# To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A_{1} , A_{2} , A_{3} , .... are located at equal distances on the ray AX and the point B is joined to

a. A_{12}

b. A_{11}

c. A_{10}

d. A_{9}

**Solution:**

It is given that

__Line segment__ AB is divided in the ratio 4:7

A: B = 4: 7

Construct a ray AX which makes an __acute angle__ BAX

So the minimum number of points which are located at equal distances on the ray is

AX = A + B = 4 + 7 = 11

Here A_{1}, A_{2}, A_{3} …… are located at equal distances on the ray AX and the point B is joined to A_{11}.

Therefore, point B is joined to A_{11}.

**✦ Try This: **To divide the line segment AB in the ratio 3 : 4, a ray AX is drawn such 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:

It is given that

Line segment AB is divided in the ratio 3: 4

Consider the parts of the line as 3x and 4x

We know that

3x = x + x + x

Where 3x contains 3 equal parts

4x = x + x + x + x

Where 4x contains 4 equal parts

So 7 equal parts are required.

Therefore, the minimum number of these marks is 7.

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

**NCERT Exemplar Class 10 Maths Exercise 10.1 Problem 2**

## To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A_{1} , A_{2} , A_{3} , .... are located at equal distances on the ray AX and the point B is joined to a. A_{12}, b. A_{11}, c. A_{10}, d. A_{9}

**Summary:**

To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A_{1} , A_{2} , A_{3} , .... are located at equal distances on the ray AX and the point B is joined to A_{11}

**☛ Related Questions:**

- 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 . . . .
- To construct a triangle similar to a given ∆ABC with its sides 8/5 of the corresponding sides of ∆AB . . . .

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