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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
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