Question

Plzzz telll me the solution ...... of this question .... . Plzzz help me.....

Answer 1

HEY MATE HERE IS YOUR ANSWER
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SOLUTION :-


Area of llgm EFGH = 1/2 Area llgm ABCD


CONSTRUCTION : JOIN HF


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HF II AB



HF is also II to DC



HFCD is a llgm and HCF is a triangle with same base and between the same parallels are equal in area.



Area of ⛛ HEF = 1/2 area HFCD - - - (1)


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➡️ ABFH is a parallelogram and ⛛ HGF the common base and between parallel sides are equal in area.



Area of ⛛ HGF = 1/2 area of llgm ABFH - - (2)


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Adding equation (1) and (2)



Area of ⛛ HEF + HGF


Area of llgm 1/2 HFCD = 1/2 ABFH


Area of HEFG = 1/2 ABCD


➡️ 1/2 ( HFCD + ABFH)


➡️ ar (EFGH) = 1/2 ar (ABCD)


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HENCE, PROVED
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