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Difficulty: HardTriangles: Properties, Perimeter, and Area

In the xyxy-plane, triangle PQRPQR has vertices at P(0,0)P(0, 0), Q(8,0)Q(8, 0), and R(2,6)R(2, 6). Point SS lies on segment PQPQ such that segment RSRS divides triangle PQRPQR into two regions of equal area. Point TT lies on segment QRQR such that segment STST is parallel to segment PRPR. What is the area of triangle QSTQST?

Answer: 6

Answer

6
The area of triangle PQRPQR is calculated using base PQ=8PQ = 8 and height h=6h = 6, yielding 12×8×6=24\frac{1}{2} \times 8 \times 6 = 24. Since segment RSRS divides triangle PQRPQR into two regions of equal area that share the height from vertex RR, point SS must be the midpoint of PQPQ, making QS=4QS = 4. Because segment STST is parallel to segment PRPR, triangle QSTQST is similar to triangle QPRQPR with a side length ratio of QSQP=48=12\frac{QS}{QP} = \frac{4}{8} = \frac{1}{2}. The ratio of the areas of similar triangles is the square of the side ratio, (12)2=14\left(\frac{1}{2}\right)^2 = \frac{1}{4}. Multiplying the total area 2424 by 14\frac{1}{4} gives the area of triangle QSTQST as 66.

Step-by-Step Solution

1
Calculate the area of the main triangle PQRPQR.
Area(PQR)=12×8×6=24\text{Area}(PQR) = \frac{1}{2} \times 8 \times 6 = 24.
The base PQPQ lies along the x-axis with length 8−0=88 - 0 = 8, and the perpendicular height from vertex R(2,6)R(2,6) to the base is 66.
2
Determine the length of segment QSQS.
QS=4QS = 4.
Line segment RSRS splits △PQR\triangle PQR into two smaller triangles, △PSR\triangle PSR and △QSR\triangle QSR, which share the same altitude from vertex RR. For their areas to be equal, their base lengths PSPS and SQSQ must be equal. Therefore, SS is the midpoint of PQPQ, giving QS=82=4QS = \frac{8}{2} = 4.
3
Establish the similarity relationship and scale factor between △QST\triangle QST and △QPR\triangle QPR.
△QST∼△QPR\triangle QST \sim \triangle QPR with scale factor k=12k = \frac{1}{2}.
Because segment STST is parallel to segment PRPR, corresponding angles are equal (∠QST=∠QPR\angle QST = \angle QPR and ∠QTS=∠QRP\angle QTS = \angle QRP). Thus, △QST\triangle QST is similar to △QPR\triangle QPR. The ratio of corresponding side lengths is QSQP=48=12\frac{QS}{QP} = \frac{4}{8} = \frac{1}{2}.
4
Compute the area of triangle QSTQST.
Area(QST)=6\text{Area}(QST) = 6.
The ratio of the areas of similar triangles is equal to the square of their linear scale factor: Area(QST)=(12)2×Area(PQR)=14×24=6\text{Area}(QST) = \left(\frac{1}{2}\right)^2 \times \text{Area}(PQR) = \frac{1}{4} \times 24 = 6.

Key Concept

Area of triangles, midpoint area partitioning, and area ratio scaling in similar triangles
Estimated Time:2m 0s
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Topics

Triangles: Properties, Perimeter, and Area

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