Area of the triangle with the sides a, b and c is
where s is half of the perimeter
Problem 1 :
Find the area of the triangle with a = 12 yards, b = 16 yards and c = 24 yards. Round to the nearest square yard.
Solution :
Finding half of the perimeter :
Finding area :
Area of the triangles is approximately 85 square yards.
Problem 2 :
Find the area of the triangle with a = 4 feet, b = 4 feet and c = 2 feet. Round to the nearest square yard.
Solution :
Finding half of the perimeter :
Finding area :
Area of the triangles is approximately 4 square feet.
Problem 3 :
The sides of a triangular field are 41 m, 40 m and 9 m. Find the number of rose beds that can be prepared in the field, if each rose bed, on an average needs 900 cm^{2} space.
Solution :
a = 41 m, b = 40 m and c = 9 m
Finding semi perimeter :
Finding area :
Area of the field = 180 square meter.
Each rose bed can occupy 900 cm^{2} space.
Converting in m^{2},
900 cm^{2 }= 900/10000
= 0.09 m^{2}
Number of rose beds can be prepared = 180/0.09
= 2000 rose beds.
Problem 4 :
Calculate the area of the shaded region given below.
Solution :
Area of the shaped region
= (area of triangles whose sides are 120 m, 122 m and 22 m)

(area of triangles whose sides are 22 m, 24 m and 26 m)
Area of large triangle :
Area of small triangle :
= 1320  246
Area of the shaded region = 1074 square meter.
Problem 5 :
Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of $7 per m^{2}.
Solution :
Area of triangular field :
Cost of laying grass in the triangular field = 7 x 900
= $6300
Problem 6 :
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m and 15 m. The advertisements yield an earning of $200 per m^{2} a year. A company hired one of its walls for 6 months. How much rent did it pay?
Solution :
Area of the wall in which advertisement to be placed
= √s(sa)(sb)(sc)
s = (a + b + c)/2
a = 13 m, b = 14 m and c = 15 m
s = (13 + 14 + 15)/2
= 42/2
s = 21
Area = √21(2113)(2114)(2115)
= √21 x 8 x 7 x 6
= √7056
= 84 m^{2}
Cost of advertisement per month = 200/12
= 16.66 per month per m^{2}
Rent for 6 month for 84 m^{2 }= 16.66 x 84
= $1394.54
Problem 7 :
Find the semi perimeter of a triangle having the length of its sides as 20 cm,15 cm and 9 cm.
Solution :
Semi perimeter o the scalene triangle = s
s = (a + b + c)/2
Here a = 20 cm, b = 15 cm and c = 9 cm
s = (20 + 15 + 9) / 2
= 44/2
= 22 cm
Problem 8 :
The edges of a triangular board are 6 cm, 8 cm and 10 cm. Find the cost of painting it at `0.09 per cm^{2}
Solution :
s = (a + b + c)/2
Here a = 6 cm, b = 8 cm and c = 10 cm
s = (6 + 8 + 10) / 2
= 24/2
= 12 cm
Area = √12(126)(128)(1210)
= √12 x 6 x 4 x 2
= √576
= 24 cm^{2}
Cost of painting = 0.09
Required cost = 0.09 x 24
= $2.16
Problem 9 :
The perimeter of a triangle is 54 cm and its sides are in the ratio 5:6:7. Find the area of the triangle.
Solution :
Perimeter of the triangle = 54 cm
From the given ratio, side lengths of the triangle are 5x, 6x and 7x
5x + 6x + 7x = 54
18x = 54
x = 3
Side lengths are 5(3), 6(3) and 7(3)
that is,
15 cm, 18 cm and 21 cm
s = (15 + 18 + 21) / 2
= 54/2
= 27 cm
Area = √27(2715)(2718)(2721)
= √27 x 12 x 9 x 6
= √17496
= 132.27 cm^{2}
Problem 10 :
Find the perimeter of an equilateral triangle whose area is equal to that of a triangle with sides 21 cm, 16 cm and 13 cm.
Solution :
Let us find the area of triangle whose side lengths are 21 cm, 16 cm and 13 cm.
a = 21 cm, b = 16 cm and c = 13 cm
s = (21 + 16 + 13) / 2
= 50/2
= 25 cm
Area = √25(2521)(2516)(2513)
= √25 x 4 x 9 x 12
= √10800
= 103.92 cm^{2}
Approximately = 104 cm^{2}
Area of equilateral triangle = 104 cm^{2}
(√3/4) a^{2} = 104
a^{2} = 104 (4/√3)
a^{2} = 104 (4/√3)
a^{2} = 240.18
a = 15.49
So, the required side length is 15.49 cm.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM