ABCD is a square while points P and Q lie on sides BC and CDrespectively such that APQ is an equilateral triangle. What is the ratio of the area of squareABCD to that of equilateral triangle APO?Solution: 4th option. 2 + root 3 / root 3

Answer»

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:

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

Can someone pls solve this?In the following diagram, ABCD is a square while points P and Q lie on sides BC and CDrespectively such that APQ is an equilateral triangle. What is the ratio of the area of squareABCD to that of equilateral triangle APO?Solution: 4th option. 2 + root 3 / root 3

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Can someone pls solve this?In the following diagram, ABCD is a square while points P and Q lie on sides BC and CDrespectively such that APQ is an equilateral triangle. What is the ratio of the area of squareABCD to that of equilateral triangle APO?Solution: 4th option. 2 + root 3 / root 3​

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Can someone pls solve this?In the following diagram, ABCD is a square while points P and Q lie on sides BC and CDrespectively such that APQ is an equilateral triangle. What is the ratio of the area of squareABCD to that of equilateral triangle APO?Solution: 4th option. 2 + root 3 / root 3