In the adjoining figure, ΔQPR is right-angled at Q in which QR = 6 cm and PQ = 7 cm. Find the area of ΔQSR, given that PS is parallel to QR.
Options:
A .  
21cm2
B .  
20cm2
C .  
10cm2
D .  
11cm2
Answer: Option A : A Area of right angled triangle PQR = 12 x Base x Height =12 x QR x PQ =12 x 6 x 7 =21cm2 ΔQPR and ΔQSR lie on same base QR and are between same parallels hence, their areas are equal. Area ΔQPR = Area ΔQSR ∴AreaofΔQSR=21cm2
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