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Constructions — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsConstructions5 Marks
Question
Draw a triangle $\triangle\text{ABC}$ in which AB = 4cm, BC = 6cm and AC = 9cm. Construct a triangle similar to $\triangle\text{ABC}$ with scale factor $\frac{3}{2}.$ Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal.
✓

Answer

Scale factor $\frac{3}{2}>1$ So, the resulting figure will be greater than $\triangle\text{ABC}.$
Steps of construction:
1. Draw line segment $B C=6 cm$.
2. From $B$ as centre, draw an arc $A_1$, of 6 cm .
3. From Cas centre, draw another arc $A_2$, of 9 cm .
4. Arcs $A_1$ and $A_2$ intersect at $A$. Join $A$ to $B$ and $C$.
5. Make an acute angle of $\angle CBX$ on other side of $A$.
6. Make the equispaced marks $B_1, B_2, B_3$ with compass.
7. Join $B_2 C$ and draw $B_3 C^{\prime} \| B_2 C$, where $C^{\prime}$ is on $B C$ produced.
8. Draw CA || C'A', where $A^{\prime}$ is on $B A$ produced.
$\therefore\ \triangle\text{A}'\text{BC}'-\triangle\text{ABC}$ with scale factor $\frac{3}{2}.$
Justification:In $\triangle \text{BB}_{3}\text{C}'$ and $\triangle\text{BB}_{2}\text{C}$
$\angle\text{B} = \angle\text{B}$ [Common]
$\text{B}_{3}\text{C}'\ ||\ \text{B}_{2}\text{C}$ [By construction]
$\therefore \ \angle\text{BB}_{2}\text{C} = \angle\text{BB}_{3}\text{C}'$ [Corresponding angles]
$\therefore \triangle \text{BB}_{3}\text{C}' \sim\triangle\text{BB}_{2}\text{C}$ [By AA criterion of similarity]
$\Rightarrow \frac{\text{BC}'}{\text{CB}} = \frac{\text{BB}_{3}}{\text{BB}_{2}} = \frac{\text{3x}}{\text{2x}} = \frac{3}{2}$ $[\because\ \text{BB}_{1} = \text{B}_{1}\text{B}_{2} = \text{B}_{2}\text{B}_{3} = \text{x}]$
$\Rightarrow\frac{\text{BC}'}{\text{BC}} = \frac{3}{2}$
In $\triangle\text{ABC}$ and $\triangle\text{A}'\text{BC}',$
$\angle\text{B} = \angle\text{B}$ [Common]
$\because \text{A}'\text{C}'\ ||\ \text{AC}$
$\therefore\angle\text{A}'\text{C}'\text{B} = \angle\text{ACB}$ [Corresponding angles]
$\therefore \triangle\text{ABC} \sim\triangle\text{A}'\text{BC}'$ [By AA criterion of similarity]
$\Rightarrow \frac{\text{A}'\text{C}'}{\text{AC}} = \frac{\text{A}'\text{B}'}{\text{AB}} = \frac{\text{C}'\text{B}'}{\text{BC}}$
$\Rightarrow\frac{\text{A}'\text{C}'}{\text{AC}} = \frac{\text{A}'\text{B}'}{\text{AB}} = \frac{3}{2}$
Hence, proved.

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