CBSE Class 10 Mathematics Chapter 11: 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. From point A, draw a ray AX making any acute angle with AB.
2. Step 2: Locate 13 points (since 5 + 8 = 13) on ray AX such that all the segments are equal in length. Let these points be $A_1, A_2, A_3, \dots, A_{13}$, where $AA_1 = A_1A_2 = \dots = A_{12}A_{13}$.
3. Step 3: Join the point $B$ to the 13th point on the ray, i.e., join $BA_{13}$.
4. Step 4: From the 5th point ($A_5$) on the ray AX, draw a line parallel to $BA_{13}$. This line will intersect the line segment AB at a point, let's call it C. To draw this parallel line, construct an angle equal to $\angle AA_{13}B$ at $A_5$ such that it forms a line parallel to $BA_{13}$.

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

Justification of the construction:

According to the construction, the line segment $A_5C$ is drawn parallel to $BA_{13}$.

In $\triangle AA_{13}B$, since $A_5C \parallel BA_{13}$, by the Basic Proportionality Theorem, we have:

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

From Step 2, we have located 13 points such that $AA_1 = A_1A_2 = \dots = A_{12}A_{13}$. The segment $AA_5$ consists of 5 equal parts, and the segment $A_5A_{13}$ consists of 8 equal parts (from $A_5$ to $A_{13}$).

Therefore, $\frac{AA_5}{A_5A_{13}} = \frac{5}{8}$.

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.

Step 1: Construct the first 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. Thus, $\triangle ABC$ is the required triangle with sides AB = 4 cm, AC = 5 cm, and BC = 6 cm.

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

1. From vertex A, draw a ray AX making an acute angle with the base AB.
2. Locate 3 points ($A_1, A_2, A_3$) on the ray AX such that $AA_1 = A_1A_2 = A_2A_3$. The number of points is the denominator of the scale factor (2/3).
3. Join $A_3$ to B.
4. From $A_2$ (the point corresponding to the numerator 2), draw a line parallel to $A_3B$. Let this line intersect AB at point D.
5. From point D, draw a line parallel to BC, intersecting AC at point E.

Then, $\triangle ADE$ is the required triangle similar to $\triangle ABC$, with its sides being 2/3 of the corresponding sides of $\triangle ABC$.

Justification of the construction:

By construction, $A_2D \parallel A_3B$. In $\triangle AA_3B$, by the Basic Proportionality Theorem:

$$\backslash frac\{AD\}\{AB\}\; =\; \backslash frac\{AA\_2\}\{AA\_3\}$$

Since $AA_1 = A_1A_2 = A_2A_3$, we have $AA_2 = 2$ parts and $AA_3 = 3$ parts. So, $\frac{AA_2}{AA_3} = \frac{2}{3}$.

Therefore, $\frac{AD}{AB} = \frac{2}{3}$.

Also by construction, $DE \parallel BC$. Therefore, $\triangle ADE \sim \triangle ABC$ (by AA similarity criterion, as $\angle DAE = \angle BAC$ and $\angle ADE = \angle ABC$ because they are corresponding angles).

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

$$\backslash frac\{AD\}\{AB\}\; =\; \backslash frac\{AE\}\{AC\}\; =\; \backslash frac\{DE\}\{BC\}\; =\; \backslash frac\{2\}\{3\}$$

This justifies that $\triangle ADE$ is similar to $\triangle ABC$ and its sides are 2/3 of the corresponding sides of $\triangle ABC$.