Question

To divide a line segment $AB$ in the ratio $5:6$, draw a ray $AX$ such that $\angle BAX$ is an acute angle, then draw a ray $BY$ parallel to $AX$ and the points $A_1,A_2,A_3........$and $B_1,B_2,B_3.....$are located at equal distances on ray $AX$ and $BY$, respectively. Then the points joined are:

• A. $A_{5}and{B}_6$
• B. $A_{6}and{B}_5$
• C. $A_{4}and{B}_5$
• D. $A_{5}and{B}_4$

Answer 1

The correct option is A

$A_{5}and{B}_6$

Let's find the point of intersection of points:

According to the given details, we know that

A line segment AB in the ratio $5:6$.

So, $A:B=5:6$

Listed below are the steps of construction :

1. Draw a ray $AX$, an acute angle $BAX$.

2. Draw a ray $BY\parallel AX$, angle $ABY$ = angle $BAX$.

3. Now, position the points $A_1,A_2,A_{3,}{A}_{4}and{A}_5$ on $AX$ and $B_1,B_2,B_3,B_4,B_{5}and{B}_6$.

(Because $A:B=5:6$)

4. Join $A_5{B}_6$.

$A_5{B}_6$ intersect $AB$ at a point $C$.

$AB:BC=5:6$