Advertisements
Advertisements
प्रश्न
Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm ∠A = 90° and BC = CD = 52 cm.
Advertisements
उत्तर
The figure can be drawn as follows :
Here ABD is a right triangle. So the area will be :
ΔABD = `1/2(24) (32)`
= 384
Again
BD = `sqrt(24^2 + 32^2)`
= `8sqrt(3^2 + 4^2)`
= 8 ( 5 )
= 40
Now BCD is an isosceles triangle and BP is perpendicular to BD, therefore
DP =`1/2"BD"`
= `1/2(40)`
= 20
From the right triangle DPC we have
PC = sqrt(52^2 - 20^2)`
= `4sqrt(13^2 - 5^2)`
= 4(12)
= 48
So
ΔDPC =`1/2 (40) (48)`
= 960
Hence the area of the quadrilateral will be :
ΔABD + ΔDPC = 960 + 384
= 1344 cm2
APPEARS IN
संबंधित प्रश्न
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
The diagonals of a quadrilateral are 16 cm and 13 cm. If they intersect each other at right angles; find the area of the quadrilateral.
Trapezium given below; find its area.
The perimeter of a rhombus is 46 cm. If the height of the rhombus is 8 cm; find its area.
The rate for a 1.20 m wide carpet is Rs. 40 per meter; find the cost of covering a hall 45 m long and 32 m wide with this carpet. Also, find the cost of carpeting the same hall if the carpet, 80 wide, is at Rs. 25. Per meter.
A triangle and a parallelogram have the same base and the same area. If the side of the triangle is 26 cm, 28 cm, and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
In the following, find the value of ‘a’ for which the given points are collinear
(2, 3), (4, a) and (6, – 3)
Find the area of the quadrilateral whose vertices are at (– 9, – 2), (– 8, – 4), (2, 2) and (1, – 3)
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (– 4, – 2), (– 3, k), (3, – 2) and (2, 3)
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC. Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
