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705-805 (Hard)|   Geometry|      
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In the given figure, XY is parallel to AC and divides the triangle △AB 12 Jun 2020, 10:51
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35% (02:10) correct 65% (01:45) wrong based on 34 sessions

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In the given figure, XY is parallel to AC and divides the triangle △ABC into two parts of equal area. What is the ratio AX:AB ?

A. \(\frac{2 - \sqrt{2}}{2}\)

B. \(\frac{1}{2}\)

C. \(\frac{\sqrt{2}}{2}\)

D. \(\frac{1 + \sqrt{2}}{2}\)

E. \(\frac{2 + \sqrt{2}}{2}\)

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A
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In the given figure, XY is parallel to AC and divides the triangle △AB 12 Jun 2020, 11:29
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Ans should be A

As AC||XY,

So, Triangle BXY is Similar to Triangle BAC (Triangles have equal angles)

For similar Triangle ratio of the area of a triangle is equal to the ratio of the square of the corresponding side

Area of Triangle BXY / Area of Triangle BAC= 1/2

Area of Triangle BXY/Area of Triangle BAC= BX/BA=> (BA-AX)/BA => 1- AX/AB

Equating above 2

=> AX/AB= 1 - 1/Sqrt (2) => (2- Sqrt(2))/ 2
=>
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In the given figure, XY is parallel to AC and divides the triangle △AB 12 Jun 2020, 12:44
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Similar triangle property.
(Ratio of side)^2 = Ratio of area.
∆BXY & ∆ABC
(BX/AB)^2 = 1/2
BX/AB = 1/√2
Subtract 1 from both sides.
BX/AB - 1 = 1/√2 - 1
BX-AB/AB = 1-√2/√2
-AX/AB = 1-√2/√2
AX/AB = √2-1/√2
Rationalization
AX/AB = √2(√2-1)/√2*√2
2-√2/2 = AX/AB

Answer is A

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In the given figure, XY is parallel to AC and divides the triangle △AB 13 Jun 2020, 00:45
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Bunuel

In the given figure, XY is parallel to AC and divides the triangle △ABC into two parts of equal area. What is the ratio AX:AB ?

A. \(\frac{2 - \sqrt{2}}{2}\)

B. \(\frac{1}{2}\)

C. \(\frac{\sqrt{2}}{2}\)

D. \(\frac{1 + \sqrt{2}}{2}\)

E. \(\frac{2 + \sqrt{2}}{2}\)

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Solution


    • Since line XY divides the triangle ABC into two parts of equal area,
      o So, Area of ABC = Area of BXY + Area of XYCA = 2*Area of BXY
      \(⟹ \frac{Area \space of \space BXY}{Area \space of \space ABC }= \frac{1}{2} \)
    • In triangle BXY and BAC,
      o \(∠XBY = ∠ABC\) [ common in both triangles]
      o \(∠BXY = ∠BAC\) [corresponding angles as XY is parallel to AC]
      o \(∠BYX = ∠BCA\) [corresponding angles as XY is parallel to AC]
      o So, triangle BXY and BAC are similar triangles.
    • If two triangles are similar,
      o \(the \space ratio \space of \space their \space areas = (the \space ratio \space of \space their \space corresponding \space sides) ^2\)
      o Therefore, in triangles BXY and BAC,
         \(\frac{Area \space of \space BXY}{Area \space of \space ABC} = (\frac{BX }{AB})^2 = \frac{1}{2}\)
        \(⟹ \frac{AB – AX}{AB} = \frac{1}{\sqrt{2}}\)
        \(⟹\frac{AX}{AB} = 1 - \frac{1}{\sqrt{2}} = 1 - \frac{\sqrt{2}}{2}\)
        \(⟹ \frac{AX}{AB} = \frac{2 -\sqrt{2}}{2}\)
Thus, the correct answer is Option A.
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In the given figure, XY is parallel to AC and divides the triangle △AB 14 Aug 2024, 07:21
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