Advertisements
Advertisements
प्रश्न
Find the area of the following fields. All dimensions are in metres.
Advertisements
उत्तर
Area of the given figure
= Area of ΔDCF + Area of ΔEGD + Area of trapezium FCBH + Area of ΔEGA + Area of ΔAHB
∴ Area of ΔDCF = `1/2` × Base × Height
= `1/2 xx 100 xx 100`
= `10000/2`
= 5000 m2
Area of ΔEGD = `1/2` × Base × Height
= `1/2 xx 120 xx 180`
= 60 × 180
= 10800 m2
Area of trapezium = `1/2` × [Sum of parallel sides] × Height
= `1/2 xx [100 + 50] xx 110`
= `1/2 xx 150 xx 110`
= 75 × 110
= 8250 m2
Area of ΔEGA = `1/2` × Base × Height
= `1/2 xx 120 xx 80`
= 60 × 80
= 4800 m2
Area of ΔAHB = `1/2` × Base × Height
= `1/2 xx 50 xx 50`
= 25 × 50
= 1250 m2
Thus, the area of the complete figure
= 5000 + 10800 + 8250 + 4800 + 1250
= 30100 m2
APPEARS IN
संबंधित प्रश्न
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.
Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as
the sum of the areas of two triangles and one rectangle.
Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as the difference of the area of a rectangle and the sum of the areas of two triangles.
If the area of a trapezium is 28 cm2 and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.
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:
The following figure shows the cross-section ABCD of a swimming pool which is a trapezium in shape.
If the width DC, of the swimming pool, is 6.4 m, depth (AD) at the shallow end is 80 cm and depth (BC) at the deepest end is 2.4 m, find its area of the cross-section.
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
A trapezium with 3 equal sides and one side double the equal side can be divided into ______ equilateral triangles of ______ area.
The area of a trapezium become 4 times if its height gets doubled.
