In \triangle ABC, angle B is a right angle, and line segment BD is an altitude drawn to si… — GRE General Test

In $\triangle ABC$ , angle $B$ is a right angle, and line segment $BD$ is an altitude drawn to side $AC$ with point $D$ lying on $AC$ . If $AD = 4$ units and $DC = 16$ units, what is the area, in square units, of $\triangle ABC$ ?

Answer: 80 square units

Answer

The area of triangle ABC is 80 square units.

In right triangle

ABC

with right angle at

B

, altitude

BD

drawn to hypotenuse

AC

divides the hypotenuse into segments

AD

and

DC

. By the geometric mean theorem,

BD^2 = AD \times DC = 4 \times 16 = 64

, which gives

BD = 8

units. The length of hypotenuse

AC

AD + DC = 4 + 16 = 20

units. The area of triangle

ABC

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 8 = 80

square units.

Step-by-Step Solution

Calculate the height (altitude)

BD

of the triangle.

BD = 8

units.

In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that

BD^2 = AD \times DC

. Substituting the given values yields

BD^2 = 4 \times 16 = 64

, so

BD = \sqrt{64} = 8

Calculate the total length of hypotenuse

AC

AC = 20

units.

Since point

D

lies on segment

AC

, the total length is the sum of its parts:

AC = AD + DC = 4 + 16 = 20

Calculate the area of

\triangle ABC

Area = 80 square units.

Using the triangle area formula

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

, with base

AC = 20

and altitude

BD = 8

, we get

\text{Area} = \frac{1}{2} \times 20 \times 8 = 80

Key Concept

Geometric mean theorem for right triangle altitude and area of a triangle

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic