Advertisements
Advertisements
प्रश्न
☐ 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.
Advertisements
उत्तर
Draw a perpendicular from Q to line MR. Where it meets the line MR, name it point N.
MN = PQ = 7 cm
In ΔPMS,
PM² + SM² = PS²
⇒ 4² + 3² = PS²
⇒ PS² = 16 + 9
⇒ PS² = 25
⇒ PS = 5cm
PQRS is an isosceles trapezium so, PS = QR = 5 cm
PM = QN = 4 cm
So, NR = SM = 3 cm
SR = SM + MN + NR
= 3 + 7 + 3
= 13 cm
Area of trapezium PQRS = `1/2 xx` (sum of parallel sides) × height
=`1/2 xx (7 + 13) xx 4`
= 40 cm²
संबंधित 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
Find the area, in square metres, of the trapezium whose bases and altitude is as under:
bases = 28 cm and 3 dm, altitude = 25 cm
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 height of a trapezium, the sum of the lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm2.
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?
The area of a trapezium is 279 sq.cm and the distance between its two parallel sides is 18 cm. If one of its parallel sides is longer than the other side by 5 cm, find the lengths of its parallel sides.
Find the missing values.
| Height 'h' | Parallel side 'a` | Parallel side 'b` | Area |
| 13 cm | 28 cm | 492 sq.cm |
The area of the trapezium, if the parallel sides are measuring 8 cm and 10 cm and the height 5 cm is
In a trapezium if the sum of the parallel sides is 10 cm and the area is 140 sq.cm, then the height is
