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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.
Solution :
Given: A triangle with side lengths $a = 5$ cm, $b = 6$ cm, and $c = 7$ cm. A scale factor of $k = \frac{7}{5}$ for the construction of a similar triangle.
To Find: Construct a triangle similar to the given triangle with sides $\frac{7}{5}$ times the corresponding sides of the first triangle and provide the geometric justification.
Step 1: Construction of the initial triangle $\triangle ABC$
1. Draw a line segment $BC = 5$ cm using a ruler.
2. With $B$ as the center and radius $7$ cm, draw an arc.
3. With $C$ as the center and radius $6$ cm, draw another arc intersecting the previous arc at point $A$.
4. Join $AB$ and $AC$. Thus, $\triangle ABC$ is constructed.
Step 2: Construction of the similar triangle $\triangle A'BC'$
1. Draw a ray $BX$ making an acute angle with $BC$ at point $B$.
2. Locate $7$ points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.
3. Join $B_5$ to $C$.
4. Draw a line through $B_7$ parallel to $B_5C$ 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
To justify the construction, we must prove that $\triangle ABC \sim \triangle A'BC'$ and that the ratio of their sides is $\frac{7}{5}$.
By construction, $B_7C' \parallel B_5C$.
In $\triangle BB_7C'$, by the Basic Proportionality Theorem (Thales Theorem):
$\frac{BC'}{BC} = \frac{BB_7}{BB_5} = \frac{7}{5}$ [Since $BB_7 = 7$ units and $BB_5 = 5$ units]
Also, since $A'C' \parallel AC$, $\triangle ABC \sim \triangle A'BC'$ by AA similarity criterion.
Therefore, the ratios of the corresponding sides are equal:
$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{7}{5}$
This confirms that the sides of $\triangle A'BC'$ are $\frac{7}{5}$ times the corresponding sides of $\triangle ABC$.
Final Answer: The triangle $\triangle A'BC'$ has been constructed such that its sides are $\frac{7}{5}$ of the sides of $\triangle ABC$, justified by the Basic Proportionality Theorem and AA similarity.
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.
- 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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