Advertisements
Advertisements
प्रश्न
The sunshade of a window is in the form of isosceles trapezium whose parallel sides are 81 cm and 64 cm and the distance between them is 6 cm. Find the cost of painting the surface at the rate of ₹ 2 per sq.cm
Advertisements
उत्तर
Given the parallel sides a = 81 cm, b = 64 cm
Distance between ‘a’ and ‘b’ is height h = 6 cm
Area of the trapezium = `1/2` × h(a + b) sq.units
= `1/2` × 6 × (81 + 64)
= 3 × 145 cm2
= 435 cm2
Cost of painting 1 cm2 = ₹ 2
Cost of painting 435 cm2 = ₹ 435 × 2
= ₹ 870
Cost of painting = ₹ 870.
APPEARS IN
संबंधित प्रश्न
The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Find the area, in square metres, of the trapezium whose bases and altitude is as under:
bases = 28 cm and 3 dm, altitude = 25 cm
The area of a trapezium is 1586 cm2 and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.
Find the area enclosed by each of the following figures [Fig. 20.49 (i)-(iii)] as the sum of the areas of a rectangle and a trapezium:
Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.
Find the area of the trapezium ABCD in which AB || DC, AB = 18 cm, ∠B = ∠C = 90°, CD = 12 cm and AD = 10 cm.
Find the missing values.
| Height 'h' | Parallel side 'a` | Parallel side 'b` | Area |
| 10 m | 12 m | 20 m |
The area of the trapezium, if the parallel sides are measuring 8 cm and 10 cm and the height 5 cm is
When the non-parallel sides of a trapezium are equal then it is known as
