Question

a triangle and a rectangle have the following conditions. the base length of an isosceles triangle is 10 meters shorter than twice the leg length. the width of a rectangle is equal to the leg length of the triangle. the length of the rectangle is 10 meters longer than twice its width. what measurements of the sides of the triangle and the sides of the rectangle will make their perimeters the same? remember for problems 6 and 7, solve for x. $\frac{2}{5}(x - 10)=\frac{1}{25}x + 5$

Explanation:

Step1: Define variables for triangle - rectangle problem

Let the leg length of the isosceles triangle be \(x\) meters. Then the base of the triangle \(b = 2x- 10\) meters, the width of the rectangle \(w=x\) meters, and the length of the rectangle \(l=2x + 10\) meters.
The perimeter of the isosceles triangle \(P_{t}=2x+(2x - 10)=4x-10\). The perimeter of the rectangle \(P_{r}=2(x+(2x + 10))=2(3x + 10)=6x+20\).

Step2: Set perimeters equal

Set \(P_{t}=P_{r}\), so \(4x-10=6x + 20\).
Subtract \(4x\) from both sides: \(-10=2x+20\).
Subtract 20 from both sides: \(-30 = 2x\).
Divide both sides by 2: \(x = 20\).
Then \(b=2x-10=2\times20 - 10=30\), \(w=x = 20\), \(l=2x+10=2\times20+10 = 50\)

Step3: Solve the equation \(\frac{2}{5}(x - 10)=\frac{1}{25}x+5\)

First, distribute on the left - hand side: \(\frac{2}{5}x-4=\frac{1}{25}x + 5\).
Subtract \(\frac{1}{25}x\) from both sides: \(\frac{2}{5}x-\frac{1}{25}x-4=5\).
Get a common denominator for \(x\) terms: \(\frac{10}{25}x-\frac{1}{25}x-4=5\), \(\frac{9}{25}x-4=5\).
Add 4 to both sides: \(\frac{9}{25}x=9\).
Multiply both sides by \(\frac{25}{9}\): \(x = 25\)

Answer:

1. For the triangle - rectangle problem: Leg length of the triangle \(x = 20\) meters, base of the triangle \(b=30\) meters, width of the rectangle \(w = 20\) meters, length of the rectangle \(l=50\) meters.
2. For the equation \(\frac{2}{5}(x - 10)=\frac{1}{25}x+5\): \(x = 25\)