Question

triangle ABC xy parallel to AC and xy divides the triangle into two parts of equal area find the ratio of a x by x b

Answer 1

Answer:

AX/AB = (2-√2)/2

Step-by-step explanation:

Theorem  : The  ratio  of  the  areas of two similar triangles is equal to the square of the ratio of their corresponding  sides.

It is given that,

In Triangle ABC,  XY parallel to AC and XY divides the triangle into two parts of equal area.

The triangle ABC is attached with this answer

ar (ABC) =  2 ar (XBY)

We have XY ||  AC, therefore  <BXY = <A  and<BYX = <C (Corresponding angles)

So, ΔABC ~Δ XBY

ar (ABC)/ar (XBY) =(AB/XB)^2

ar (ABC)/ar (XBY) = 2/1     [ since ar (ABC) =  2 ar (XBY)]

Therefore, (AB/XB )^2=  2/1

AB/XB  = √2/1

XB/AB = 1/√2

1- XB/AB = 1-1/√2

(AB - XB)/AB =(√2 -1)/√2

Therefore, AX/AB = (√2 -1)/√2 = (2-√2)/2

Answer 2

Answer:

AX/AB = (2-√2)/2

Step-by-step explanation:

Theorem  : The  ratio  of  the  areas of two similar triangles is equal to the square of the ratio of their corresponding  sides.

In Triangle ABC,  XY parallel to AC and XY divides the triangle into two parts of equal area.

The triangle ABC is attached with this answer

ar (ABC) =  2 ar (XBY)

We have XY ||  AC, therefore  <BXY = <A  and<BYX = <C (Corresponding angles)

So, ΔABC ~Δ XBY

ar (ABC)/ar (XBY) =(AB/XB)^2

ar (ABC)/ar (XBY) = 2/1     [ since ar (ABC) =  2 ar (XBY)]

Therefore, (AB/XB )^2=  2/1

AB/XB  = √2/1

XB/AB = 1/√2

1- XB/AB = 1-1/√2

(AB - XB)/AB =(√2 -1)/√2

Therefore, AX/AB = (√2 -1)/√2 = (2-√2)/2

ANKIT