In acute triangle PQR, an altitude PS is drawn from vertex P perpendicular to side QR at p… — ACT

In acute triangle $PQR$ , an altitude $PS$ is drawn from vertex $P$ perpendicular to side $QR$ at point $S$ . The measure of $\angle PQS$ is $60^\circ$ , the length of segment $PQ$ is $24$ , and the length of segment $PR$ is $39$ . What is the length of side $QR$ ?

Answer: 45

Answer

The length of side

QR

45

The altitude divides the acute triangle into two right triangles. In the first right triangle,

\triangle PQS

, the angles are

30^\circ

60^\circ

, and

90^\circ

, with a hypotenuse of

24

. This makes the adjacent leg

QS = 12

and the shared altitude

PS = 12\sqrt{3}

. In the second right triangle,

\triangle PRS

, the hypotenuse is

39

and one leg is

12\sqrt{3}

. Using the Pythagorean Theorem, we find the other leg

SR = \sqrt{39^2 - (12\sqrt{3})^2} = \sqrt{1521 - 432} = \sqrt{1089} = 33

. Summing the two segments gives the total length of side

QR = 12 + 33 = 45

Step-by-Step Solution

Identify the two right triangles formed by the altitude.

The altitude

PS

divides

\triangle PQR

into two adjacent right triangles:

\triangle PQS

and

\triangle PRS

, which share the side

PS

Establishing these right triangles allows us to apply right-triangle trigonometric ratios and the Pythagorean Theorem.

Use the properties of the

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle

\triangle PQS

to find

QS

and

PS

QS = 12

and

PS = 12\sqrt{3}

In a

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle, the leg opposite the

30^\circ

angle is half the hypotenuse, and the leg opposite the

60^\circ

angle is

\sqrt{3}

times the shorter leg. Here, hypotenuse

PQ = 24

, so

QS = 12

and

PS = 12\sqrt{3}

Apply the Pythagorean Theorem to

\triangle PRS

to find

SR

SR = 33

In right triangle

\triangle PRS

, the hypotenuse is

PR = 39

. By the Pythagorean Theorem,

PS^2 + SR^2 = PR^2

. Squaring the sides gives

(12\sqrt{3})^2 + SR^2 = 39^2 \implies 432 + SR^2 = 1521

. Solving for

SR

gives

SR^2 = 1089 \implies SR = 33

Sum the segments

QS

and

SR

to find the total length of

QR

QR = 45

Because

\triangle PQR

is an acute triangle, the altitude

PS

lands at a point

S

on the segment

QR

, meaning

QR = QS + SR

. Adding the lengths gives

12 + 33 = 45

Key Concept

Applying properties of

30^\circ\text{-}60^\circ\text{-}90^\circ

special right triangles and the Pythagorean Theorem in multi-step geometric figures.

Estimated Time:2m 30s

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