NMTC 2014(Junior level)

Geometry Level 3

In the given figure, PQ=42 cmPQ=42\text{ cm}. QRQR is the tangent to the semicircle at QQ. If the difference of the areas of region AA and BB is 357 cm2357\text{ cm}^2, then the base QRQR of the right triangle PQRPQR is (in cm)(Take π=227\pi=\frac{22}{7})

Note: This problem has appeared in nmtc 2014 junior level.

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