Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
DC = 6.4 m
AD = 80 cm = `80/100` = 0.8 m
BC = 2.4 m
Area = ?
Area of the cross-section = Area of trapezium ABCD
= `1/2 ("Sum of parallel sides") xx "height"`
= `1/2 (0.8 + 2.4) xx 6.4`
= `1/2 xx 3.2 xx 6.4`
= 10.24 m2
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.
Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m2 and whose height is 280 cm.
Find the area of Fig. as the sum of the areas of two trapezium and a rectangle.
The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom and the area of cross-section is 72 m2 determine its depth.
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 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.
There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some another way of finding its areas?
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 |
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
