Perfect Math 10 ( 2074BS)

The lateral surface area of pyramid = 2 ¦PCD + 2 ¦PBC

=2× 1 × CD × PN + 2 × 1 × BC × PM
2 2

= 10 × 15 + 18 × 13

= 150 + 234 = 384cm2

Hence, the lateral surface area of the prism is 384cm2.

Example 8: P

Find the lateral surface area and volume of the given equilateral triangular- 24cm
based pyramid.
A
Solution: OM Mensuration
B 14 3cm C
Here, height of pyramid, PO = 24cm
Pyramid 97
As, AM = CM, and CM = 1 × 14 3 = 7 3 cm
2

From right-angled triangle BMC,

BC2 = BM2 + MC2

or, (14 3 )2 = BM2 + (7 3 )2
or, BM2 = 588 – 147 = 441
² BM = 21cm
Now, BO : OM = 2 : 1
BO = 14cm and OM = 7cm
From right-angled triangle POM, PM2 = PO2 + OM2

= 242 + 72
= 576 + 49
= 625
² PM = 25cm

Area of base ¦ABC = 3 × (14 3 )2 = 3 × 196 × 3
4 4

= 147 3 cm2

Volume of pyramid (V) = 1 Ah = 1 × 147 3 × 24
3 3

= 1176 3 cm3

Lateral surface area = 3¦PAC

=3× 1 × AC × PM
2

= 3 × 14 3 × 25
2

= 525 3 cm2

Hence, the lateral surface area of the pyramid is 525 3 cm2.