In∆ABC area of ∆ABC=48 sq cm. Line XY is drawn parallel to side BC dividing AB in the ratio 3:5.If A-X-B and A-Y-C. Find the area of ∆BXY.
Answer (1)
31st Mar, 2020
Hi,
Theorem : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
3 ar (ABC) = 5 ar (XBY)
We have XY || AC, therefore <BXY = <A and<BYX = <C (Corresponding angles)
So, ABC ~ XBY
ar (ABC)/ar (XBY) =(AB/XB)^2
ar (ABC)/ar (XBY)=5/3
Therefore, (AB/XB )^2= 5/3
Theorem : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
3 ar (ABC) = 5 ar (XBY)
We have XY || AC, therefore <BXY = <A and<BYX = <C (Corresponding angles)
So, ABC ~ XBY
ar (ABC)/ar (XBY) =(AB/XB)^2
ar (ABC)/ar (XBY)=5/3
Therefore, (AB/XB )^2= 5/3
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