Question

"Question 13 ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY). [Hint: Join CX.]

Class 9 - Math - Areas of Parallelograms and Triangles Page 164"

Answer 1

Two Triangles on the same base and between the same parallels are equal in area.

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Given,

ABCD is a trapezium with AB || DC & XY || AC

Construction,

Join CX .

 

To Prove,

ar(ADX) = ar(ACY)

Proof:

ar(△ADX) = ar(△AXC) — (i)

(On the same base AX and between the same parallels AB and CD)
also,

ar(△ AXC)=ar(△ ACY) — (ii)

 (On the same base AC and between the same parallels XY and AC.)

From (i) and (ii),

ar(△ADX)=ar(△ACY)

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Hope this will help you...

 

Answer 2

Answer:

Area (ΔADX) = Area (ΔACX) (i)

Area (ΔADX) = Area (ΔACX) (i)ΔACY and ΔACX lie on the same base AC and are between the same parallels AC and XY

Area (ΔADX) = Area (ΔACX) (i)ΔACY and ΔACX lie on the same base AC and are between the same parallels AC and XYArea (ΔACY) = Area (ACX) (ii)

Area (ΔADX) = Area (ΔACX) (i)ΔACY and ΔACX lie on the same base AC and are between the same parallels AC and XYArea (ΔACY) = Area (ACX) (ii)From equations (i) and (ii), we obtain

Area (ΔADX) = Area (ΔACX) (i)ΔACY and ΔACX lie on the same base AC and are between the same parallels AC and XYArea (ΔACY) = Area (ACX) (ii)From equations (i) and (ii), we obtainArea (ΔADX) = Area (ΔACY)