Question

find the height of an isosceles triangle each of whose equal side measures 45 cm and the bases 72 CM also find its area

Answer 1

Answer in attachment:☺

Answer 2

The height of triangle is 27 cm and area is 972 sq.cm.

Step-by-step explanation:

We are given an isosceles triangle each of whose equal side measures 45 cm and the bases 72 CM

So, a = 45 cm

b = 45 cm

c = 72 cm

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

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

Substitute the values in the formula :$S=\frac{45+45+72}{2}=81$

Area of triangle = $\sqrt{81(81-72)(81-45)(81-45)}=972 sq.cm.$

Base of triangle = 72 cm

Area of triangle = $\frac{1}{2} \times base \times Height = \frac{1}{2} \times 72 \times Height$

So, $972 = \frac{1}{2} \times 72 \times Height \\\frac{972 \times 2}{72} = Height$

27 = Height

Hence The height of triangle is 27 cm and area is 972 sq.cm.

#Learn more:

Find the area of a triangle

sides- 9,40,41​

https://brainly.in/question/14834556