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ICSE Class 9 Answered

Prove that a line joining the mid points of the parallel sides of a trapezium divides it into two equal parts
Asked by aswani1617 | 18 Sep, 2018, 23:02: PM
answered-by-expert Expert Answer
G
 
Given : ABCD is a Trapezium
Point H and G are the mid point of parallel sides.
 
Construction: AE and BF are the height of the trapezium.
 
To prove : Area of begin mathsize 16px style square end styleAHGD = Area of begin mathsize 16px style square end styleHBCG
 
Let us assume that The given Trapezium is an isosceles Trapezium, with AD = BC
 
Here,
AH = HB and DG = GC      ....since Point H and G are the mid-point of AB and CD.
 
 begin mathsize 16px style square end styleAHGE and  begin mathsize 16px style square end styleBHGF are rectagle.
(since AE = BF ..since AB || CD and  AE and BF are perpendicular to DG and GC respectively)
 
EG = AH .......(sides of a reactangle)
HB = GF ......(side of a rectangle)
Therefore Area of a begin mathsize 16px style square end styleAHGE =  Area of a begin mathsize 16px style square end styleBHGF ....(i)
 
In ΔADE and ΔBFC
 
AE = BF
AD = BC
and angleAED =  angleBFC = 900   
Therefore ΔADE  approximately equal to ΔBFC   ...(RHS )
Therefore ΔADE = ΔBFC.......(ii)
 
Area of a begin mathsize 16px style square end styleAHGD = ΔADE +begin mathsize 16px style square end styleAHGE and 
Area of a begin mathsize 16px style square end styleBHCG = ΔBFC + begin mathsize 16px style square end styleBHGF 
from (i) and (ii), we get
Area of begin mathsize 16px style square end styleAHGD = Area of begin mathsize 16px style square end styleHBCG
Answered by Yasmeen Khan | 19 Sep, 2018, 14:37: PM

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