Question

1. the area of a trapezoid is found by finding the product of one - half of the sum of the bases and the height. the equation is given by a = 1/2(b1 + b2)h. solve for the second base. 11. the area of a triangle is found by taking one - half of the product of the base and height. the equation is given by a = 1/2bh. solve for the base.

Explanation:

Step1: Isolate the term with \(b_2\) for trapezoid formula

Starting with \(A=\frac{1}{2}(b_1 + b_2)h\), first multiply both sides by \(2\) to get \(2A=(b_1 + b_2)h\).

Step2: Divide by \(h\)

Then \(\frac{2A}{h}=b_1 + b_2\).

Step3: Solve for \(b_2\)

Subtract \(b_1\) from both sides, so \(b_2=\frac{2A}{h}-b_1\).

For the triangle formula \(A = \frac{1}{2}bh\), to solve for \(b\):

Step1: Multiply both sides by 2

We have \(2A=bh\).

Step2: Divide by \(h\)

So \(b=\frac{2A}{h}\).

Answer:

For the trapezoid, \(b_2=\frac{2A}{h}-b_1\); for the triangle, \(b=\frac{2A}{h}\)