# Word Problems Involving Area of Quadrilaterals and Triangles

In this article, the focus is on teaching you how to solve word problems involving the area of quadrilaterals and triangles.

## A step-by-step guide to word problems involving the area of quadrilaterals and triangles

The word problem is a way of learning the area of quadrilaterals and triangles better.

The area of a parallelogram: \(A=base×height\)

The area of a triangle: \(A=\frac{1}{2}(base×height)\)

The area of a trapezoid: \(A= \frac{1}{2}×\)(Sum of parallel sides)\(×\)(perpendicular distance in between the parallel sides). \(\frac{1}{2}×(AB ̅+DC) ̅×(d)\)

**Word Problems Involving Area of Quadrilaterals and Triangles – Examples 1**

Solve.

A triangle has an area of 64 square inches and a height of 16 inches. What is the length of the triangle’s base?**Solution:**

Since the area of a triangle is \(\frac{1}{2}\)(base\(×\)height), then you must divide 64 by 16 and then multiply the product by 2 to find the length of its base.

\(64÷16=4\)

\(4×2=8\)

So, the length of the triangle’s base is 8 in.

**Word Problems Involving Area of Quadrilaterals and Triangles – Examples 2**

Solve.

The area of a parallelogram is \(484 ft^2\) and its height is 22 ft. What is the base length?**Solution:**

Since the area of a parallelogram is base×height, so you have to divide 484 by 22 to find the base length.

\(484÷22=22\)

So, the base length of the parallelogram is 22 ft.

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