Raj is having a triangular open space in his plot. He divided the land into two parts by drawing the boundary. Where PQ||BC. P,Q are the midpoint of side AB, AC . Other measures are given as. Height of triangle is 12 cm, AP = 7 cm, BC = 10 cm, AC = 11cm
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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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