If a, b and c are the sides of a triangle, then
the area of a triangle = √s(s − a)(s − b)(s − c) sq.units.
where s = (a + b + c)/2
‘s’ is the semi-perimeter (that is half of the perimeter) of the triangle.
Question 1 :
The perimeter of a triangular plot is 600 m. If the sides are in the ratio 5 : 12 : 13, then find the area of the plot.
Solution :
The sides of triangle be 5x, 12x and 13x.
Perimeter of triangular plot = 600
5x + 12x + 13x = 600
30x = 600
x = 600/30 = 20
5x = 5(20) = 100
12x = 12(20) = 240
13x = 13(20) = 260
s = 600/2 = 300
s - a = 300 - 100 = 200
s - b = 300 - 240 = 60
s - c = 300 - 260 = 40
√s(s − a)(s − b)(s − c) = √300(200)(60)(40)
= √(3 ⋅ 10 ⋅ 10 ⋅ 2 ⋅ 10 ⋅ 10 ⋅ 3 ⋅ 2 ⋅ 10 ⋅ 2 ⋅ 2 ⋅10)
= 12000 m2
Question 2 :
Find the area of an equilateral triangle whose perimeter is 180 cm.
Solution :
Perimeter of equilateral triangle = 180 cm
3a = 180
a = 180/3 = 60 cm
Area of equilateral triangle = (√3/4)a2
= (√3/4)602
= 900√3
= 900(1.732)
= 1558.8 cm2
Question 3 :
An advertisement board is in the form of an isosceles triangle with perimeter 36 m and each of the equal sides are 13 m. Find the cost of painting it at ₹ 17.50 per square metre.
Solution :
Let "x" be the unknown side of the triangle.
Then sides of triangle are x, 13 and 13.
perimeter = 36 m
x + 13 + 13 = 36
x + 26 = 36
x = 36 - 26 = 10 m
a = 10, b = 13 and c = 13
height = √132 - 52
= √(169 - 25)
= √144
height = 12 cm
Area of triangle = (1/2) x base x height
= (1/2) x 10 x 12
= 60 cm2
Cost of of painting it at ₹ 17.50 per square metre
Required cost = 60 (17.50)
= ₹ 1050
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