Question

The sides of a triangle are 51 cm 52 cm and 53 cm. Find:
Length of the perpendicular to the side of length 52 em from its opposite vertex.​

Answer 1

It seems minor mistake in the Question ;

Correct Question:

The sides of a triangle are 51 cm 52 cm and 53 cm. Find:Length of the perpendicular to the side of length 52 cm from its opposite vertex

Answer :

• Length of the perpendicular to the side of length 52 cm from its opposite vertex = 45

Given :

• Sides of the triangle are 51,52 and 53.

To find :

• Length of the perpendicular to the side of length 52 cm from its opposite vertex =?

Step-by-step explanation:

The given side of triangle are 51,52 and 53.

The semi-perimeter s of given triangle with sides a = 51 , b = 52 , c = 53 is given as

S = a + b + c/2

Substituting the values in the above formula, we get,

= 51 + 52 + 53 / 2

= 156/2

= 78

Now, using Heron's formula, the area Δ of given triangle is given as

Δ = √[s(s-a) (s-b) (s-c)]

= √[78 (78 - 51)(78 - 52) (78 - 53)]

= √ (78 x 27 x 26 x 25)

= √1,368,900

= 1170

If h is the length of perpendicular to the side c = 52 drawn from opposite vertex then the area of given triangle

Δ = 1/2(52)(h)

1170 = 1/2 (52)(h)

h = 1170 x 2 / 52

h = 45

Hence

Length of the perpendicular to the side of length 52 cm from its opposite vertex = 45

Answer 2

GIVEN :

• sides of triangle = 51 cm , 52 cm , 53 cm

TO FIND :

• Length of perpendicular on side of length 52 cm

SOLUTION :

Semi-perimeter of triangle

$\sf s = \frac{a + b + c}{2} \\ \\ \sf s = \frac{51 + 52 + 53}{2} \\ \\ \boxed{ \sf s = 78 }$

$\sf Area = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \sf Area = \sqrt{78(78 - 51)(78 - 52)(78 - 53)} \\ \\ \sf Area = \sqrt{78 \times 27 \times 26 \times 25} \\ \\ \sf \boxed{ \sf Area = 1170 \: {cm}^{2} }$

Using formula :

$\sf Area = \frac{1}{2} \times b \times h \\ \\ \sf 1170 = \frac{1}{2} \times 52 \times h \\ \\ \sf h = \frac{1170 \times 2}{52} \\ \\ \large \boxed{\sf \orange{h = 45 \: cm}}$

Length of perpendicular = 45 cm