Question

An equilateral triangle has sides of length(3k-2)units. A regular pentagon(5 sides) has sides of (2k+0.5)units. If the perimeter of the triangle is twice the perimeter of the pentagon, find the dimensions of each shape.

Answer 1

The length of side of triangle is 5 units and length of side of pentagon is 1.5 units.

SOLUTION:

Given, An equilateral triangle has sides of length $(3k-2)$ units.  

Then, perimeter of the triangle will be $3 \times(3 k-2)=9 k-6$

A regular pentagon (5 sides) has sides of $(2k+0.5)$ units.  

Then, perimeter of the pentagon will be $5 \times(2 k+0.5)=10 k+2.5$

We have to find the dimensions of each shape . Now, the perimeter of the triangle is twice the perimeter of the pentagon,  

So, $\text {perimeter of triangle} = 2\times \text{perimeter of pentagon}$

$\begin{array}{l}{\rightarrow 9 k-6=2(10 k+2.5)} \\\\ {\rightarrow 9 k-6=20 k+5} \\\\ {\rightarrow 20 k-9 k=-6-5} \\\\ {\rightarrow 11 k=-11} \\\\ {\rightarrow k=-1}\end{array}$

Then, length of side of triangle $= 3(-1)-2 = - 5$

Length of side of pentagon $= 2(-1) + 0.5 = - 1.5$

We have to neglect, negative sign as lengths can’t be negative. Even if we change the sign above all conditions are satisfied.