Advertisements
Advertisements
प्रश्न
Using Heron’s formula, find the area of a triangle whose sides are 1.8 m, 8 m, 8.2 m
Advertisements
उत्तर
Here a = 1.8 m, b = 8 m, c = 8.2 m
s = `("a" + "b" + "c")/2`
= `((1.8 + 8 + 8.2)"m")/2`
= `18/2`
= 9 m
s – a = 9 – 1.8 = 7.2 m
s – b = 9 – 8 = 1 m
s – c = 9 – 8.2 m = 0.8
Area of triangle
= `sqrt("s"("s" - "a")("s" - "b")("s" - "c"))`
= `sqrt(9 xx 7.2 xx 1 xx 0.8)`
= `sqrt(9 xx 5.76)`
= `sqrt9 xx sqrt(576/100)`
= `3 xx 24/10`
= 3 × 2.4
= 7.2 m2
∴ Area of the triangle = 7.2 sq.m
APPEARS IN
संबंधित प्रश्न
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122m, 22m, and 120m (see the given figure). The advertisements yield an earning of ₹ 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.
Some measures are given in the adjacent figure, find the area of ☐ABCD.
Find the area of the triangle formed by the points
(–10, –4), (–8, –1) and (–3, –5)
Find the area of triangle AGF
Find the area of the unshaded region
The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs 3 per m2 is Rs 918.
How much paper of each shade is needed to make a kite given in the following figure, in which ABCD is a square with diagonal 44 cm.
The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.
