The Sides of a Triangle Are 11 M, 60 M and 61 M. the Altitude to the Smallest Side is - Mathematics

MCQ

The sides of a triangle are 11 m, 60 m and 61 m. The altitude to the smallest side is

• 11 m

• 66 m

•  50 m

• 60 m

Solution

The area of a triangle having sides aband s as semi-perimeter is given by,

A = sqrt(s(s-a)(s-b)(s-c)), where

s = (a+b+c)/2

We need to find the altitude to the smallest side

Therefore the area of a triangle having sides 11 m, 60 m and 61 m is given by

a = 11 m ; b = 60 m ; c = 61 m

s = (a+b+c)/2

s = (11+60+61)/2

s = 132/2

s = 66 m

A = sqrt(66(66-11)(66-60)(66-61))

A = sqrt(66(55)(6)(5))

A = sqrt(108900)

A = 330 m2

The area of a triangle having base AC and height is given by

" Area " (A) = 1/2 ("Base" xx "Height")

"Area " (A) = 1/2 (ACxx p )

We have to find the height corresponding to the smallest side of the triangle. Here smallest side is 11 m

AC = 11 m

330 = 1/2 (11 xx p)

330xx2=(11xxp)

` p = (330xx2)/11

p = 60 m

Concept: Application of Heron’s Formula in Finding Areas of Quadrilaterals
Is there an error in this question or solution?

APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 17 Heron’s Formula
Exercise 17.4 | Q 6 | Page 25

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