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.

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 [tex](3k-2)[/tex] units.  Then, perimeter of the triangle will be [tex]3 \times(3 k-2)=9 k-6[/tex]A regular pentagon (5 sides) has sides of [tex](2k+0.5)[/tex] units.  Then, perimeter of the pentagon will be [tex]5 \times(2 k+0.5)=10 k+2.5[/tex]We have to find the dimensions of each shape . Now, the perimeter of the triangle is twice the perimeter of the pentagon,  So, [tex]\text {perimeter of triangle} = 2\times \text{perimeter of pentagon}[/tex][tex]\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}[/tex]Then, length of side of triangle [tex]= 3(-1)-2 = - 5[/tex] Length of side of pentagon [tex]= 2(-1) + 0.5 = - 1.5[/tex] We have to neglect, negative sign as lengths can’t be negative. Even if we change the sign above all conditions are satisfied.