In - ABC - XY is parallel to AC and Area - BXY - Area - ACY X -1- 2- Then the ratio AX - p equals-3-1-33 -3-3-33-3-3-33-1-3

The correct option is B 3 √(3)3 nbsp;Area (Δ BXY):Area (□ ACYX)=1:2 ⇒ nbsp; nbsp;Area (Δ BXY):Area (Δ ABC)=1:3 ⇒BXAB=1√(3) ⇒BXAX=1√(3) 1 ⇒ABAX=√(3)√(3) 1 ...

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Standard XII

Mathematics

Properties of Determinants

In Δ ABC , XY...

Question

In $Δ A B C,$ $X Y$ is parallel to $A C$ and $Area (Δ B X Y) : Area (□ A C Y X) = 1 : 2.$ Then the ratio $\frac{A X}{A B}$ equals

\frac{\sqrt{3} + 1}{3}

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\frac{3 - \sqrt{3}}{3}

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\frac{3 + \sqrt{3}}{3}

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\frac{\sqrt{3} - 1}{3}

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Solution

The correct option is B $\frac{3 - \sqrt{3}}{3}$
$Area (Δ B X Y) : Area (□ A C Y X) = 1 : 2 \Rightarrow Area (Δ B X Y) : Area (Δ A B C) = 1 : 3 \Rightarrow \frac{B X}{A B} = \frac{1}{\sqrt{3}} \Rightarrow \frac{B X}{A X} = \frac{1}{\sqrt{3} - 1} \Rightarrow \frac{A B}{A X} = \frac{\sqrt{3}}{\sqrt{3} - 1} \Rightarrow \frac{A X}{A B} = \frac{\sqrt{3} - 1}{\sqrt{3}} \Rightarrow \frac{A X}{A B} = \frac{3 - \sqrt{3}}{3}$

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Similar questions

Q. If

P = 5^{33} + 5^{33} + 5^{33} + 5^{33} + 5^{33}

and

Q = 3^{33} + 3^{33} + 3^{33}

and

P Q = 15^{x}

, then x equal to

Q. The sum

1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

is equal to :

Q. The sum

1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

is equal to :

Q. If the value of

{(1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \dots upto \infty)}^{{log}_{(0.25)} ⎛ ⎝ \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots upto \infty ⎞ ⎠}

l

, then

l^{2}

is equal to

Given a = 6 + √3
⇒ (1/a) = 1 / (6 + √3)
Multiply and divide by (6 − √3)
⇒ (1/a) = (6− √3) / [(6 + √3)(6 − √3)]
⇒ (1/a) = (6 − √3) / [36 − 3]
⇒ (1/a) = (6− √3)/33
∴ a - (1/a) = (6 + √3) -(6 − √3)/33
= [33(6 + √3) - (6 - √3)]/33
= [32*6 + 34*√3]/33
= [192 + 34√3]/33

Answer : [192 + 34√3]/33

Why does 1/a come above

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In ΔABC, XY is parallel to AC and Area (ΔBXY):Area (□ACYX)=1:2. Then the ratio AXAB equals

The correct option is B 3−√33 Area (ΔBXY):Area (□ACYX)=1:2⇒Area (ΔBXY):Area (ΔABC)=1:3⇒BXAB=1√3⇒BXAX=1√3−1⇒ABAX=√3√3−1⇒AXAB=√3−1√3⇒AXAB=3−√33

In $Δ A B C,$ $X Y$ is parallel to $A C$ and $Area (Δ B X Y) : Area (□ A C Y X) = 1 : 2.$ Then the ratio $\frac{A X}{A B}$ equals