Advertisements
Advertisements
प्रश्न
Trapezium given below; find its area.
Advertisements
उत्तर
In ΔCMB
CB2 = CM2 + MB2
(10)2 = CM2 + (6)2
100 - 36 = CM2
`sqrt64` = CM
CM = 8 cm
Area of ABCD = `1/2 xx ( 8 + 14 ) xx 8`
= 22 × 4
= 88 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.
Find the area of a quadrilateral one of whose diagonals is 30 cm long and the perpendiculars from the other two vertices are 19 cm and 11 cm respectively.
The area of a rectangular is 640 m2. Taking its length as x cm; find in terms of x, the width of the rectangle. If the perimeter of the rectangle is 104 m; find its dimensions.
Trapezium given below; find its area.
The length and the breadth of a rectangle are 6 cm and 4 cm respectively. Find the height of a triangle whose base is 6 cm and the area is 3 times that of the rectangle.
The width of a rectangular room is `4/7`of its length, x, and its perimeter is y. Write an equation connecting x and y. Find the length of the room when the perimeter is 4400 cm.
The length of a rectangular verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.
(i) Taking x as the breadth of the verandah, write an equation in x that represents the above statement
(ii) Solve the equation obtained in (i) above and hence find the dimensions of the verandah.
Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of the two squares.
A quadrilateral field of unequal has a longer diagonal with 140m. The perpendiculars from opposite vertives upon this diagonal are 20m and 14m. Find the area of the field.
In the following, find the value of ‘a’ for which the given points are collinear
(a, 2 – 2a), (– a + 1, 2a) and (– 4 – a, 6 – 2a)
