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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.
Solution :
Given: A right-angled triangle $ABC$ with sides $AB = 3\text{ cm}$ and $BC = 4\text{ cm}$, where $\angle B = 90^\circ$.
To Find: Construct a triangle $A'BC'$ similar to $\triangle ABC$ such that its sides are $\frac{5}{3}$ times the corresponding sides of $\triangle ABC$.
Step 1: Construction of the original triangle $\triangle ABC$
1. Draw a line segment $BC = 4\text{ cm}$.
2. At point $B$, construct an angle of $90^\circ$ using a compass and ruler.
3. Cut an arc of $3\text{ cm}$ on the perpendicular ray from $B$ to mark point $A$.
4. Join $AC$. Thus, $\triangle ABC$ is formed.
Step 2: Construction of the similar triangle
1. Draw an acute angle $\angle CBX$ below the line $BC$.
2. Mark 5 points (since the numerator of $\frac{5}{3}$ is 5) $B_1, B_2, B_3, B_4, B_5$ on ray $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.
3. Join $B_3$ to $C$ (since the denominator is 3).
4. Draw a line through $B_5$ parallel to $B_3C$ intersecting the extended line segment $BC$ at $C'$.
5. Draw a line through $C'$ parallel to $AC$ intersecting the extended line segment $BA$ at $A'$.
6. $\triangle A'BC'$ is the required triangle.
Step 3: Justification
By construction, $B_5C' \parallel B_3C$.
In $\triangle BB_5C'$, by Basic Proportionality Theorem (Thales Theorem):
$\frac{BC'}{BC} = \frac{BB_5}{BB_3} = \frac{5}{3}$ [Since $BB_5 = 5$ units and $BB_3 = 3$ units].
Also, since $A'C' \parallel AC$, $\triangle ABC \sim \triangle A'BC'$ by AA similarity criterion.
Therefore, the ratio of corresponding sides is:
$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{5}{3}$.
This confirms that the sides of $\triangle A'BC'$ are $\frac{5}{3}$ times the corresponding sides of $\triangle ABC$.
Final Answer: The triangle $A'BC'$ has been constructed such that its sides are $\frac{5}{3}$ times the sides of $\triangle ABC$, with $A'B = 5\text{ cm}$ and $BC' = 6.67\text{ cm}$ approximately.
More Questions from Class 10 Mathematics Constructions EXERCISE 11.1
- 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.
- 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 $\triangle ABC$.
CBSE Solutions for Class 10 Mathematics Constructions
Chapters in CBSE - Class 10 Mathematics
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