If ST is parallel to QR, and PS = SQ = 3 cm, PT = 4 cm and QR = 10 cm, what is the area of trapezoid STQR?

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Solution

The correct option is D 22
Option (d).
Triangle PST & PQR are similar. Thus, $\frac{P S}{P Q}$ = $\frac{P T}{P R}$ $\Rightarrow$ PR=8

Thus, PQR is a right angled triangle with sides-6,8,10

Area of trapezoid STQR= Area of triangle PQR- Area of triangle PST = $\frac{1}{2}$ 68 - $\frac{1}{2}$ 34. Answer is 18 cm $^{2}$

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Similar questions

P S = 4 c m, S R = 2 c m, P T = 3 c m

Q T = 5 c m

S T

Q R = 5.8 c m

{PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}

{PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}

In the given figure, ST is parallel to QR. PS=3cm, PQ=8cm TR=10cm PT=?

△ P Q R, S T ∥ Q R

\frac{P S}{S Q} = \frac{3}{5}

P R = 6

P T

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