The base of a right pyramid is a square of si | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Area of base $= 10 	imes 10 = 100 	ext{ cm}^2$.Perimeter of base $= 4 	imes 10 = 40 	ext{ cm}$.The slant height $(l)$ of the pyramid is calcul

Vedclass · Accepted Answer

$360$

Vedclass · Answer

(C) Area of base $= 10 	imes 10 = 100 	ext{ cm}^2$.Perimeter of base $= 4 	imes 10 = 40 	ext{ cm}$.The slant height $(l)$ of the pyramid is calculated using the height $(h = 12 	ext{ cm})$ and half the side of the base $(a/2 = 5 	ext{ cm})$:$l = \sqrt{(a/2)^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 	ext{ cm}$.Curved Surface Area $(C.S.A.)$ of the pyramid $= \frac{1}{2} 	imes 	ext{Perimeter of base} 	imes 	ext{slant height}$$C.S.A. = \frac{1}{2} 	imes 40 	imes 13 = 20 	imes 13 = 260 	ext{ cm}^2$.Total Surface Area $(T.S.A.)$ $= C.S.A. + 	ext{Area of base}$$T.S.A. = 260 	ext{ cm}^2 + 100 	ext{ cm}^2 = 360 	ext{ cm}^2$.

The base of a right pyramid is a square of side $10 \text{ cm}$. If the height of the pyramid is $12 \text{ cm}$,then its total surface area is ....... $\text{cm}^2$.

Similar Questions

$A$ $4\, cm$ cube is cut into $1\, cm$ cubes. What is the ratio of the total surface area of all the small cubes to the surface area of the large cube?

The area of three adjacent faces of a rectangular box are $p, q$ and $r$ square $cm$. The volume of the box is given by

$A$ cylinder with base radius $8\, cm$ and height $2\, cm$ is melted to form a cone of height $6\, cm$. The radius of the cone will be (in $cm$):

$A$ right cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is

$A$ ball of lead,$4\, cm$ in diameter,is covered with gold. If the volume of the gold and lead are equal,then the thickness of gold is approximately (in $cm$).
$[$ Given $\sqrt[3]{2} = 1.259 ]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module