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Difficulty: MediumPythagorean Theorem and Special Right Triangles

A robot starting at point PP travels due east for 44 meters to point QQ. At point QQ, the robot turns 60∘60^\circ counterclockwise from its original direction and travels in a straight line for 88 meters, stopping at point RR. What is the straight-line distance, in meters, between point PP and point RR?

  1. A
    434\sqrt{3}
  2. B
    454\sqrt{5}
  3. 474\sqrt{7}Answer
  4. D
    8+438 + 4\sqrt{3}
  5. E
    1212

Answer

The straight-line distance between point PP and point RR is 474\sqrt{7} meters.
Dropping a vertical perpendicular from RR to the extended line PQPQ forms a 30∘30^\circ-60∘60^\circ-90∘90^\circ right triangle QSRQSR with hypotenuse 88. The base extension QSQS equals 44 and the altitude RSRS equals 434\sqrt{3}. Applying the Pythagorean theorem to the right triangle PSRPSR with legs PS=4+4=8PS = 4 + 4 = 8 and RS=43RS = 4\sqrt{3} gives PR=82+(43)2=64+48=112=47PR = \sqrt{8^2 + (4\sqrt{3})^2} = \sqrt{64 + 48} = \sqrt{112} = 4\sqrt{7} meters.

Step-by-Step Solution

1
Construct a right triangle by extending segment PQPQ past QQ to a point SS directly below RR, dropping altitude RS⊥PSRS \perp PS.
Triangle QSRQSR is formed with ∠RQS=60∘\angle RQS = 60^\circ and ∠QSR=90∘\angle QSR = 90^\circ.
Decomposing the angled path into perpendicular horizontal and vertical components allows the use of special right triangles and the Pythagorean theorem.
2
Use 30∘30^\circ-60∘60^\circ-90∘90^\circ side ratios (1:3:21 : \sqrt{3} : 2) in triangle QSRQSR where hypotenuse QR=8QR = 8.
Horizontal segment QS=8×12=4QS = 8 \times \frac{1}{2} = 4 meters, and vertical altitude RS=8×32=43RS = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3} meters.
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 short leg.
3
Calculate the total horizontal distance PSPS and vertical distance RSRS for the large right triangle PSRPSR.
Total base PS=PQ+QS=4+4=8PS = PQ + QS = 4 + 4 = 8 meters, and height RS=43RS = 4\sqrt{3} meters.
Point PP, point QQ, and point SS are collinear on the horizontal line.
4
Apply the Pythagorean theorem to right triangle PSRPSR to find distance PRPR.
PR=PS2+RS2=82+(43)2=64+48=112=47PR = \sqrt{PS^2 + RS^2} = \sqrt{8^2 + (4\sqrt{3})^2} = \sqrt{64 + 48} = \sqrt{112} = 4\sqrt{7} meters.
The straight-line distance PRPR is the hypotenuse of right triangle PSRPSR.

Key Concept

Combining 30∘30^\circ-60∘60^\circ-90∘90^\circ special right triangles with the Pythagorean theorem to solve multi-step 2D path/distance problems.
Estimated Time:1m 45s
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Topics

Pythagorean Theorem and Special Right Triangles

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