Q 24. Find the area of a trapezium whose parallel sides are
25cm, 13cm and the other sides are 15cm each.
Answers
Given:- ABCD Is a trapezium
AB = 25 cm
DC = 13 cm
AD & BC = 15 cm
Construction:- Draw CE || AD
To Find :- Area of trapezium ABCD
Solution :- ADCE is a parallelogram ( AD || CE & AE || CD).
∴ AE = DC = 13 cm ( Opposite side of parallelogram are equal)
BE = AB - AE
BE = 25 - 13
BE = 12 cm
In ∆ BCE
S = a + b + c/2
S = 15 + 15 + 12 /2
S = 21
Area of ∆ BCE = √ s( s - a)(s - b)( s - c )
Area of ∆ BCE = √ 21(21-15)(21-15)(21-12)
Area of ∆ BCE = √ 21 × 6 × 6 × 9
Area of ∆ BCE = 18√21 cm^2 -----1
h is the height of ∆ BCE
Area of BCE = 1/2 ( Base × Height )
= 1/2(12)(h)
= 6h -----2
From 1 & 2
6h = 18√21
=> h = 3√21 cm
The height of trapezium ABCD is equal to height of ∆ BCE.
Area of trapezium = 1/2 ( AB + CD ) × h
= 1/2 (25 + 13) × 3√21cm^2
= 57√21 cm^2
Answer:
12500 Joules is the required energy.
Given:-
Time taken ,t = 10 h
Number of Devices ,n = 5
Power of each device ,P = 250 W
To Find:-
Energy Consumed by Devices,E
Solution:-
As we have to calculate the Energy Consumed by devices in 10 hours .
Energy is defined as the Product of Power and Time.
• E = P× n × t
Where,
P denote Power of device
n denote number of devices
t denote time
Substitute the value we get
→ E = 250×5 × 10
→ E = 1250 × 10
→ E = 12500 J or 12500 Wh = 12.5 kWh
Hence , the energy consumed by the devices is 12500 Joules .