The parallel sides of an isosceles trapezium are 48 cm and 24 cm. Each of its non-parallel side is 26 cm long. Find the area of trapezium.
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3
ln trapezium ABCD
taking triangle ACD
p^2+b^2=h^2
p=26,b=48
676+2304=h^2
h=54.5
area=p+b+h
=26+48+54.5
=128.5
Taking triangle ACB
same as above
butp=26
b=24
by solving we get h=35.3
area=p+b+h
=26+24+35.3
=85.3
According to question
Adding ar(ACD)&ar(ACB)
we get area of trapezium ABCD=213.8 Ans
taking triangle ACD
p^2+b^2=h^2
p=26,b=48
676+2304=h^2
h=54.5
area=p+b+h
=26+48+54.5
=128.5
Taking triangle ACB
same as above
butp=26
b=24
by solving we get h=35.3
area=p+b+h
=26+24+35.3
=85.3
According to question
Adding ar(ACD)&ar(ACB)
we get area of trapezium ABCD=213.8 Ans
pratyushdubey19:
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But i check in answer sheet there is something else answer
what is answer
may it is right answer that l give ATMe
Answered by
4
Let ABCD be the given trapezium in which AB =48 cm, DC = 24 cm, BC = 26 cm, AD = 15 cm.
Through C, draw CF||AD.
Now, FB = AB - AF = 48 - 24 = 24 cm.
In triangle FBC, FC = BC = 26 cm
So, it is an isosceles triangle.
Also,CE is perpendicular to FB. So, E is the mid - point of FB.
So,
FE = 1/2FB = 1/2 *24 =12 cm
Ina right-angled triangle CEF,
CF^2 = FE^2 + CE^2 [by pythagoras theorem]
26^2 = 12^2 +CE^2
CE^2 = 676 - 144
CE^2 = 532
CE = √532
Area of trapezium = 1/2(AB+DC)*CE
=1/2(48 + 24)*√532
= 1/2 * 72 *√532
= 36√532
Through C, draw CF||AD.
Now, FB = AB - AF = 48 - 24 = 24 cm.
In triangle FBC, FC = BC = 26 cm
So, it is an isosceles triangle.
Also,CE is perpendicular to FB. So, E is the mid - point of FB.
So,
FE = 1/2FB = 1/2 *24 =12 cm
Ina right-angled triangle CEF,
CF^2 = FE^2 + CE^2 [by pythagoras theorem]
26^2 = 12^2 +CE^2
CE^2 = 676 - 144
CE^2 = 532
CE = √532
Area of trapezium = 1/2(AB+DC)*CE
=1/2(48 + 24)*√532
= 1/2 * 72 *√532
= 36√532
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