Given
ABC be a triangle in which sides AB = 35 cm, BC = 54 cm, CA = 61 cm
Find out
We have to determine the a;altitude of the triangle
Solution
AreaofΔABC=√s(s−a)(s−b)(s−c)
where s is the perimeter
s = (35+54+61)/2 = 75 cm.
=√75(75−35)(75−54)(75−61)
=√75×40×21×14
=√25×3×4×2×5×7×3×7×2
=5×2×2×3×7√5=420√5cm2
Also we know that
AreaofΔABC=12×AB×Altitude
⇒12×35×CD=420√5
⇒CD=420×2√535
∴CD=24√5
Hence, the length of altitude is 24√5cm