Hi, there-

The Problem:
A triangle has an 8-in base and a height of 4-in. A second triangle has three-fourths the area and 
half the height of the first triangle. What is the base of the second triangle?

Solution:
The formula for the area of a triangle is A=(1/2)*b*h where b is its base and h is its height.

We see that the first triangle has a base of 16 square inches (sq. in.) because,
A=(1/2)(8)(4)=16

The second triangle has three-fourths the area of the first triangle. Its area is 12 sq. in. because
(3/4)(16)=12

The height of the second triangle is 2-inches because it's half of 4-inches.
(1/2)(4)=2

Now we know the area of the second triangle and its height. Substitute 12 for A and 2 for h into the
area formula.
A=(1/2)*b*h
12=(1/2)*b*(2)

Solve for b.
b=12

The base of the second triangle is 12 inches.

It's always a good idea to check your work against the conditions of original problem:

"A second triangle has three-fourths the area ... of the first triangle."
[12] is three-fourths of [16]. TRUE!

"A second triangle has ... half the height of the first triangle."
[2] is half of [4]. TRUE!

That's it. Please email me if you have questions or comments about the solution. I'm happy to 
explain further and I always appreciate feedback.

Ms.Figgy
math.in.the.vortex@gmail.com