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Zorluk: ZorPythagorean Theorem and Special Right Triangles

In acute triangle PQRPQR, an altitude PSPS is drawn from vertex PP perpendicular to side QRQR at point SS. The measure of ∠PQS\angle PQS is 60∘60^\circ, the length of segment PQPQ is 2424, and the length of segment PRPR is 3939. What is the length of side QRQR?

Cevap: 45

Cevap

The length of side QRQR is 4545.
The altitude divides the acute triangle into two right triangles. In the first right triangle, △PQS\triangle PQS, the angles are 30∘30^\circ, 60∘60^\circ, and 90∘90^\circ, with a hypotenuse of 2424. This makes the adjacent leg QS=12QS = 12 and the shared altitude PS=123PS = 12\sqrt{3}. In the second right triangle, △PRS\triangle PRS, the hypotenuse is 3939 and one leg is 12312\sqrt{3}. Using the Pythagorean Theorem, we find the other leg SR=392−(123)2=1521−432=1089=33SR = \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=45QR = 12 + 33 = 45.

Adım Adım Çözüm

1
Identify the two right triangles formed by the altitude.
The altitude PSPS divides △PQR\triangle PQR into two adjacent right triangles: △PQS\triangle PQS and △PRS\triangle PRS, which share the side PSPS.
Establishing these right triangles allows us to apply right-triangle trigonometric ratios and the Pythagorean Theorem.
2
Use the properties of the 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ triangle △PQS\triangle PQS to find QSQS and PSPS.
QS=12QS = 12 and PS=123PS = 12\sqrt{3}.
In a 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ triangle, the leg opposite the 30∘30^\circ angle is half the hypotenuse, and the leg opposite the 60∘60^\circ angle is 3\sqrt{3} times the shorter leg. Here, hypotenuse PQ=24PQ = 24, so QS=12QS = 12 and PS=123PS = 12\sqrt{3}.
3
Apply the Pythagorean Theorem to △PRS\triangle PRS to find SRSR.
SR=33SR = 33.
In right triangle △PRS\triangle PRS, the hypotenuse is PR=39PR = 39. By the Pythagorean Theorem, PS2+SR2=PR2PS^2 + SR^2 = PR^2. Squaring the sides gives (123)2+SR2=392  ⟹  432+SR2=1521(12\sqrt{3})^2 + SR^2 = 39^2 \implies 432 + SR^2 = 1521. Solving for SRSR gives SR2=1089  ⟹  SR=33SR^2 = 1089 \implies SR = 33.
4
Sum the segments QSQS and SRSR to find the total length of QRQR.
QR=45QR = 45.
Because △PQR\triangle PQR is an acute triangle, the altitude PSPS lands at a point SS on the segment QRQR, meaning QR=QS+SRQR = QS + SR. Adding the lengths gives 12+33=4512 + 33 = 45.

Anahtar Kavram

Applying properties of 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ special right triangles and the Pythagorean Theorem in multi-step geometric figures.
Tahmini Süre:2m 30s
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