Question

ABCD is a parallelogram,x and y are midpoints of BC and CD,prove that ar(AXY)=3/8 ar(ABCD)

Answer 1

TO PROVE THAT:

ar(tri AXY) = 3/8 ar (parallelogram ABCD)

PROOF: XY // DB & XY = DB/2 ( by mid point theorem )

=> tri CYX ~ tri CDB ( by AAA similarity theorem)

So, ar( tri CYX)/ar( tri CDB) = 1:4 ( by area similar triangle theorem states that, the ratio of the areas of two similar triangles = the ratio of the squares of their corresponding sides.)

=> if ar( tri CYX) = a = 1/2 * b * h/2 = bh/4

=> ar( tri CDB) = 4a = 4* bh/4 = bh

=> ar( ADB) = 4a = bh

=> ar( // gm ABCD) = 8a = 8* bh/4 = 2bh

& ar( trapezium YDBX) = 4a- a= 3a = 3bh/4

So, ar( pentagon YDABX) = 4a+ 3a = 7a = 7bh/4

Now, ar( tri AXY) ={ ar( pentagon) }— {( ar tri ADY+ ar triABX)}

={ 7bh/4} — { (1/2* b * h) + (1/2 * 2b * h/2) }

= 7 bh/4 — bh = 3bh/4

ar( tri AXY ) = 3bh/4

=> ar(tri AXY) / ar( parallelogram ABCD)

= (3bh/4 ) ÷ 2bh

= 3/8

=> ar(tri ABD) = 3/8 ( ar // gm ABCD)