Question

Find the area of the triangle whose sides are 42 CM 34 cm and 20 CM in length and find the height corresponding to the longest side

Answer 1

Answer: h=16cm

Step-by-step explanation:

sides of the triangle are A=42, B=34 , C=20

according to hero's formula, area of the triangle A=undrroot of S(S-A) (S-B) (S-C)

where S= A+B+C/2 = 42+34+20/2=48

now,

A=underroot of  48(48-42) (48-34) (48-20)

A=unedrroot of 48(6) (14) (28)

A=unedrroot of (6 x 8) (6) (14) (14 x 2)

A=14 x 6 x 4

A=336 cm2

let the height corresponding to longest side (42cm) is H

area = 1/2 x base x height

1/2 x 42 x H = 336

H=336 x 2/42 = 16cm

H=16CM

Answer 2

Given:-

• Sides of the Triangle = 42cm, 34cm and 20cm.

To Find:-

• Length of the Height corresponding to the longest side.

Formula Used:-

• ${\boxed{\bf{Heron's Formula : Area= \sqrt{S(S-a)(S-b)(S-c)}}}}$

• ${\boxed{\bf{Area\:of\: Triangle= \dfrac{1}{2}\times base \times height}}}$

Solution:-

For Area of Triangle:-

Using Heron's Formula,

Firstly,

$\sf :\implies\:S =\dfrac{a+b+c}{2}$

Here,

• S = Semi Perimeter

• a = 42 cm

• b = 34 cm

• c = 20 cm

Putting values,

$\sf :\implies\:S =\dfrac{42+34+20}{2}$

$\sf :\implies\:S =\dfrac{96}{2}$

$\sf :\implies\:S =48\:cm$

Now,

$\sf :\implies\:Area= \sqrt{S(S-a)(S-b)(S-c)}$

Putting values,

$\sf :\implies\:Area= \sqrt{48(48-42)(48-34)(48-20)}$

$\sf :\implies\:Area= \sqrt{48\times 6\times 14\times 28}$

$\sf :\implies\:Area= \sqrt{1,12,896}$

$\sf :\implies\:Area= 336\:cm^2$

For Height corresponding to the longest side:-

• Longest Side = 42cm

$\sf :\implies\:Area\:of\: Triangle= \dfrac{1}{2}\times base \times height$

$\sf :\implies\:336= \dfrac{1}{2}\times 42 \times height$

$\sf :\implies\:336= 21 \times height$

$\sf :\implies\: Height= \dfrac{336}{21}$

$\sf :\implies\: Height= 16\:cm$

Hence, The Length of the Height Corresponding to the longest side is 16cm.

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