In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA.
Solution:-
The square of any ratio between any two corresponding sides of two similar triangles is equal to the ratio of their areas.
From the question it is given that, PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm²
From the figure,
Consider the ∆AOQ and ∆BOP,
∠OAQ = ∠OBP … [both angles are equal to 90o]
∠AOQ = ∠BOP … [because vertically opposite angles are equal]
Therefore, ∆AOQ ~ ∆BOP
Then, area of ∆AOQ/area of ∆BOP = OQ2/PO2
Area of ∆AOQ/120 = 92/62
Area of ∆AOQ/120 = 81/36
Area of ∆AOQ = (81 × 120)/36
Area of ∆AOQ = 270 cm2
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