Heron’s Formula (Part -2)

1. Find the area of triangle two sides of which 9 cm and 5 cm and the perimeter is 21 cm.

Sol. Given a = 9 cm b = 5 cm c = ? P = 21 cm

P = a + b + c

21 = 9 + 5 + c

21 = 14 + c

C = 21 – 14

C = 7 cm

S = (a + b + c) / 2

S = ( 9 + 5 + 7 ) / 2

S = 21 / 2

By Heron’s Formula

Area = √ s ( s – a ) ( s – b ) ( s – c )

Area = √ ( 21 / 2 ) ( 21 / 2 – 9 ) ( 21 / 2 – 5 ) ( 21 / 2 – 7 )

Area = 1 / 2 * 1 / 2 √ 21 ( 21 – 18 ) ( 21 – 10 ) ( 21 – 14 )

Area = 1 / 4 √ 21 * 3 * 11 * 7

Area = ¼ √ 7 * 3 * 3 * 11 * 7

Area = ( 21 / 4 ) * √11

Area = ( 21 * 3.317 ) / 4

Area = 17.41 cm^2

2. Find the area of triangle two sides of which 8 cm and 6 cm and the perimeter is 22 cm.

Sol. Given a = 8 cm b = 6 cm c = ? P = 22 cm

P = a + b + c

22 = 8 + 6 + c

22 = 14 + c

C = 22 – 14

C = 8 cm

S = (a + b + c) / 2

S = ( 8 + 6 + 8 ) / 2

S = 22 / 2 => S = 11

By Heron’s Formula

Area = √ s ( s – a ) ( s – b ) ( s – c )

Area = √ 11 ( 11 – 8 ) ( 11 – 6 ) ( 11 – 8 )

Area = √ 11 * 3 * 5 * 3

Area = 3 √ 55

Area = 7.42 * 3

Area = 22.25 cm^2

3. Side of a triangle are in the ratio of 18 : 10 : 14 and its perimeter is 210 cm . Find its area .

Sol. Given P = 210 cm

Let the sides are 18 x , 10 x , and 14 x

So , 18 x + 10 x + 14 x = 210

42 x = 210

X = 210 / 42

X = 5

So the sides are 90 cm , 50 cm and 70 cm

S = (a + b + c) / 2

S = ( 90 + 50 + 70 ) / 2

S = 210 / 2 => S = 105

By Heron’s Formula

Area = √ s ( s – a ) ( s – b ) ( s – c )

Area = √ 105 ( 105 – 90 ) ( 105 – 50 ) ( 105 – 70 )

Area = √ 105 * 15 * 55 * 35

Area = √ 3 * 5 * 7 * 3 * 5 * 11 * 5 * 5 * 7

Area = 3 * 5 * 5 * 7 √ 11

Area = 525 * 3.317

Area = 1741.23 cm^2

maths ncert solutions

4. Side of a triangle are in the ratio of 9 : 5 : 7 and its perimeter is 210 cm . Find its area .

Sol. Given P = 210 cm

Let the sides are 9 x , 5 x , and 7 x

So , 9 x + 5 x + 7 x = 210

21 x = 210

X = 210 / 21

X = 10

So the sides are 90 cm , 50 cm and 70 cm

S = (a + b + c) / 2

S = ( 90 + 50 + 70 ) / 2

S = 210 / 2 => S = 105

By Heron’s Formula

Area = √ s ( s – a ) ( s – b ) ( s – c )

Area = √ 105 ( 105 – 90 ) ( 105 – 50 ) ( 105 – 70 )

Area = √ 105 * 15 * 55 * 35

Area = √ 3 * 5 * 7 * 3 * 5 * 11 * 5 * 5 * 7

Area = 3 * 5 * 5 * 7 √ 11

Area = 525 * 3.317

Area = 1741.23 cm^2

5. An isosceles triangle has perimeter 25 cm and each of the equal side is 8 cm . Find the area of triangle.

Sol. Given P = 25 cm

Two equal sides of triangle = a = b = 8 cm

We know that

P = a + b + c

25 = 8 + 8 + c

25 = 16 + c

C = 25 – 16

C = 9 cm

S = (a + b + c) / 2

S = ( 8 + 8 + 9 ) / 2

S = 25 / 2

By Heron’s Formula

Area = √ s ( s – a ) ( s – b ) ( s – c )

Area = √ ( 25 / 2 ) ( 25 / 2 – 8 ) (25 / 2 – 8 ) ( 25 / 2 – 9 )

Area = ( 1 / 2 * 1 / 2 ) √ 25 ( 25 – 16 ) ( 25 - 16 ) ( 25 - 18 )

Area = 1 / 4 √ 25 * 9 * 9 * 7

Area = ( 3 * 3 * 5 ) / 4 √ 7

Area = ( 45 * 2.65 ) / 4

Area = 29.765 cm^2

6. An isosceles triangle has perimeter 25 cm and each of the equal side is 9 cm . Find the area of triangle.

Sol. Given P = 25 cm

Two equal sides of triangle = a = b = 9 cm

We know that

P = a + b + c

25 = 9 + 9 + c

25 = 18 + c

C = 25 – 18

C = 7 cm

S = (a + b + c) / 2

S = ( 9 + 9 + 7 ) / 2

S = 25 / 2

By Heron’s Formula

Area = √ s ( s – a ) ( s – b ) ( s – c )

Area = √ ( 25 / 2 ) ( 25 / 2 – 9 ) (25 / 2 – 9 ) ( 25 / 2 – 7 )

Area = ( 1 / 2 * 1 / 2 ) √ 25 ( 25 – 18 ) ( 25 - 18 ) ( 25 - 14 )

Area = 1 / 4 √ 25 * 7 * 7 * 11

Area = ( 5 * 7 ) / 4 √ 11

Area = ( 35 * 3.317 ) / 4

Area = 29 cm^2

Milan Tomic

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