Mensuration – Exercise 4.6.3 – Class IX

 

 

2. 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 cm²

TSA = 2B + Ph = 2a² +  Ph = 2(2.5)² + 65 = 77.5cm²

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3. 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 = 234m²

 

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)m²

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4. 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.

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5. The TSA of a triangular based prism is 570 cm² and its height is 22 cm. if the area of its base is 24 cm² . Find the perimeter of the base.

Solution:

TSA = 576 cm²

h = 22 cm.

B = 24 cm²

P = ?

TSA = 2B + Ph

576 = 2×24 + Px22

576 = 48 + 22P

22P = 576 -48 = 528

P = ⁵²⁸/₂₂

P = 24 cm

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