What is the ratio $\frac{A C}{B C}$ for the line segment AB following the construction method below?
Step 1. A ray is extended from A and 30 arcs of equal lengths are cut, cutting the ray at $A_{1}, A_{2}, . . ., A_{30}$
Step 2. A line is drawn from $A_{30}$ to $B$ and a line parallel to $A_{30} B$ is drawn, passing through the point $A_{17}$ and meet AB at C.

17:30

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17:13

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Solution

The correct option is B
17:13

Here the total number of arcs is equal to m+n in the ratio m:n.

The triangles $△$ $A A_{17} C$ and $△$ $A A_{30} B$ are similar.

Hence, $\frac{A C}{A B}$ = $\frac{A A_{17}}{A A_{30}} = \frac{17}{30}$

$\frac{B C}{A B}$ = $\frac{A B - A C}{A B}$
$\frac{B C}{A B}$ = 1 - $\frac{17}{30}$ = $\frac{13}{30}$

Hence, $\frac{A C}{B C}$ = $\frac{17}{13} \to 17 : 13$

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Similar questions

What is the ratio $\frac{A B}{B C}$ for the line segment AB following the construction method below?
Step 1. A ray is extended from A and 30 arcs of equal lengths are cut, cutting the ray at $A_{1}, A_{2}, . . ., A_{30}$
Step 2. A line is drawn from $A_{30}$ to $B$ and a line parallel to $A_{30} B$ is drawn, passing through the point $A_{17}$ .

What is the ratio $\frac{A C}{B C}$ for the following construction method:
A ray is extended from A and 30 arcs of equal lengths are cut, cutting the ray at $A_{1}, A_{2} . . . . . . . A_{30}$ .
A line is drawn from $A_{30}$ to B and a line parallel to $A_{30} B$ is drawn, passing through the point $A_{17}$ .

What is the ratio $\frac{A C}{B C}$ for the following construction:
A line segment AB is drawn.
A single ray is extended from A and 12 arcs of equal lengths are cut, cutting the ray at $A_{1}, A_{2} \dots A_{12}$ .
A line is drawn from $A_{12}$ to B and a line parallel to $A_{12} B$ is drawn, passing through the point $A_{6}$ and cutting AB at C.

What is the ratio $\frac{A C}{B C}$ for the following construction method?

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The correct option is B 17:13 Here the total number of arcs is equal to m+n in the ratio m:n. The triangles △ AA17C and △ AA30B are similar. Hence, ACAB = AA17AA30=1730 BCAB = AB−ACAB BCAB = 1 - 1730 = 1330 Hence, ACBC = 1713→17:13