InterviewSolution
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A pyramid is with a triangular base ABC where AB = 25 cm, BC = 24 cm and CA = 7 cm. The point D is vertically above the point C where CD = 34 cm. If the area of ΔABD is 40 sq.cm, then what is the total surface area of the pyramid?1). 244 sq.cm2). 145 sq.cm3). 651 sq.cm4). 354 sq.cm |
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Answer» Solution : The SIDES of triangle ABC are $25$ cm, $24$cm and $7$ cm . Triangle $ABC$ is a right angle triangle as $25$ cm, $24$cm and $7$ cm are Pythagoras triplets. A pyramid is made up of four triangles namely BASE triangle $ABC$, triangle $BCD$, triangle $ACD$ and triangle $ABD$. Triangles $BCD$ and $ACD$ are ALSO right angle triangles. Area of triangle $ABC = (1/2 )× 24 × 7 = 84$ sq.cm Area of triangle $BCD = (1/2 )× 24 × 34 = 408$sq.cm Area of triangle $ACD = (1/2 )× 7 × 34 = 119$ sq.cm Given : Area of triangle $ABD = 40$ sq.cm Total surface area of the pyramid $= ( 84 + 408 + 119 + 40) $sq.cm = 651 sq.cm The correct option is 3). 651 sq.cm
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