Question

find the area of triangle two sides of which are 18 cm and 10cm and the perimeter is 42 cm

Answer 1

Given,
Side a=18cm
Side b=10cm
Perimeter=42cm=a+b+c
:. Putting value
42=18+10+c
42=28+c
42-28=c
14=c
Now,
S=(a+b+c)/2
:. Putting value
S=42/2
S=21
Now according to Heron's formula-
Area of a triangle
=√{s(s-a)(s-b)(s-c)}
:. Putting value
=√{21(21-18)(21-10)(21-14)}
=√{21(3)(11)(7)}
=√4851
=21√11cm²
(ÆÑẞWĒRẞ)

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Answer 2

Answer:

$=>21\sqrt{11}cm^2$

Step-by-step explanation:

As per the data given in the question, we have to find the area of the triangle for the two sides.

The sides of triangle given: a = $18$ cm, b = $10$ cm

Perimeter is $42$ cm

As we know that,

The perimeter of the triangle = ($a + b + c$)

=> $42 = 18 + 10 + c$

=> $42 = 28 + c$

=> $c = 42 - 28$

=> $c = 14 cm$

So, Semi Perimeter

s = ($a + b + c$)

$=>\frac{42}{2}=21cm$

Now, by using Heron's Formula,

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

$=\sqrt{21 \times 3 \times 11 \times 7}$

$=>21\sqrt{11}cm^2$

Hence, the area of a triangle is $21\sqrt{11}cm^2$