Login

please answer:

the sides of a triangle are 120m,170m and 250m. Find the length of the altitude of the triangle corresponding to the longest side.

$G i v e n : T h e s i d e s o f a Δ a r e 120 m, 170 m & 250 m s = \frac{120 + 170 + 250}{2} = \frac{540}{2} = 270 N o w, A r e a o f Δ = \sqrt{s (s - a) (s - b) (s - c)} [u \sin g h e r o n f o r m u l a] = \sqrt{270 (270 - 120) (270 - 170) (270 - 250)} = \sqrt{270 \times 150 \times 100 \times 20} = \sqrt{81 \times 1000000} = 9000 m^{2} ................ (i) A l s o, A r e a o f Δ = \frac{1}{2} \times b \times h, w h e r e b = b a s e & h = c o r r e s p o n d i n g a l t i t u d e 9000 = \frac{1}{2} \times 250 \times h [u \sin g (i)] \Rightarrow h = \frac{9000 \times 2}{250} = 72 m$

5

View Full Answer

Copyright © 2024 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?