Two parallel sides DC and AB of a Trapezium are 12 cm and 36 cm respectively. If non parallel sides are each equal to 15 cm find the area of a Trapezium
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Let ABCD be the trapezium & AB = 36 cm, DC = 12 cm and AD = 12cm.
Now, EB = (AB-AE) = (AB-DC)
= 36 - 12
= 24cm.
In triangle EBC, CE = BC = 15cm.
CF is perpendicular to AB.
F is the midpoint of EB.
EF = 1/2 * AB
= 1/2 * EB
= 1/2 * 24
= 12cm.
In right- angled triangle CFE, CF = 15cm, EF = 12cm.
By Pythagoras theorem, we have
CF = root CE^2 - EF^2
= (15^2 - 12^2)
= root 225 - 144
= root 81
= 9.
The distance between the parallel sides = 9cm.
Therefore, Area of the trapezium = 1/2 * (sum of parallel sides) * (distance)
= 1/2 * (12+36) * 9
= 432/2
= 216 cm^2.
Now, EB = (AB-AE) = (AB-DC)
= 36 - 12
= 24cm.
In triangle EBC, CE = BC = 15cm.
CF is perpendicular to AB.
F is the midpoint of EB.
EF = 1/2 * AB
= 1/2 * EB
= 1/2 * 24
= 12cm.
In right- angled triangle CFE, CF = 15cm, EF = 12cm.
By Pythagoras theorem, we have
CF = root CE^2 - EF^2
= (15^2 - 12^2)
= root 225 - 144
= root 81
= 9.
The distance between the parallel sides = 9cm.
Therefore, Area of the trapezium = 1/2 * (sum of parallel sides) * (distance)
= 1/2 * (12+36) * 9
= 432/2
= 216 cm^2.
Answered by
9
➩Given:
(i) ABCD is a Trapezium
where DC||AB
(ii) AD=CB=15cm
(iii) DC=12cm
(iv) AB=36cm
➩To Find:
The Area of the Trapezium ABCD
➩Solution:
If DC||AB (given)
then DC||AE (parts of parallel lines)
Draw a line CE||AD
Now
AD||CE
DC||AE
Thus, AECD is a Parallelogram.
AD=CE (Opposite sides of parallelogram are always equal)
∴CE=15cm
Similarly,
DC=AE (Opposite sides of parallelogram are always equal)
∴AE=12cm
AE+EB=36cm
12+EB=36
EB=36-12
EB=24cm
Draw an Altitude CF⊥EB (Construction)
EF=FB (when an Altitude falls on a line ,it besects it or divide it into two equal parts)
EF+FB=24
FB+FB=24(as EF=FB)
2FB=24
FB=12cm
Now in △CFB
we can use Pythagoras theorem as ∠CFB=90°
CF=9cm
Also CF is the Height of the Trapezium ABCD
Area of Trapezium=
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