Question

18. Raj is having a triangular open space in his plot. He divided the land into two parts by
drawing the boundary PQ, where PO is parallel to BC.
P is the mid-point of side AB. Other measures are given as:
Height of triangle is 12cm, AP=7cm, BC=10cm, AC=11cm. Find area of triangle ABC​

Answer 1

Given : A triangular open space in his plot.

He divided the land into two parts by drawing the boundary PQ, where PQ || BC.

P is the mid point of side AB

Height of triangle is 12 cm

AP = 7 cm, BC = 10 cm AC = 11 cm.

To Find :  

Area of ∆ ABC

Solution:

Height of triangle is 12 cm

Base BC = 10 cm

Area of ΔABC = (1/2) * base * height

= (1/2) * BC  * height

= (1/2) * 10 * 12

= 60 cm²

Area of ΔABC = 60 cm²

2}Length of side PQ = BC/2  = 10/2 = 5 cm

3}Length of AQ is =   AQ = AC/2  = 11/2 = 5.5 cm

4}Length of PB =  AP  = 7 cm

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