A cool geometry problem!

Geometry Level 2

In the adjoining figure, X Y XY is parallel to A C AC . If X Y XY divides the triangle into two halves with equal area, compute A X A B \dfrac{AX}{AB} .

1 2 \frac{1}{\sqrt{2}} 2 + 1 2 \frac{\sqrt{2}+1}{\sqrt{2}} 2 − 1 2 \frac{\sqrt{2}-1}{\sqrt{2}} 1 2 \frac{1}{2}

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Rayyan Shahid by Rayyan Shahid , 20, India

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2 solutions

Triangles Δ B X Y \Delta BXY and Δ B A C \Delta BAC are similar. In order for X Y XY to divide the larger triangle into two equal parts, the corresponding dimensions of Δ B X Y \Delta BXY will be 1 2 \frac{1}{\sqrt{2}} those of Δ B A C \Delta BAC .

Thus B X A B = 1 2 ⟹ A B − A X A B = 1 2 ⟹ A X A B = 1 − 1 2 = 2 − 1 2 \frac{BX}{AB} = \frac{1}{\sqrt{2}} \Longrightarrow \frac{AB - AX}{AB} = \frac{1}{\sqrt{2}} \Longrightarrow \frac{AX}{AB} = 1 - \frac{1}{\sqrt{2}} = \boxed{\frac{\sqrt{2} - 1}{\sqrt{2}}} .

12 Helpful 4 Interesting 3 Brilliant 6 Confused

Can you tell how did you get 1 2 \dfrac{1}{\sqrt{2}} ?

Syed Hamza Khalid - 1 year, 8 months ago

For Δ B X Y \Delta BXY to have half the area, each of it's dimensions (e.g., base and height) must be 1 / 2 1/\sqrt{2} those of the corresponding dimensions of Δ B A C \Delta BAC , so that when the area (e.g., base x height) is calculated the result is 1 2 × 1 2 = 1 2 \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2} that of the larger triangle.

Brian Charlesworth - 1 year, 8 months ago
Avn Bha
Nov 22, 2014

solve by area theorem!! .this question is there in class 10 NCERT book

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Right bro. Its in the NCERT of class 10.

satvik pandey - 6 years, 6 months ago

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