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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.
Solution :
Given: A triangle $ABC$ with side $BC = 6$ cm, $AB = 5$ cm, and $\angle ABC = 60^{\circ}$.
To Construct: A triangle $A'BC'$ similar to $\triangle ABC$ such that its sides are $\frac{3}{4}$ of the corresponding sides of $\triangle ABC$.
Visual Representation of the Construction:
Steps of Construction:
Step 1: Draw a line segment $BC = 6$ cm.
Step 2: At point $B$, construct an angle of $60^{\circ}$ using a compass and ruler. Draw a ray $BY$ such that $\angle CBY = 60^{\circ}$.
Step 3: From ray $BY$, cut off a segment $BA = 5$ cm. Join $AC$. Thus, $\triangle ABC$ is constructed.
Step 4: Draw an acute angle $\angle CBX$ below the line segment $BC$.
Step 5: Locate 4 points $B_1, B_2, B_3, B_4$ on ray $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
Step 6: Join $B_4$ to $C$.
Step 7: Through $B_3$, draw a line parallel to $B_4C$ intersecting $BC$ at $C'$.
Step 8: Through $C'$, draw a line parallel to $CA$ intersecting $AB$ at $A'$.
Step 9: $\triangle A'BC'$ is the required triangle.
Justification:
To justify the construction, we must prove that $\triangle A'BC' \sim \triangle ABC$ and the ratio of their sides is $\frac{3}{4}$.
1. By construction, $B_3C' \parallel B_4C$.
2. In $\triangle BB_4C$, since $B_3C' \parallel B_4C$, by the Basic Proportionality Theorem (Thales Theorem):
$\frac{BC'}{BC} = \frac{BB_3}{BB_4}$ [Since $B_3$ is the 3rd point and $B_4$ is the 4th point on the ray]
$\frac{BC'}{BC} = \frac{3}{4}$ --- (Equation 1)
3. Since $A'C' \parallel AC$, in $\triangle ABC$, $\triangle A'BC' \sim \triangle ABC$ by AA similarity criterion.
4. Therefore, the ratio of corresponding sides is equal:
$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$
5. Substituting the value from Equation 1:
$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{4}$
Final Answer: The constructed triangle $A'BC'$ has sides exactly $\frac{3}{4}$ of the sides of $\triangle ABC$, satisfying the condition of 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.
- 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.
- 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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