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Difficulty: HardLaw of Sines and Law of Cosines

A triangular region PQRPQR has side lengths PQ=8PQ = 8 meters and QR=10QR = 10 meters, and the measure of angle ∠PQR\angle PQR is 120∘120^\circ. A straight walkway QMQM is constructed from vertex QQ to a point MM on side PRPR such that QMQM bisects ∠PQR\angle PQR. What is the length, in meters, of the walkway QMQM?

  1. 409\frac{40}{9}Answer
  2. B
    809\frac{80}{9}
  3. C
    209\frac{20}{9}
  4. D
    403\frac{40}{3}
  5. E
    434\sqrt{3}

Answer

409\frac{40}{9} meters
The total area of triangle PQRPQR is equal to the sum of the areas of triangles PQMPQM and MQRMQR. Using the formula Area=12absin⁡(θ)\text{Area} = \frac{1}{2}ab\sin(\theta), the equation 12(8)(10)sin⁡(120∘)=12(8)(x)sin⁡(60∘)+12(10)(x)sin⁡(60∘)\frac{1}{2}(8)(10)\sin(120^\circ) = \frac{1}{2}(8)(x)\sin(60^\circ) + \frac{1}{2}(10)(x)\sin(60^\circ) simplifies to 203=9x3220\sqrt{3} = \frac{9x\sqrt{3}}{2}, which gives x=409x = \frac{40}{9}.

Step-by-Step Solution

1
Express the area of the entire triangle PQRPQR using the sine area formula.
Area(△PQR)=12⋅PQ⋅QR⋅sin⁡(120∘)=12⋅8⋅10⋅32=203\text{Area}(\triangle PQR) = \frac{1}{2} \cdot PQ \cdot QR \cdot \sin(120^\circ) = \frac{1}{2} \cdot 8 \cdot 10 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3} square meters.
The area of a triangle given two sides and the included angle is 12absin⁡(C)\frac{1}{2}ab\sin(C).
2
Express the sum of the areas of the two smaller triangles △PQM\triangle PQM and △MQR\triangle MQR created by the angle bisector QM=xQM = x.
Since QMQM bisects ∠PQR=120∘\angle PQR = 120^\circ, ∠PQM=60∘\angle PQM = 60^\circ and ∠MQR=60∘\angle MQR = 60^\circ. Area(△PQM)=12⋅8⋅x⋅sin⁡(60∘)=2x3\text{Area}(\triangle PQM) = \frac{1}{2} \cdot 8 \cdot x \cdot \sin(60^\circ) = 2x\sqrt{3}. Area(△MQR)=12⋅10⋅x⋅sin⁡(60∘)=5x32\text{Area}(\triangle MQR) = \frac{1}{2} \cdot 10 \cdot x \cdot \sin(60^\circ) = \frac{5x\sqrt{3}}{2}.
An angle bisector divides the total angle into two equal 60∘60^\circ angles.
3
Equate the total area to the sum of the partial areas and solve for xx.
203=2x3+5x32  ⟹  20=2x+5x2  ⟹  20=9x2  ⟹  x=40920\sqrt{3} = 2x\sqrt{3} + \frac{5x\sqrt{3}}{2} \implies 20 = 2x + \frac{5x}{2} \implies 20 = \frac{9x}{2} \implies x = \frac{40}{9}.
The total area of the figure is equal to the sum of its non-overlapping component areas.

Key Concept

Area of Triangles using Sine and Angle Bisector Properties
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Topics

Law of Sines and Law of Cosines

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