Question

In the figure, ABC is an equilateral triangle and PQRS is a square of side $6$ cm. By how many $c{m}^2$ in the area of the triangle more than the square?

• A. $\frac{21}{\sqrt{3}}$
• B. $21$
• C. $\sqrt{21}$
• D. $63$

Answer 1

The correct option is A $\frac{21}{\sqrt{3}}$

Hint: Use the fact that $△ABC$ is similar to $△APQ$.

Step 1 : Calcuaion of the height of the triangle ABC.

The $△ABC$ is similar to $△APQ$ and therefore, $\frac{AP}{AB}=\frac{PQ}{BC}$.

Since the length of the sides of the square is 6, $PQ=6$ and hence $\frac{AP}{AB}=\frac{PQ}{BC}$ becomes$\frac{AP}{AB}=\frac{6}{BC}$.

The height of an equilateral triangle is given by $\frac{\sqrt{3}a}{2}$, where $a$ is the length of the side of the equilateraltriangle.

Substitute 6 for $a$ in $\frac{\sqrt{3}a}{2}$ to calculate the height of $△APQ$.

$\begin{array}{c}\frac{\sqrt{3}\cdot 6}{2}=3\sqrt{3}\end{array}$

Add 6 and $3\sqrt{3}$ to obtain the height of $△ABC$.

$6+3\sqrt{3}$

Step 2: Calculation of the length of the side $BS$ and $RC$.

Since each interior angle in an equilateral triangle is ${60}^{\circ }$, $\angle PBS={60}^{\circ }$.

The tangent of angle $\theta$ is given by $\tan \theta =\frac{a}{b}$, where $a$ and $b$ are the lengths of the side opposite and side adjacent.

For $\angle PBS$, $\tan {60}^{\circ }=\frac{PS}{BS}$.

Substitute 6 for $PS$ in $\tan {60}^{\circ }=\frac{PS}{BS}$ and then solve for $BS$.

$\begin{array}{cc}\tan {60}^{\circ }& =\frac{6}{BS}\\ \sqrt{3}& =\frac{6}{BS}\\ BS& =\frac{6}{\sqrt{3}}\\ BS& =2\sqrt{3}\end{array}$

In a similar manner, it can be shown that $RC=2\sqrt{3}\text{cm}$.

Step 3: Calculation of the length of the side of triangle.

Add the lengths of $BS$, $SR$, and $RC$ to obtain the length of the sides of $△ABC$.

$2\sqrt{3}+6+2\sqrt{3}=6+4\sqrt{3}$

Step 4: Calculation of the area of the triangle.

The area of an equilateral triangle is given by $\frac{\sqrt{3}{a}^2}{4}$, where $a$ is the length of the side of the equilateraltriangle.

Substitute $6+4\sqrt{3}$ for $a$ in $\frac{\sqrt{3}{a}^2}{4}$ to calculate the area of $△ABC$.

$\begin{array}{cc}\frac{\sqrt{3}(6+4\sqrt{3}{)}^2}{4}& =\frac{\sqrt{3}(36+48\sqrt{3}+48)}{4}\\ & =\frac{\sqrt{3}(84+48\sqrt{3})}{4}\\ & =36+21\sqrt{3}\end{array}$

Step 5: Calculation of the difference between the area of the triangle and the area of the square.

The formula to calculate the area of a square is $A=a^2$, where $a$ is the length of the sides of the square.

The area of the square $PQRS$ is $6^2=36{\text{cm}}^2$.

Subtract 36 from $36+21\sqrt{3}$ to obtain the difference between the area of triangle $ABC$ and square $PQRS$.

$36+21\sqrt{3}-36=21\sqrt{3}$

Final step: The area of the triangle is $21\sqrt{3}{\text{cm}}^2$ more than the square.