Question

12. If a line segment AB is to be divided in the ratio 5:8 internally, we draw a ray AX such

that ∠BAX is an acute angle. What will be the minimum number of points to be

located at equal distances on ray AX?


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Answer 1

$\sf\small\underline\green{To\: Find:-}$

$\implies{\rm }$number of points to be located at equal distances on ray AX

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

$\implies{\rm }$13 points are required  To divide a line segment AB in the ratio 5:8

$\implies{\rm }$To divide a line segment AB in the ratio 5:8,

$\sf\small\underline\green{(i)}$ Draw a line segment AB  of some length

$\sf\small\underline\green{(ii)}$ Draw a line segment AX such that ∠BAX is an acute angle

$\sf\small\underline\green{(iii)}$ Take 13 point on AX of Equal length one by one ( consecutively)

$\sf\small\underline\green{(iv)}$Join 13th Point with B as a straight line

$\sf\small\underline\green{(v)}$Draw a line parallel to line drawn in step 4 such that it passes through 5th point of step 3 and intersect AB at M

$\implies{\rm }$M divides AB in to  5 : 8 Ratio.

∴13 points are required$.$

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