Chapter 3.6, Problem 6PSA

To determine

To Find: The values of $x$ and $y$ from the given figure.

Answer to Problem 6PSA

  →$x=\text{ 7 , }y=\text{ 63 }$

Explanation of Solution

Given information: In the given figure we have an equilateral triangle $\Delta HJK$ where $HJ\text{ = 15 , }JK\text{ = }\frac{1}{3}y\text{ - 6 , }HK\text{ = }x+\text{ 8 }$

Concept used: Measurement of the all the sides of an equilateral triangle is equal.

Calculation: According to the given question we have:

  $HJ\text{ = 15 , }JK\text{ = }\frac{1}{3}y\text{ - 6 , }HK\text{ = }x+\text{ 8 }$ , also $\Delta HJK$ is an equilateral triangle thus:

  $\begin{array}{l}HJ\text{ = }JK\text{ = }HK\\ \Rightarrow \text{ 15 = }\frac{1}{3}\text{ y - 6 = }x\text{ + 8}\end{array}$

Or we can say: $HJ\text{ = }JK\text{ and }HJ\text{ = }HK\mathrm{.................................................}\text{(i)}$

  $\Rightarrow \text{ 15 = }\frac{1}{3}\text{ y - 6 and }x\text{ + 8 = 15 }\text{.}$

Solving for $y$ using $HJ=JK$

  $\begin{array}{l}\frac{1}{3}\text{ y - 6 = 15}\\ \Rightarrow \frac{1}{3}\text{ y = 15 + 6}\\ \Rightarrow \frac{1}{3}\text{ y }=\text{ 21}\\ \Rightarrow \text{ y = 63 }\mathrm{.....................................................................}\text{(ii)}\end{array}$

Solving for $x$ using $HJ=HK$

  $\begin{array}{l}x\text{ + 8 = 15}\\ \Rightarrow x\text{ = 15 - 8}\\ \Rightarrow x\text{ = 7 }\mathrm{....................................................................}\text{(iii)}\end{array}$

  $$ From (ii) and (iii) we get : $x=\text{ 7 , }y=\text{ 63 }$

Conclusion:

  $x=\text{ 7 , }y=\text{ 63 }$