Advertisements
Advertisements
प्रश्न
Find the area, in square metres, of the trapezium whose bases and altitude is as under:
bases = 12 dm and 20 dm, altitude = 10 dm
Advertisements
उत्तर
Given:
Bases:
\[12 dm =\frac{12}{10}m = 1.2 m\]
And, 20 dm \[=\frac{20}{10}m=2 m\]
Altitude = 10 dm \[=\frac{10}{10}m = 1 m\]
Area of trapezium \[=\frac{1}{2}\] x (Sum of the bases) x (Altitude)
\[ = \frac{1}{2} \times (1 . 2 + 2) m \times (1) m\]
\[ = 1 . 6 \times m\times m\]
\[ {=1.6 m}^2 \]
संबंधित प्रश्न
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 area of a trapezium is 91 cm2 and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.
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.
In Fig. 20.38, a parallelogram is drawn in a trapezium, the area of the parallelogram is 80 cm2, find the area of the trapezium.
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.
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 |
| 16 cm | 15 cm | 360 sq.cm |
A window is in the form of trapezium whose parallel sides are 105 cm and 50 cm respectively and the distance between the parallel sides is 60 cm. Find the cost of the glass used to cover the window at the rate of ₹ 15 per 100 sq.cm
The table top is in the shape of a trapezium with measurements given in the figure. Find the cost of the glass used to cover the table at the rate of ₹ 6 per 10 sq.cm
