# The Sides of a Triangle Are 50 Cm, 78 Cm and 112 Cm. the Smallest Altitude is - Mathematics

MCQ

The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is

• 20 cm

• 30 cm

• 40 cm

• 50 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

Therefore the area of a triangle, say A having sides 50 cm, 78 cm and 112 cm is given by

a = 50 cm ; b = 78 cm ;  c = 112 cm

s = (50+78+112)/2

s = 240/2

s = 120 cm

A = sqrt(120 (120-50)(120-78)(120-112))

A = sqrt((120(70)(42)(8))

A = sqrt(2822400)

A = 1680 cm2

The area of a triangle, having p as the altitude will be,

$\text{ Area} = \frac{1}{2} \times \text{ base } \times \text{ height }$

Where, A = 1680  cm2

We have to find the smallest altitude, so will substitute the value of the base AC with the length of each side one by one and find the smallest altitude distance i.e. p

Case 1

AC = 50 cm

1680 = 1/2 (50 xx p)

1680 xx 2 = 50 xx p

p = (1680 xx 2)/50

p = 67.2   cm

Case 2

AC = 78  cm

1680 = 1/2 (78 xx p)

1680 xx 2 = 78 xx p

p = (1680xx2)/78

p = 43 cm

Case 3

Ac = 112 cm

1680 =1/2 (112 xx p)

1680 xx 2 = 112 xx p

p = (1680 xx 2 )/112

p = 30 cm

Is there an error in this question or solution?
Chapter 17: Heron’s Formula - Exercise 17.4 [Page 24]

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 17 Heron’s Formula
Exercise 17.4 | Q 5 | Page 24

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