CBSE Class 10 Mathematics Chapter 26: Constructions NCERT Solutions

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

1. Step 1: Draw a line segment AB of length 7.6 cm.
2. Step 2: Draw a ray AX making any acute angle with the line segment AB.
3. Step 3: Locate points on ray AX such that they are equally spaced. Since the ratio is 5:8, we need to mark a total of 5 + 8 = 13 points. Let these points be A₁, A₂, ..., A₁₃, where AA₁ = A₁A₂ = ... = A₁₂A₁₃.
4. Step 4: Join the point B to the 13th point on the ray, i.e., join BA₁₃.
5. Step 5: From the point A₅ (which corresponds to the first part of the ratio 5), draw a line parallel to BA₁₃. This line should intersect AB at a point, let's call it C. This can be done by constructing an angle ∠A₅CA equal to ∠BA₁₃A.

Point C divides the line segment AB in the ratio 5:8. Upon measuring, the length of AC is approximately 2.9 cm and the length of CB is approximately 4.7 cm.

Justification:

By construction, we have drawn A₅C parallel to BA₁₃. According to the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

In triangle ABA₁₃, since A₅C is parallel to BA₁₃, we can write:

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

From Step 3, we have marked 13 equally spaced points on AX. Therefore, the segment AA₅ consists of 5 equal parts, and the segment A₅A₁₃ consists of 8 equal parts (13 - 5 = 8).

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

Comparing the two equations, we get:

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

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

Part 1: Constructing the original triangle (let's call it ΔABC)

1. Step 1: Draw a line segment AB of length 4 cm.
2. Step 2: With A as the center, draw an arc of radius 5 cm.
3. Step 3: With B as the center, draw an arc of radius 6 cm.
4. Step 4: The point where these two arcs intersect is point C. Join AC and BC.

Now, ΔABC has sides AB = 4 cm, AC = 5 cm, and BC = 6 cm.

Part 2: Constructing a similar triangle (let's call it ΔA'B'C') with sides 2/3 of ΔABC

1. Step 1: Draw a ray BX making an acute angle with the base AB of ΔABC.
2. Step 2: Locate points on ray BX such that they are equally spaced. Since the scale factor is 2/3, we need to mark a total of 3 points. Let these points be B₁, B₂, B₃, where BB₁ = B₁B₂ = B₂B₃.
3. Step 3: Join the point B₃ to A.
4. Step 4: From the point B₂ (corresponding to the numerator 2 in the scale factor 2/3), draw a line parallel to AB₃. This line should intersect AB at a point, let's call it A'.
5. Step 5: From point A', draw a line parallel to BC, intersecting AC at point C'.

The triangle ΔA'B'C' is similar to ΔABC, and its sides are 2/3 the length of the corresponding sides of ΔABC.

Justification:

By construction, A'B is parallel to AB₃. By the Basic Proportionality Theorem in ΔABB₃, we have:

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

Also by construction, A'C' is parallel to AC. Therefore, ΔA'B'C' is similar to ΔABC (by AA similarity criterion, as ∠BAC = ∠B'A'C' and ∠BCA = ∠B'C'A' due to parallel lines).

Since the triangles are similar, the ratio of their corresponding sides is equal:

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

This confirms that the sides of the constructed triangle ΔA'B'C' are 2/3 of the corresponding sides of the original triangle ΔABC.