Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5

Solution:

Steps of Construction

1. Construct a line segment AB = 7 cm

2. Construct a ray AX which makes an acute angle ∠BAX

3. Along the line AX mark 3 + 5 = 8 points

A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈ where

AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇ = A₇A₈

4. Let us join A₈B

5. From the point A₃ construct A₃C||A₈B which meets AB at C

It makes an angle equal to ∠BA₈ at A₃

6. C is the point on AB which divides it in the ratio 3: 5

7. So AC: BC = 3: 5

Verification:

Consider AA₁ = A₁A₂ = A₂A₃ = A₇A₈ = x

In triangle ABA₈ we know that A₃C||A₈B

Here

AC/CB = AA₃/A₃A₈ = 3x/5x = 3/5

So AC:CB = 3: 5

Therefore, the point P which divides it in the ratio 3:5 is found.

✦ Try This: Draw a line segment of length 5 cm. Find a point P on it which divides it in the ratio 1:3.

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

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NCERT Exemplar Class 10 Maths Exercise 10.3 Problem 1

Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5

Summary:

A line segment is a part of a line that has two endpoints and a fixed length. A line segment of length 7 cm is drawn. A point P on it which divides it in the ratio 3: 5 is found

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