1.

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

Answer»

Solution :

GIVEN 

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

 

 



Discussion

No Comment Found

Related InterviewSolutions