**The side of a square based prism is 2.5 cm and its height is 6.5 cm . Find its lateral and total surface area.**

Solution:

a = 6.5 c

h = 6.5cm

LSA = Ph = Psquare x h

4ah = 4 x 2.5 x 6.5

= 65 cm^{2}

TSA = 2B + Ph = 2a^{2} + Ph = 2(2.5)^{2} + 65 = 77.5cm^{2}

**The perimeter of a regular triangle based prism is 18 m and its height is 13 m. find its LSA and TSA.**

Solution:

Regular triangular based prsim(equilateral triangle)

P Δ = 18m

h = 13m

LSA ? and TSA=?

LSA = Ph = 18 x 13 = 234m^{2}

We have B = area of base

A = P/3 = 18/3 = 6cm

Then, B = ^{√3 x 6 x 6}/_{2} = 18√3

TSA = 2B + Ph = 2×18√3 + 234 = (36√3 + 234)m^{2}

**The shaper of a pole is in the form of square based prism. If its side is 2 feet and latest surface area is 160 square feet. Find the height of the pole**

Solution:

LSA = 160 sq. ft.

a = 2ft

h= ?

LSA = Ph (since P = 4a)

160 = 4ah = 4 x 2 x h

h = ^{160}/_{8} = 20 ft.

**The TSA of a triangular based prism is 570 cm**^{2 }and its height is 22 cm. if the area of its base is 24 cm^{2}. Find the perimeter of the base.

Solution:

TSA = 576 cm^{2}

h = 22 cm.

B = 24 cm^{2}

P = ?

TSA = 2B + Ph

576 = 2×24 + Px22

576 = 48 + 22P

22P = 576 -48 = 528

P = ^{528}/_{22}

P = 24 cm

## 1 thought on “Mensuration – Exercise 4.6.3 – Class IX”

Comments are closed.