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
Find the area, in square metres, of the trapezium whose bases and altitude is as under:
bases = 150 cm and 30 dm, altitude = 9 dm.
Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to the given base is 9 cm long.
Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm.
Find the height of a trapezium, the sum of the lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm2.
The area of a trapezium is 960 cm2. If the parallel sides are 34 cm and 46 cm, find the distance between them.
Top surface of a table is trapezium in shape. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium.
Find the area of the trapezium ABCD in which AB || DC, AB = 18 cm, ∠B = ∠C = 90°, CD = 12 cm and AD = 10 cm.
The area of a trapezium with equal non-parallel sides is 168 m2. If the lengths of the parallel sides are 36 m and 20 m, find the length of the non-parallel sides.
Find the area of the shaded portion in the following figure.
