In a trapezium ABCD,AD=2cm,XY=3.2cm and AZ=3cm.Find area of trapezium ABCD. Prove that the area of trapezium=XY×AZ.
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Area of trapezium = (1/2) * sum of ∥ sides * height
p
let AZ meets XY at L and draw ⊥ dronlm D to BC at M and cuts XY at N
as AX = BX
so according to BPT theorem XL=2BZ, NY=2LC
Area of trapezium=(1/2)[AD+BC]AZ
=(1/2)[2+BZ+ZM+MC]*3
=(1/2)[2+XL/2 +2+NY/2]*3
=(1/2)[4+(XL+NY)/2 ]*3
=(1/2)[4+(3.2-2)/2]*3
=(1/2)[4+0.6]*3
=(1/2)[4.6]*3
=3.2*3
=XY*AZ=9.6cm²
MARK AS BRAINLIEST IF YOU GOT ANSWER AFTER CONSTRUCTION
p
let AZ meets XY at L and draw ⊥ dronlm D to BC at M and cuts XY at N
as AX = BX
so according to BPT theorem XL=2BZ, NY=2LC
Area of trapezium=(1/2)[AD+BC]AZ
=(1/2)[2+BZ+ZM+MC]*3
=(1/2)[2+XL/2 +2+NY/2]*3
=(1/2)[4+(XL+NY)/2 ]*3
=(1/2)[4+(3.2-2)/2]*3
=(1/2)[4+0.6]*3
=(1/2)[4.6]*3
=3.2*3
=XY*AZ=9.6cm²
MARK AS BRAINLIEST IF YOU GOT ANSWER AFTER CONSTRUCTION
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