Question

Maths
Class 10
Triangles

Ques -: prefer attachment !


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Answer 1

Appropriate Question :-

In triangle ABC, X and Y are points on sides AB and BC respectively such that XY || AC, such that AX : AB = 2 - √2 : 2, then ratio of area of triangle BXY and BAC is

a) 1 : √2

b) 1 : 2

c) 1 : 4

d) 2 : 3

$\large\underline{\sf{Solution-}}$

Given that, in triangle ABC,

X and Y are points on sides AB and BC respectively such that XY || AC.

$\sf \: AX : AB \: = \: 2 - \sqrt{2} : 2$

$\rm \: \dfrac{AX}{AB} = \dfrac{2 - \sqrt{2} }{ 2 }$

can be further rewritten as

$\rm \: \dfrac{AB - BX}{AB} = \dfrac{2 - \sqrt{2} }{2}$

$\rm \: \dfrac{AB}{AB} - \dfrac{BX}{AB} = \dfrac{2 - \sqrt{2} }{ 2 }$

$\rm \: 1 - \dfrac{BX}{AB} = \dfrac{2 - \sqrt{2} }{ 2 }$

$\rm \: - \dfrac{BX}{AB} = \dfrac{2 - \sqrt{2} }{ 2 } - 1$

$\rm \: - \dfrac{BX}{AB} = \dfrac{2 - \sqrt{2} - 2}{ 2 }$

$\rm \: - \dfrac{BX}{AB} = \dfrac{- \sqrt{2}}{ 2 }$

$\rm\implies \: \: \boxed{\sf{ \: \: \rm \: \dfrac{BX}{AB} = \dfrac{ \sqrt{2}}{ 2 } \: \: }} - - - - (1)$

Now, Consider

$\rm \: In \: \triangle\:BXY \: and \: \triangle\:BAC$

$\rm \: \: \angle\:BXY \: = \: \angle\:BAC \: \: \{alternate \: interior \: angles \}$

$\rm \: \: \angle\:BYX \: = \: \angle\:BCA \: \: \{alternate \: interior \: angles \}$

So,

$\rm\implies \: \: \triangle\:BXY \: \sim \: \triangle\:BAC \: \: \{AA \: similarity \}$

We know,

Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

So, using this theorem, we have

$\rm \: \dfrac{ar(\triangle\:BXY)}{ar(\triangle\:BAC)} = \dfrac{ {BX}^{2} }{ {AB}^{2} }$

$\rm \: \dfrac{ar(\triangle\:BXY)}{ar(\triangle\:BAC)} = {\bigg[\dfrac{BX}{AB} \bigg]}^{2}$

$\rm \: \dfrac{ar(\triangle\:BXY)}{ar(\triangle\:BAC)} = {\bigg[\dfrac{ \sqrt{2} }{2} \bigg]}^{2}$

$\rm \: \dfrac{ar(\triangle\:BXY)}{ar(\triangle\:BAC)} = \dfrac{2}{4}$

$\rm \: \dfrac{ar(\triangle\:BXY)}{ar(\triangle\:BAC)} = \dfrac{1}{2}$

$\rm\implies \:ar(\triangle\:BXY) : ar(\triangle\:BAC) = 1 : 2$

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Additional Information

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Basic Proportionality Theorem :-

This theorem states that : If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answer 2

question is : In triangle ABC, X and Y are points on sides AB and BC respectively such that XY || AC, such that AX : AB = 2 - √2 : 2, then ratio of area of triangle BXY and BAC is

the answer is : 1 : 2