Question

a triangle has sides 35cm 54cm and 61cm long. find its area. also find the smallest of its altitudes ( using heron's formula) please help !!

Answer 1

s = 35 + 54 + 61 ÷ 2 => 119.5
area = under root s× s-a ×s-b ×s- c
u could found the area by rhis method and the hight by 1/2 × 61 × h = area of triangle

Answer 2

Answer:

The area of the triangle is 939.14 cm.sq. and the smallest altitude is 30.79 cm.

Step-by-step explanation:

Given : A triangle has sides 35 cm, 54 cm and 61 cm long.

To find : The area of the triangle and the smallest of its altitudes?

Solution :

Using Heron's formula,

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Where,

$s=\frac{a+b+c}{2}$

a=35 ,b=54 , c=61

Substitute the value to find s,

$s=\frac{35+54+61}{2}$

$s=\frac{150}{2}$

$s=75$

Substitute in the area,

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

$A=\sqrt{75(40)(21)(14)}$

$A=\sqrt{882000}$

$A=939.14$

To have smallest altitude is with longest base.

So, taking base = 61 cm.

The area of the triangle is $A=\frac{1}{2}\times b\times h$

b is the base b=61 cm

h is the altitude

Area A=939.14 cm.sq.

Substitute in the formula,

$939.14=\frac{1}{2}\times 61\times h$

$h=\frac{939.14\times 2}{61}$

$h=30.79$

Therefore, The area of the triangle is 939.14 cm.sq. and the smallest altitude is 30.79 cm.