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# 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. 11

d. 12

**Solution:**

Let us construct the figure

Here AX is divided into 5 + 7 = 12 parts

Construct a line __parallel__ to BX from line number 5 to AB

Then the line will divide both AB and AX in the ratio 5: 7

So the right option is d.

Therefore, the minimum number of points is 12.

**✦ Try This: **To divide the line segment AB in the ratio 2 : 3, a ray AX is drawn such that ∠BAX is acute, AX is then marked at equal intervals. Find the minimum number of these marks.

It is given that

__Line segment__ AB is divided in the ratio 2: 3

Consider the parts of the line as 2x and 3x

We know that

2x = x + x

Where 2x contains 2 equal parts

3x = x + x + x

Where 3x contains 3 equal parts

So 5 equal parts are required.

Therefore, the minimum number of these marks is 5.

**☛ Also Check:** NCERT Solutions for Class 10 Maths Chapter 11

**NCERT Exemplar Class 10 Maths Exercise 10.1 Problem 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. 11, d. 12

**Summary:**

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 12

**☛ Related Questions:**

- To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that ∠BAX is an acute ang . . . .
- To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then . . . .
- To construct a triangle similar to a given ∆ABC with its sides 3/7 of the corresponding sides of ∆AB . . . .

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