Question

Draw a line segment of length $7.6cm$and divide it in the ratio $5:8$. Measure the two parts. Give the justification of the construction

Answer 1

Construction of the required line segment

Step 1: Draw line segment $AB$ with the length dimension of $7.6cm$.

Step 2: Draw a ray $\overrightarrow{AX}$ that shapes an acute angle with line segment $AB$.

Step 3: Divide the ray $\overrightarrow{AX}$ into $13$ equal parts ($5+8$, based on the ratio given)

Name the points as$A_1,A_2,A_3,A_4\dots \dots ..A_{13}$such that it becomes $A{A}_1=A_1{A}_2=A_2{A}_3$and so on.

Step 4: Join the line segment and the ray by a line segment $B{A}_{13}$.

Step 5: Now through the point $A_5$, draw a line parallel to $B{A}_{13}$ which makes an angle equal to $\angle A{A}_{13}B$. This line intersects the line $AB$ at point $C$.

The point $C$ divides line segment $AB$ of $7.6cm$ in the required ratio of $5:8$.

Step 6: Now, mark the lengths of the line $AC$ and $CB$. It comes out that the measures of $AC$ and $CB$ are $2.9cm$and $4.7cm$respectively.

Justification:

By construction, we have

$C{A}_5\parallel B{A}_{13}$

From Basic proportionality theorem for the triangle $A{A}_{13}B,$ we get

$\frac{AC}{CB}=\frac{A{A}_5}{A_5{A}_{13}}...\left(i\right)$

From the diagram, we can observe that $A{A}_5$ and$A_5{A}_{13}$contain $5$ and $8$ equal divisions of line segments respectively.

Therefore, it becomes

$\frac{A{A}_5}{A_5{A}_{13}}=\frac{5}{8}...\left(ii\right)$

On comparing the equations $\left(i\right)$and $\left(ii\right)$ , we get

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

Hence, Justified.