# The Sides of a Triangle Are 11 Cm, 15 Cm and 16 Cm. the Altitude to the Largest Side is - Mathematics

MCQ

The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is

#### Options

• $30\sqrt{7} cm$

• $\frac{15\sqrt{7}}{2}cm$

• $\frac{15\sqrt{7}}{4}cm$

• 30 cm

#### 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 corresponding to the longest side

Therefore the area of a triangle having sides 11 cm, 15 cm and 16 cm is given by

a = 11 m ; b = 15 cm ; c = 16 cm

s = (a+b+c)/2

s =(11+15+16)/2

s = 42/2

s = 21 cm

A = sqrt(21(21-11)(21-15)(21-6))

A = sqrt(21(10)(6)(5)

A = sqrt(6300)

A = 30 sqrt(7)   cm^2

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

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

"Area(A)" = 1/2 (AC xx p)

We have to find the height corresponding to the longest side of the triangle.Here longest side is 16 cm, that is AC=16 cm

30 sqrt(7) = 1/2 (16 xx p)

30 sqrt(7) xx 2 = (16 xx p)

 p = (30sqrt(7) xx 2) /16

p = (15 sqrt(7) )/4 cm

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 7 | Page 25

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