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Q1:
In each of the following, give the justification of the construction also:
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution :
Given: A line segment $AB$ of length $7.6\text{ cm}$.
To Find: Divide the line segment $AB$ in the ratio $5 : 8$ and measure the lengths of the two parts.
Visual Representation:
Steps of Construction:
Step 1: Draw a line segment $AB = 7.6\text{ cm}$ using a ruler.
Step 2: Draw any ray $AX$ making an acute angle with $AB$.
Step 3: Locate $5 + 8 = 13$ points $A_1, A_2, \dots, A_{13}$ on $AX$ such that $AA_1 = A_1A_2 = \dots = A_{12}A_{13}$.
Step 4: Join $BA_{13}$.
Step 5: Through the point $A_5$, draw a line parallel to $BA_{13}$ (by making an angle equal to $\angle AA_{13}B$ at $A_5$) to intersect $AB$ at a point $P$.
Step 6: The point $P$ divides $AB$ in the ratio $5 : 8$.
Justification:
In $\triangle ABA_{13}$, $A_5P$ is parallel to $A_{13}B$ (by construction).
By the Basic Proportionality Theorem (Thales Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Therefore, $\frac{AP}{PB} = \frac{AA_5}{A_5A_{13}}$
Since $AA_5$ contains $5$ equal parts and $A_5A_{13}$ contains $8$ equal parts, we have:
$\frac{AP}{PB} = \frac{5}{8}$
This justifies that point $P$ divides $AB$ in the ratio $5 : 8$.
Measurement:
The total length is $7.6\text{ cm}$. The ratio is $5 : 8$.
Sum of ratio parts = $5 + 8 = 13$.
Length of first part $AP = \frac{5}{13} \times 7.6\text{ cm} \approx 2.92\text{ cm}$.
Length of second part $PB = \frac{8}{13} \times 7.6\text{ cm} \approx 4.68\text{ cm}$.
Final Answer: The line segment is divided into two parts measuring approximately 2.92 cm and 4.68 cm.
More Questions from Class 10 Mathematics Constructions EXERCISE 11.1
- Q2: In each of the following, give the justification of the construction also: Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle.
- Q3: In each of the following, give the justification of the construction also: Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.
- Q4: In each of the following, give the justification of the construction also: Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $1\frac{1}{2}$ times the corresponding sides of the isosceles triangle.
- Q5: In each of the following, give the justification of the construction also: Draw a triangle ABC with side $BC = 6$ cm, $AB = 5$ cm and $\angle ABC = 60^{\circ}$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle ABC.
- Q6: In each of the following, give the justification of the construction also: Draw a triangle ABC with side $BC = 7$ cm, $\angle B = 45^{\circ}$, $\angle A = 105^{\circ}$. Then, construct a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $\triangle ABC$.
- Q7: In each of the following, give the justification of the construction also: Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.
CBSE Solutions for Class 10 Mathematics Constructions
Chapters in CBSE - Class 10 Mathematics
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