Question

Question 3
To divide a line segment AB in the ratio 5:6, draw AX such that $\angle$ BAX is an acute angle, the 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 to 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

Given, a line segment AB and we have to divide it in the ratio 5 : 6


Steps of constructions
Draw a ray AX making an acute $\angle$BAX
Draw a ray By parallel to AX by making $\angle$ABY equal to $\angle$BAX.
Now locate the points $A_1,A_2,A_3,A_4$ and $A_5$ ( m = 5 ) on AX and $B_1,B_2,B_3,B_4,B_5$ and $B_6$ (n=6) such that all the points are at equal distance from each other.
Join $B_6{A}_5$ let it intersect AB at a point C.
Then AC : BC = 5 : 6