Advertisements
Advertisements
प्रश्न
Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Advertisements
उत्तर
AB = DE
EC = BC − BE
EC = (48 − 40) m
EC = 8 m
In Δ DEC,
(DE)2 + (EC)2 = (D C)2
(DE)2 + (8)2 = (17)2
(DE)2 = (17)2 − (8)2
DE = 289 − 64
DE = 225 m2
DE = `sqrt225` m
DE = 15 m
AB = 15 m
Area of trapezium, ABCD
`= 1/2` × (sum of parallel sides) × Height
`= 1/2` × (AD + BC) × AB
`1/2` × (40 + 48) ×15
= 660m2
APPEARS IN
संबंधित प्रश्न
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 sum of the lengths of the bases of a trapezium whose area is 4.2 m2 and whose height is 280 cm.
Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Find the area of the field shown in Fig. 20.39 by dividing it into a square, a rectangle and a trapezium.
Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm.
☐ PQRS is an isosceles trapezium l(PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm, Distance between two parallel sides is 4 cm, find the area of ☐ PQRS.

Find the missing values.
| Height 'h' | Parallel side 'a` | Parallel side 'b` | Area |
| 19 m | 16 m | 323 sq.m |
Find the missing values.
| Height 'h' | Parallel side 'a` | Parallel side 'b` | Area |
| 16 cm | 15 cm | 360 sq.cm |
Arivu has a land ABCD with the measurements given in the figure. If a portion ABED is used for cultivation (where E is the mid-point of DC), find the cultivated area.

The areas of two circles are in the ratio 49 : 64. Find the ratio of their circumferences.
