Question

Find the area of the triangle whose side are 42cm, 34cm, and 20cm in length. Hence, find the height corresponding to the longest side ​

Answer 1

Given:-

•Sides of a triangle are 42cm,34cm and 20cm

To find:-

•Area of the triangle

•Height corresponding to the longest side

Solution:-

Let a=42cm,b=34cm,c=20cm

=>Perimeter of the triangle=(42+34+20)cm

=>96cm

=>s=1/2(a+b+c)

=>s=1/2(96)

=>s=48cm

=>(s-a)=48-42=6cm

=>(s-b)=48-34=14cm

=>(s-c)=48-20=28cm

By,Heron's formula,the area of the given triangle is----

$= > \sqrt{s(s - a)(s - b)(s - c)}$

$= > \sqrt{48 \times 6 \times 14 \times 28}$

$= > \sqrt{112896}$

$= > 336 {cm}^{2}$

The longest of the triangle is 42cm and let the corresponding height =h cm

Then,area=(1/2×base×height)

=>(1/2×42×h)=336

=>h=336/21

=>h=16cm

Hence,the height corresponding to the longest side is 16cm.

Answer 2

Answer:

Semiperimeter (s) = (42 + 34 + 20 ) / 2

So it is = 96 / 2 = 48 cm.

Using herons formula = (√s )(√s-a )(√s-b) (√s-c)

By substituting s and a , b , c i.e the given sides.

We have √48 × √6 × √14 × √28

So √ (4 × 6 × 2) × √6 × √(7 × 2) × √(7 × 4)

Simplifying the numbers we have area = 4 × 6 × 2 × 7 = 336 cm ²

Hence area is 336 cm²

Now height corresponding to longest side implies that base is 42 cm and area remains same

So area of triangle is 1/ 2 b × h

336 = 1/ 2 × 42 × H

HENCE H = 16 cm

So height is 16 cm .