**Find the LSA and TSA of the following pyramid.**

**a) Square based pyramid with base 6 cm. and slant height 14 cm.**

**b) Triangular based pyramid with base 12cm, slant height 20 cm.**

Solution:

- a) Square based pyramid with base 6 cm. and slant height 14 cm.

A = 6; L = 14

LSA = ^{1}/_{2} pL

L = ^{1}/_{2} x 4axL

= ^{1}/_{2} x 4 x 6 x 14 = 168m^{2}

TSA = B + LSA = B + ^{1}/_{2} p

= a^{2} + 168 = 6^{2} + 168 = 36 + 168 = 204 cm^{2}

(b) Triangular based pyramid with base 12cm, slant height 20 cm.

LSA = ^{1}/_{2} pxL

L = ^{1}/_{2} x 3 x axL = ^{1}/_{2} x 3 x 12 x 20 = 3 x 12 x 10 = 360cm^{2}

TSA = B + ^{1}/_{2} p = ^{√3}/_{4} x 12 x 12 + 360

= 36√3 + 360

= 36(√3+10) cm^{2}

**Find the difference between the lateral and TSA of a square based pyramid with side 2.5 cm and slant height 8 cm.**

Solution:

LSA = ?

TSA = ?

a = 2.5cm

L = 8cm

LSA = ^{1}/_{2} PL

L = ^{1}/_{2} x 4axL =^{1}/_{2} x 4 x 2.5 x 8

=^{1}/_{2} x 4 x 2.5 x 8

L= 40 cm^{2}

TSA = B + LSA = B + ^{1}/_{2} P

= a^{2} + 40

= (2.5)^{2} + 40

= 46.25cm^{2}

Difference between TSA and LSA = TSA – LSA

=46.25 – 40 = 6.25cm^{2}

**Find the perimeter of a triangular based pyramid with slant height 18 m. and LSA = 360 m**^{2}

Solution:

LSA = ^{1}/_{2} pL

L = ^{1}/_{2} x p x L

360 = ^{1}/_{2} x p x 18

p = ^{360×2}/_{18} = 40m

**The base and height of a square based pyramid is 10 cm, and 12 cm. respectively. Find its TSA.**

Solution:

a = 10 cm

h = 12 cm

TSA = ?

L^{2} = H^{2} + (^{A}/_{2})^{2}

L^{2} = 144 + 25 = 169

L = 13cm

TSA = B + ^{1}/_{2} PL

= a^{2} + ^{1}/_{2} PL

=(10)^{2} + ^{1}/_{2}x 4^{2} x 10 x 13

= 100 + 260

= 360cm^{2}

**The roof of a temple is in the shape of square based pyramid. The edge of the base is 7.5 m. and the slant height is 7 m. find the area of metal sheet required to cover the top of roof.**

Solution:

a = 7.5 cm

L = 7m

LSA = ?

LSA = ^{1}/_{2} PL = ^{1}/_{2} x 4x a x L

= ^{1}/_{2} x 4^{2} x 7.5 x 7 = 14 x 7.5 = 105 m^{2}

**A square based pyramid is placed on a square based prism with the same base 14 cm. if the height of the prism is 9 cm. and height of the pyramid is 7 cm. find the TSA of the solid obtained by joining them.**

Solution:

a = 14 cm

height of prism = 9cm

Height of pyramid = 7cm

L^{2} = h^{2} + a^{2}

= 49 + 72

= 49 + 49

= 98

L = √98 = 7√2 cm

TSA of solid = LSA of prism + LSA of pyramid

= Ph + ^{1}/_{2} PL

= 4ah + ^{1}/_{2} 4a x L

= 4 x 14 x9 + ^{1}/_{2} x 4 x 14 x 7√2

= 504 + 28 x 7√2

= 504 + 196√2

= 14(36 + 14√2)

= 28(18 + 7√2) cm

## 1 thought on “Mensuration – Exercise 4.6.4 – Class IX”

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