Advertisements
Advertisements
प्रश्न
Find the area of triangle AGF
Advertisements
उत्तर
Area of a triangle = `1/2[(x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3)]`
Area of ∆AGF = `1/2[(-2.5 - 13.5 - 6) - (-13.5 - 1 - 15)]`
= `1/2[-22 - (-29.5)]`

= `1/2[-22 + 29.5]`
= `1/2 xx 7.5`
= 3.75 sq.units
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?

Some measures are given in the adjacent figure, find the area of ☐ABCD.

Find the area of the triangle formed by the points
(1, – 1), (– 4, 6) and (– 3, – 5)
If the points A(– 3, 9), B(a, b) and C(4, – 5) are collinear and if a + b = 1, then find a and b
Find the area of an equilateral triangle whose perimeter is 180 cm
The semi-perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is
A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Find the area of the parallelogram.
The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. Find the area of the triangular field.
A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs 5 per m2. Find the cost of painting.
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.

