Question

Given the equilateral triangle inscribed in a square of side s find the ratio of $\mathrm{\Delta }BCR$ to $\mathrm{\Delta }PRD$ ?

Answer 1

Hint: The main objective is to find the ratio of two triangles. $\mathrm{\Delta }BCR$ is a part of the square. We cannot assume that it is an equilateral triangle. Hence it can be found from the sides of the square. Since the square is perpendicular.

Formula used:
Area of the triangle.
$\mathrm{\Delta }={\displaystyle \frac{1}{2}}\times b\times h$
b is the base of the triangle and h is the height of the triangle.

Complete step by step answer:
Given,
To find ${\displaystyle \frac{\mathrm{\Delta }BCR}{\mathrm{\Delta }PRD}}$
$\mathrm{\Delta }BCR={\displaystyle \frac{1}{2}}\times BC\times CR$
The $\mathrm{\Delta }BCR$ is symmetric, The symmetric with respect to side $BD$ ,
The angle $CBR$ is the angle which is obtained by half the angle between the perpendicular line and the angle of $\mathrm{\Delta }BCR$.
The value of BC is s.
Since $\mathrm{\Delta }BCR$ is equilateral triangle angle, each corner has same angle such as $60{}^0$
The perpendicular angle is ${90}^0$ .
 $|\underset{―}{CBR}={\displaystyle \frac{{90}^0-{60}^0}{2}}$
Subtract the values in the numerator
$|\underset{―}{CBR}={\displaystyle \frac{{30}^0}{2}}$
Divide the numerator and denominator
$|\underset{―}{CBR}={15}^0$
To find the side of CR of $\mathrm{\Delta }BCR$ , The side of square should be multiplied by the angle of $|\underset{―}{CBR}={15}^0$
$CR=s\times \tan {15}^0$
Substitute the value of $\tan {15}^0$ in the above equation,
We know $\tan {15}^0=\tan ({45}^0-{30}^0)$
We know the formula for $\tan (A-B)={\displaystyle \frac{\mathrm{Tan}B-\mathrm{Tan}A}{1+\tan A\tan B}}$
$\tan (45-30)={\displaystyle \frac{\tan 30-\tan 45}{1+\tan 45\tan 30}}$
Substitute the value of $\tan 45=1$and $\tan 30=\sqrt{3}$
$\tan (45-30)={\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}$
$\tan 15={\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}$
Substitute $\tan {15}^0$ value in the equation $CR=s\times \tan {15}^0$
$CR=s\times {\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}$
Substitute $CR$ and $BC$ in area formula
Area of $\mathrm{\Delta }BCR={\displaystyle \frac{1}{2}}\times BC\times CR$
$\mathrm{\Delta }BCR={\displaystyle \frac{1}{2}}\times s\times {\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}\times s$
Multiply the s terms
$\mathrm{\Delta }BCR={\displaystyle \frac{1}{2}}\times s^2\times {\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}$
Now we need to calculate area of $\mathrm{\Delta }PRD$
$\mathrm{\Delta }PRD={\displaystyle \frac{1}{2}}\times DR\times DR$
Since we know that the side DR is equal to the square distance subtracts the Side CR
$DR=s-CR$
Substitute $CR=s\times {\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}$ in above equation
$DR=s-s\times {\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}$
Taking common out s
$DR=s\left(1-{\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}\right)$
Evaluate the values in right side
$$DR=s\left({\displaystyle \frac{1+\sqrt{3}-\left(\sqrt{3}-1\right)}{1+\sqrt{3}}}\right)$$
Subtract the terms
$DR=s\left({\displaystyle \frac{2}{1+\sqrt{3}}}\right)$
Substitute DR in area of $\mathrm{\Delta }PRD$
$\mathrm{\Delta }PRD={\displaystyle \frac{1}{2}}\times {\left(s\left({\displaystyle \frac{2}{1+\sqrt{3}}}\right)\right)}^2$
Square the terms
$\mathrm{\Delta }PRD={\displaystyle \frac{1}{2}}\times s^2\times {\displaystyle \frac{4}{{(\sqrt{3}+1)}^2}}$
We need to calculate
${\displaystyle \frac{\mathrm{\Delta }BCR}{\mathrm{\Delta }PRD}}={\displaystyle \frac{{\displaystyle \frac{1}{2}}\times s^2\times {\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}}}{{\displaystyle \frac{1}{2}}\times s^2\times {\displaystyle \frac{4}{{(\sqrt{3}+1)}^2}}}}$
Cancel the same terms in the numerator and denominator
$${\displaystyle \frac{\mathrm{\Delta }BCR}{\mathrm{\Delta }PRD}}={\displaystyle \frac{\sqrt{3}-1}{{\displaystyle \frac{4}{\sqrt{3}+1}}}}$$
The value in the denominator goes to numerator by multiplication
$${\displaystyle \frac{\mathrm{\Delta }BCR}{\mathrm{\Delta }PRD}}=\sqrt{3}-1\times {\displaystyle \frac{\sqrt{3}+1}{4}}$$
We know $(a-b)(a+b)=a^2-b^2$
$${\displaystyle \frac{\mathrm{\Delta }BCR}{\mathrm{\Delta }PRD}}={\displaystyle \frac{{\left(\sqrt{3}\right)}^2-1^2}{4}}$$
Squaring the terms, root and square get cancelled
$${\displaystyle \frac{\mathrm{\Delta }BCR}{\mathrm{\Delta }PRD}}={\displaystyle \frac{3-1}{4}}$$
Subtract the values in numerator
$${\displaystyle \frac{\mathrm{\Delta }BCR}{\mathrm{\Delta }PRD}}={\displaystyle \frac{2}{4}}$$
Divide the value in numerator and denominator
$${\displaystyle \frac{\mathrm{\Delta }BCR}{\mathrm{\Delta }PRD}}={\displaystyle \frac{1}{2}}$$
The ratio of $\mathrm{\Delta }BCR$ to $\mathrm{\Delta }PRD$ is $1:2$.

Note:
The side value should be taken correctly. Major mistakes can be made while choosing an angle as sometimes students forgets to take half of the angle. We should know that we can only proceed further because of the angle property of the equilateral triangle.