In triangle PQR, S lies on PQ and T lies on PR with ST parallel to QR, and PS : SQ = 3 : 5. The area of triangle PST is $18 cm^{2}$ .
Find the area of trapezium SQRT (in $cm^{2}$ ).

Question

In triangle PQR, S lies on PQ and T lies on PR with ST parallel to QR, and PS : SQ = 3 : 5. The area of triangle PST is 18 cm^2. Find the area of trapezium SQRT (in cm^2).

Accepted Answer

110. PS : PQ = 3 : 8, so area(PST) : area(PQR) = (3/8 )^2 = 9/64. From 18 area(PQR) = 9/64 we get area(PQR) = 128 cm^2, so trapezium SQRT = 128 - 18 = 110 cm^2. The option 128 is the WHOLE triangle rather than the trapezium; using PS : SQ = 3 : 5 in place of PS : PQ = 3 : 8 gives area(PQR) = 50 and a trapezium of 32.

In triangle PQR, S lies on PQ and T lies on PR with ST parallel to QR, and PS : SQ = 3 : 5. The area of triangle PST is $18 cm^{2}$ .
Find the area of trapezium SQRT (in $cm^{2}$ ).

Step-by-Step Solution

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More Class 10 Areas of Similar Triangles practice

In triangle PQR, S lies on PQ and T lies on PR with ST parallel to QR, and PS : SQ = 3 : 5. The area of triangle PST is 18cm2.Find the area of trapezium SQRT (in cm2).

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Areas of Similar Triangles practice

In triangle PQR, S lies on PQ and T lies on PR with ST parallel to QR, and PS : SQ = 3 : 5. The area of triangle PST is $18 cm^{2}$ .
Find the area of trapezium SQRT (in $cm^{2}$ ).