CBSE Class 10 Maths Chapter 11: Constructions NCERT Solutions

To divide a line segment of length 7.6 cm in the ratio 5:8, follow these steps:

1. Draw the line segment: Draw a line segment AB of length 7.6 cm.
2. Draw a ray: Construct a ray AX making any acute angle with the line segment AB.
3. Mark points on the ray: Locate and mark points A₁, A₂, ..., A₁₃ on the ray AX such that AA₁ = A₁A₂ = ... = A₁₂A₁₃. The total number of points is 5 + 8 = 13.
4. Join the endpoint: Join the point B to the 13th point on the ray, i.e., join BA₁₃.
5. Draw a parallel line: From the 5th point (A₅) on the ray AX, draw a line parallel to BA₁₃. This line will intersect AB at a point, let's call it C. This can be done by constructing an angle ∠A₅CB equal to ∠AA₁₃B.

The point C divides the line segment AB in the required ratio of 5:8. By measuring, AC ≈ 2.9 cm and CB ≈ 4.7 cm.

Justification:

According to the construction, A₅C is parallel to A₁₃B. In triangle AA₁₃B, by the Basic Proportionality Theorem (BPT), we have:

$$\backslash frac\{AC\}\{CB\}\; =\; \backslash frac\{AA\_5\}\{A\_5A\_\{13\}\}$$

Since AA₁ = A₁A₂ = ... = A₁₂A₁₃, the segment AA₁₃ is divided into 13 equal parts. Therefore, AA₅ represents 5 parts and A₅A₁₃ represents 8 parts.

$$\backslash frac\{AA\_5\}\{A\_5A\_\{13\}\}\; =\; \backslash frac\{5\}\{8\}$$

Comparing the two equations, we get:

$$\backslash frac\{AC\}\{CB\}\; =\; \backslash frac\{5\}\{8\}$$

This proves that the line segment AB is divided by point C in the ratio 5:8.

Step 1: Construct the original triangle

1. Draw a line segment AB of length 4 cm.
2. Take point A as the center and draw an arc with a radius of 5 cm.
3. Take point B as the center and draw an arc with a radius of 6 cm.
4. The point where these two arcs intersect is point C. Join AC and BC. Triangle ABC is the required triangle with sides 4 cm, 5 cm, and 6 cm.

Step 2: Construct a similar triangle with sides 2/3 of the original

1. Draw a ray BX from vertex B making an acute angle with the side AB.
2. Locate and mark three points B₁, B₂, B₃ on the ray BX such that BB₁ = B₁B₂ = B₂B₃. The number of points is determined by the denominator of the ratio 2/3.
3. Join the point B₂ (corresponding to the numerator 2) to point A.
4. Draw a line through B₃ parallel to AB₂, intersecting the extended line segment AB at a point A'.
5. From point A', draw a line parallel to AC, intersecting the extended line segment BC at point C'.

Triangle A'BC' is the required triangle, similar to triangle ABC, with sides being 2/3 of the corresponding sides of triangle ABC.

Justification:

By construction, B₃A' is parallel to AB₂. Using the Basic Proportionality Theorem in triangle ABB₃:

$$\backslash frac\{BA\text{'}\}\{AB\_2\}\; =\; \backslash frac\{BB\_3\}\{B\_2B\_3\}\; =\; \backslash frac\{3\}\{2\}$$

This implies:

$$\backslash frac\{AB\}\{BA\text{'}\}\; =\; \backslash frac\{2\}\{3\}$$

Also, by construction, A'C' is parallel to AC. Therefore, triangle A'BC' is similar to triangle ABC (by AA similarity criterion, as ∠A'BC' = ∠ABC and ∠BA'C' = ∠BAC).

The ratio of corresponding sides is:

$$\backslash frac\{A\text{'}C\text{'}\}\{AC\}\; =\; \backslash frac\{BC\text{'}\}\{BC\}\; =\; \backslash frac\{BA\text{'}\}\{BA\}\; =\; \backslash frac\{2\}\{3\}$$

This confirms that triangle A'BC' is similar to triangle ABC with sides 2/3 of the original.