The base of the pyramid is a rectangle. The length and width of t
A. 40(√5 + √61) + 240 sq. cm
B. 20(6√5 + √61) + 240 sq. cm
C. 12(4√5 + 2√61) + 180 sq. cm
D. 10[2√5 + √61] sq. cm
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Given:
Length and width of the rectangle is 24 cm and 10 cm, respectively
Height of the pyramid is 10 cm
Calculation:
AB = 24 cm, BC = 10 cm, EO = 10 cm
Area of rectangle ABCD = 24 × 10 = 240 sq. cm
To find the area of the ΔABE and ΔDEC, we have to find the value of EG.
In ΔEGO –
GO = BC/2 = 5 cm
EO = 10 cm
\(⇒ GE = \sqrt {{{10}^2} + {5^2}}\)
⇒ GE = √125
⇒ GE = 5√5 cm
Therefore, Area of ΔABE = Area of ΔDEC = 1/2 × 24 × 5√5 = 60√5 sq. cm
To find the area of the ΔBCE and ΔADE
In ΔEOF
OF = 24/2 = 12 cm
⇒ \({\rm{EF}} = \sqrt {{{10}^2} + {{12}^2}}\)
⇒ EF = √244
⇒ EF = 2√61 cm
Area of ΔBCE = Area of ΔADE = 1/2 × 10 × 2√61 = 10√61 sq.cm.
∴, the total surface area of pyramid
⇒ 2(60√5 + 10√61) + 240 = 20(6√5 + √61) + 240 sq. cm
