Question

ii. In the adjoining fig. XY LAC divides
the triangular region ABC into
two equal areas. Determine AX : AB.
3
Plz ans me properly.
Don't put useless ans plzz guys​

Answer 1

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✦ Required Answer:

♦️ GiveN:

• Here, is a triangle ABC
• XY is parallel to AC, i.e. XY||AC
• ar(️BXY) = ar(AXYC)

♦️ To FinD:

• AX : AB

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✦ How to solve ?

The above Q. is based upon similar triangles concepts, especially we will use the relationship of sides and area of the triangle. Ratio of Area of two triangles is equal to square of corresponding sides of these triangles. So, let's find, All the necessary conditions like reasons are also mentioned .

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✦ Solution:

In △ ABC and △XBY,

$\large{ \sf{ \angle \: ABC = \angle \: XBY \: (common)}} \\ \\ \large{ \sf{ \angle \: ACB = \angle \: XYB \: (since \:XY || AB \: corresponding \: \angle \: s}} \\ \\ \large{ \sf{ \triangle \: ABC \sim \: \triangle \: XBY \: (AA \: similarity)}}$

Now we know that in similar triangles, Ratio of area of triangles is equal to ratio of squares of corresponding sides, Then,

$\large{ \sf{ \frac{area \: of \triangle \: ABC}{area \: of \triangle \: XBY} = ( \frac{AB}{XB} ) {}^{2}}} \\ \\ \large{ \sf{ \frac{area \: of \triangle \: XBY + area \: of \: AXYC}{area \: of \triangle \: XBY} = ( \frac{AB}{XB} ) {}^{2}}} \\ \\ \large{ \sf{ \frac{area \: of \triangle \: XBY + area \: of \: XBY}{area \: of \triangle \: XBY} = ( \frac{AB}{XB} ) {}^{2}}} \\ \\ \sf{ \dag{ \pink{given \: ar( \triangle \: XBY) = ar(AXYC)}}} \\ \\ \large{ \sf{ \frac{2 \times area \: of \triangle \: XBY}{area \: of \triangle \: XBY} = ( \frac{AB}{XB} ) {}^{2}}} \\ \\ \large{ \sf{ 2 = ( \frac{AB}{XB}) {}^{2}}}$

Now we can write it as,

$\large{ \sf{ \frac{AB}{XB} = \sqrt{2} }} \\ \\ \large{ \sf{XB = \frac{AB}{ \sqrt{2} } }}$

So, we need to find, AX : AB

$\large{ \sf{AX + XB = AB}} \\ \\ \large{ \sf{AX + \frac{AB}{ \sqrt{2} } = AB}} \\ \\ \large{ \sf{AX = AB - \frac{AB}{ \sqrt{2} }}} \\ \\ \large{ \sf{AX = \frac{AB( \sqrt{2} - 1) }{\sqrt{2}}}} \\ \\ \large{ \sf{ \frac{AX}{AB} = \boxed{ \sf{ \red{\frac{ \sqrt{2} - 1}{\sqrt{2}} }} }}}$

$\large{ \therefore{ \underline{ \sf{ \underline{ \green{Hence, \: solved \: \dag}}}}}}$

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