In trapezoid ABCD, the bases AB and CD are parallel. The diagonals AC and BD intersect at … — ACT

In trapezoid $ABCD$ , the bases $AB$ and $CD$ are parallel. The diagonals $AC$ and $BD$ intersect at point $E$ . If the area of $\triangle ABE$ is $16\text{ cm}^2$ and the area of $\triangle CDE$ is $36\text{ cm}^2$ , what is the total area, in square centimeters, of trapezoid $ABCD$ ?

Answer: 100 cm^2

Answer

The total area of trapezoid

ABCD

100\text{ cm}^2

Triangles

ABE

and

CDE

are similar with an area ratio of

16:36

, yielding a side length ratio of

2:3

. The adjacent triangles

\triangle ADE

and

\triangle BCE

each share an altitude with the base triangles, making their areas equal to the geometric mean of the base areas:

\sqrt{16 \times 36} = 24\text{ cm}^2

each. Summing the four regions gives

16 + 36 + 24 + 24 = 100\text{ cm}^2

Step-by-Step Solution

Establish the similarity of triangles

ABE

and

CDE

\triangle ABE \sim \triangle CDE

Since bases

AB

and

CD

are parallel, alternate interior angles are congruent (

\angle EAB \cong \angle ECD

and

\angle EBA \cong \angle EDC

). By AA Similarity, the triangles are similar.

Calculate the linear scale factor between the similar triangles.

\frac{AE}{EC} = \frac{BE}{ED} = \sqrt{\frac{16}{36}} = \frac{2}{3}

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding side lengths.

Find the areas of the adjacent triangles

\triangle ADE

and

\triangle BCE

\text{Area}(\triangle ADE) = 24\text{ cm}^2

and

\text{Area}(\triangle BCE) = 24\text{ cm}^2

Triangles

ADE

and

CDE

share the same altitude from vertex

D

to diagonal

AC

, meaning the ratio of their areas is equal to the ratio of their bases:

\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{AE}{EC} = \frac{2}{3} \implies \text{Area}(\triangle ADE) = \frac{2}{3} \times 36 = 24

. By similar logic,

\frac{\text{Area}(\triangle BCE)}{\text{Area}(\triangle ABE)} = \frac{EC}{AE} = \frac{3}{2} \implies \text{Area}(\triangle BCE) = \frac{3}{2} \times 16 = 24

Sum the areas of the four individual triangles.

\text{Area}(ABCD) = 16 + 36 + 24 + 24 = 100\text{ cm}^2

The total area of the trapezoid is the sum of the areas of the four non-overlapping triangles formed by its diagonals.

Key Concept

For any trapezoid with diagonals intersecting at

E

and parallel bases forming triangles of areas

A_1

and

A_2

, the other two triangles each have an area equal to

\sqrt{A_1 A_2}

, and the total area of the trapezoid is given by

(\sqrt{A_1} + \sqrt{A_2})^2

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic