The base of a right prism is an equilateral t | vedclass

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

Question

(A) Let the side of the equilateral triangle base be $a$ and the height of the prism be $h$.Lateral surface area of the prism $= 3 	imes a 	imes h =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the side of the equilateral triangle base be $a$ and the height of the prism be $h$.Lateral surface area of the prism $= 3 	imes a 	imes h = 120 \, cm^2$.$	herefore a 	imes h = \frac{120}{3} = 40 \, cm^2 \dots (1)$Volume of the prism $=$ Area of base $	imes$ height $= \frac{\sqrt{3}}{4} a^2 	imes h = 40 \sqrt{3} \, cm^3$.$\Rightarrow a^2 	imes h = \frac{40 \sqrt{3} 	imes 4}{\sqrt{3}} = 160 \, cm^3 \dots (2)$Dividing equation $(2)$ by $(1)$:$\frac{a^2 	imes h}{a 	imes h} = \frac{160}{40}$$a = 4 \, cm$.Thus,the side of the base is $4 \, cm$.

The base of a right prism is an equilateral triangle. If the lateral surface area and volume are $120 \, cm^2$ and $40 \sqrt{3} \, cm^3$,respectively,then the side of the base of the prism is ...... $cm$.

Similar Questions

The length of a room is $12\, m$,width $8\, m$,and height $6\, m$. How many boxes will it hold if each box occupies $1.5\, m^3$ of space?

Find the curved surface area (in $cm^{2}$) of a hemisphere of diameter $28\, cm$.

$A$ right circular cylindrical tunnel of diameter $5\, m$ and length $10\, m$ is to be constructed from a sheet of iron. The area of iron sheet required will be..........$\pi\, m^2$.

The material of a cone is converted into the shape of a cylinder of equal radius. If the height of the cylinder is $5\, cm$,the height of the cone is........$cm$.

$A$ plane divides a right circular cone into two parts of equal volume. If the plane is parallel to the base,then the ratio,in which the height of the cone is divided,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module