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

Answer:

Minimum number of points to be located on the ray AX at equal distances is 13.

Hopefully it will help : )

Answer 2

Answer:

Therefore, the minimum number of points is 12.

Explanation:

Step :  1 To divide a line segment AB in the ratio 5'7 first a ray ax is drawn so that ∠ Bax is an acute angle and then at equal distances points are marked on the Ray ax such that the minimum number of these points is a 8 B 10 C

We should divide AX into 5+7=12 parts and draw a line parallel to BX from line no. 5 to AB.

Then this line will divide AX as well as AB in the ratio 5:7.

Step : 2  Summary: Point C divides the line segment AB of length 7.6 cm in the ratio of 5:8. By measurement, we find that AC = 2.9 cm and CB = 4.7 cm.Similar to this, a 15 cm line segment may be divided in two parts in the ratio 2:1 since AB is the line segment and C divides it in two parts in the ratio 2:1.

Step : 3 If CB equals x, AC equals 2x. AC plus CB equals 2x + x + 15x = 5x. Draw a ray AX with an acute angle at BAX to split a line segment AB in the ratio p: q. Then, mark points on the ray AX at equal intervals so that the least amount of these points is: Greater of p and q in (a). (B) p + q; (C) p + q-1. All fractions are ratio expressions. The denominator in 5:8 is the second portion. Accordingly, the first item contains 5 parts for every 8 parts of the second thing.

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