Question

During van mahotsav some children planted trees in a triangular region two sides of which are 18 m and 10 m and the perimeter is 42 find the area of the region

Answer 1

Let a, b, c be the sides of a triangular region.
Here, a= 18 cm , b= 10 cm
Perimeter= 42 cm (given)
Third side (c)= 42 -(18+10)
c = 42- 28 = 14 cm
Semi perimeter (s) =( a+b+c)/2
s= 42/2= 21 cm
s-a = 21-18= 3 cm………. (1)
s-b = 21 - 10= 11 cm………(2)
s-c= 21 - 14 = 7 cm…………(3)
We have to find the area of a triangular region by Heron's formula because in this triangle height is not given only three sides are given.
Heron's formula is very useful where it is not possible to find the height of the triangle.
Area of ∆ = √s(s-a)(s-b)(s-c)
Area of ∆ = √ 21 × 3 × 11 × 7 (from eq 1,2,3)
Area of ∆ = √ (3×7) × 3 × 11 × 7
Area of ∆ = √ (3×3)(7×7)(11)
Area of ∆ = 3×7√11= 21√11 cm²
Hence , the area of the triangular region = 21√11 cm².

HOPE THIS WILL HELP YOU....

Answer 2

Answer :

The area of the region is 21 √ 11 m²

Step-by-step explanation :

Let the sides of triangle be a, b, and c.

Here, a = 18, b = 10

And,

Perimeter = 42 m

⇒ a + b + c = 42

⇒ 18 + 10 + c = 42

⇒ 28 + c = 42

⇒ c = 42 - 28

⇒ c = 14 m

S = a + b + c / 2

S = 42 / 2

S = 21 m

Area of ΔABC = √s ( s - a )( s - b )( s - c )

⇒ √21 ( 21 - 18 ) ( 21 - 10 ) ( 21 - 14 )

⇒ √21 × 3 × 11 × 7

⇒ √7 × 7 × 3 × 3 × 11

⇒ 7 × 3 × √ 11

⇒ 21 √ 11 m²

Area of ΔABC = 21 √ 11 m²

Hence, the area of ΔABC is 21 √ 11 m².