In rhombus ABCD, the side length is 10 and the length of diagonal BD is 12. Line segment A… — GRE General Test

In rhombus $ABCD$ , the side length is $10$ and the length of diagonal $BD$ is $12$ . Line segment $AP$ is drawn perpendicular to side $BC$ , with point $P$ lying on segment $BC$ . What is the length of segment $AP$ ?

Answer: 9.6

Answer

9.6

The diagonals of rhombus

ABCD

intersect perpendicularly at

O

and bisect each other. Given

BD = 12

, half of the diagonal is

BO = 6

. Right triangle

AOB

has hypotenuse

AB = 10

and leg

BO = 6

, so by the Pythagorean theorem, leg

AO = \sqrt{10^2 - 6^2} = 8

. Thus, diagonal

AC = 16

. The area of rhombus

ABCD

\frac{1}{2} \times AC \times BD = \frac{1}{2} \times 16 \times 12 = 96

. The area is also equal to

\text{base} \times \text{height} = BC \times AP = 10 \times AP

. Setting

10 \times AP = 96

gives

AP = 9.6

Step-by-Step Solution

Find half the length of diagonal BD.

Segment BO = 6.

The diagonals of a rhombus bisect each other at right angles.

Apply the Pythagorean theorem to right triangle AOB to determine half of diagonal AC.

AO = 8, so diagonal AC = 16.

Triangle AOB has hypotenuse 10 and leg 6, forming a 6-8-10 Pythagorean triple.

Calculate the total area of rhombus ABCD from its diagonal lengths.

Area = 96.

The area of a rhombus equals half the product of its two diagonals.

Use the alternative area formula (base × height) to solve for altitude AP.

AP = 9.6.

Base BC = 10 and height AP give Area = 10 × AP = 96.

Key Concept

Properties of rhombus diagonals, Pythagorean theorem, and dual area formulas for quadrilaterals

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