A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also,

1.	A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.
Answer» Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by: A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}{2}\)[Heron’s Formula] a = 35, b = 54, c = 61 s = \(\frac{a+b+c}{2}\) = \(\frac{35+54+61}{2}\) = 75 A = \(\sqrt{75(75-35)(75-54)(75-61)}\) A = \(\sqrt{75\times 40 \times21 \times 14}\) = 939.15 cm² Altitude on side 35 cm: Area of triangle = \(\frac{1}2\)(Base x Altitude) 939.15 = \(\frac{1}2\)(35 x Altitude) Altitude = 53.66 cm Altitude on side 54 cm: Area of triangle =\(\frac{1}2\)(Base x Altitude) 939.15 =\(\frac{1}2\)(54 x Altitude) Altitude = 34.78 cm Altitude on side 61 cm: Area of triangle =\(\frac{1}2\)(Base x Altitude) 939.15 =\(\frac{1}2\)(61 x Altitude) Altitude = 30.79 cm Therefore smallest Altitude is: 30.79 cm

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.

Answer»

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}{2}\)[Heron’s Formula]

a = 35, b = 54, c = 61

s = \(\frac{a+b+c}{2}\) = \(\frac{35+54+61}{2}\) = 75

A = \(\sqrt{75(75-35)(75-54)(75-61)}\)

A = \(\sqrt{75\times 40 \times21 \times 14}\) = 939.15 cm²

Altitude on side 35 cm:

Area of triangle = \(\frac{1}2\)(Base x Altitude)

939.15 = \(\frac{1}2\)(35 x Altitude)

Altitude = 53.66 cm

Altitude on side 54 cm:

Area of triangle =\(\frac{1}2\)(Base x Altitude)

939.15 =\(\frac{1}2\)(54 x Altitude)

Altitude = 34.78 cm

Altitude on side 61 cm:

Area of triangle =\(\frac{1}2\)(Base x Altitude)

939.15 =\(\frac{1}2\)(61 x Altitude)

Altitude = 30.79 cm

Therefore smallest Altitude is: 30.79 cm