Line segment XY // BC of triangle ABC. Given AX: XB = 3:5, then the ratio
areas of trapezium XBCY and triangle AXY is
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Answer :-
1) in ∆AXY & ∆ABC
As XY // BC, corresponding angles are equal
ANGLE AXY = ANGLE ABC
ANGLE AYX = ANGLE ACB
∆ AXY ~ ∆ ABC
=> AX/AB = XY/BC
=> 3/8 = 18/BC
=> BC = 48 CM
2) AREA OF ∆AXY/ AREA OF ∆ABC = AX²/ AB² = 9/64
AREA OF ∆ ABC - AREA OF ∆AXY / AREA OF ∆ABC = 64-9/64
AREA OF TRAPEZIUM × BCY / AREA OF ∆ABC = 55/64
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AX
AB
XB
() In AAXY and AABC,
As XY BC, Corresponding angles are
equal
ZAXY = ZABC
LAYX = LACB
AAXY AABC
XY
AX
R BC
18
BC
BC = 48cm
(i)
Area of AAXY-AX
9
64
AB?
Area of A ABC
Area of AABC -Area of AAXY-64-9 55
64
Area of A ABC
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