Advertisements
Advertisements
प्रश्न
The perimeter of a triangle is 300 m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle ?
Advertisements
उत्तर
Given that
The perimeter of a triangle = 300 m
The sides of a triangle in the ratio 3 : 5 : 7
Let 3x, 5x, 7x be the sides of the triangle
Perimeter ⇒ 2s = a + b + c
⇒ 3x + 5x + 17x = 300
⇒ 15x = 300
⇒ x = 20m
The triangle sides are a = 3x
= 3 (20)m = 60 m
b = 5x = 5(20) m = 100m
c = 7x = 140 m
semi perimeter s = `(a+b+c+)/2`
`=(300)/2m`
`=150m`
∴The area of the triangle `=sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(150(150-60)(150-100)(150-140))`
`=sqrt(150xx10xx90xx50)`
`=sqrt(1500xx1500) 3 cm^2`
`∴Δ` le Area = 1500 `sqrt3 cm^2`
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?

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.
Find the area of triangle FED
If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is
Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0
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.
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.
A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 4 to plough 1 m2 of the field, find the total cost of ploughing the field.
In the following figure, ∆ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ∆ABC is constructed. Find the height DF of the parallelogram.

