Advertisements
Advertisements
प्रश्न
ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (see the given figure). The height AD from A to BC, is 6 cm. Find the area of ΔABC. What will be the height from C to AB i.e., CE?
Advertisements
उत्तर
Area of triangle ABC = `1/2 xx "Base" xx "Height"`
= `1/2 xx BC xx AD`
= `1/2 xx 9 xx 6`
= 27 cm2
Area of triangle ABC = `1/2 xx "Base" xx "Height"`
= `1/2 xx AB xx CE`
`27 = 1/2 xx 7.5 xx CE`
CE = `(27 xx 2)/7.5`
CE = 7.2 cm
APPEARS IN
संबंधित प्रश्न
If A(4, –6), B(3, –2) and C(5, 2) are the vertices of ∆ABC, then verify the fact that a median of a triangle ABC divides it into two triangle of equal areas.
Find the area of a triangle whose vertices are
`(at_1^2,2at_1),(at_2^2,2at_2)` and `(at_3^2,2at_3)`
Prove that the points (a, 0), (0, b) and (1, 1) are collinear if `1/a+1/b=1`
Find the area of ΔABC whose vertices are:
A(10,-6) , B (2,5) and C(-1,-3)
Find the value of k so that the area of the triangle with vertices A (k+1, 1), B(4, -3) and C(7, -k) is 6 square units
Find the area of the triangle whose vertices are (-2, 6), (3, -6), and (1, 5).
The area of a trapezium is 475 cm2 and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.
Find the missing value:
| Base | Height | Area of parallelogram |
| ______ | 8.4 cm | 48.72 cm2 |
In the following figure, if PR = 12 cm, QR = 6 cm and PL = 8 cm, then QM is ______.
Area of a triangle = `1/2` base × ______.
