[5 Mark Questions]

Question 515 Marks

Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{7}{3}$ of the corresponding sides of the triangle PQR (scale factor $\frac{7}{3}>1$ )

Answer

Given triangle $A B C$, we are required to construct another triangle whose sides are $\frac{7}{3}$ of the corresponding sides of the $\triangle A B C$.

Steps of construction:

(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 7 points $Q _1, Q _2, Q _3, Q _4, Q _5, Q _6, Q _7$ on QX .
So that $Q_1=Q_1 Q_2=Q_2 Q_3=Q_3 Q_4=Q_5 Q_6=Q_6 Q_7$

(iv) Join $Q_3$ to $R$ and draw a line through $Q_7$ parallel to $Q_3 R$ intersecting the extended line segment $Q R$ at $R^{\prime}$.
(v) Draw a line parallel to RP. Intersecting the extended line segment QP at P'.
$\therefore \triangle P ^{\prime} QR$ ' is the required triangle.

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Question 525 Marks

Construct a triangle similar to a given triangle $A B C$ with its sides equal to $\frac{6}{5}$ of the corresponding sides of the triangle $ABC$ (scale factor $\frac{6}{5}>1$ )

Answer

Given triangle $\triangle A B C$, we are required to construct another triangle whose sides are $\frac{6}{5}$ of the corresponding sides of the $\triangle A B C$.

Steps of construction:
(i) Construct an $\triangle ABC$ with any measurement.
(ii) Draw a ray $BX$ making an acute angle with $BC$.
(iii) Locate 6 points $Q _1, Q _2, Q _3, Q _4, Q _5, Q _6$ on $B X$ such that $B Q_1=Q_1 Q_2=Q_2 Q_3=Q_3 Q_4=Q_5 Q_6$
(iv) Join $Q_5$ to $C$ and draw a line through $Q_6$ parallel to $Q_5 C$ intersecting the extended line $BC$ at $C ^{\prime}$.
(v) Draw a line through $C^{\prime}$ parallel to $A C$ intersecting the extended line segment $A B$ at $A^{\prime}$.
$\therefore \triangle A ^{\prime} B C ^{\prime}$ is the required triangle.

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Question 535 Marks

Construct a triangle similar to a given triangle LMN with its sides equal to $\frac{4}{5}$ of the corresponding sides of the triangle LMN (scale factor $\frac{4}{5}<1$ )

Answer

Given a triangle LMN, we are required to construct another triangle whose sides are $\frac{4}{5}$ of the corresponding sides of the $\triangle LMN$.

Steps of Construction:
1. Construct a $\triangle LMN$ with any measurement.
2. Draw a ray $M X$ making an acute angle with $M N$ on the side opposite to the vertex $L$.
3. Locate 5 Points $Q _1, Q _2, Q _3, Q _4, Q _5$ on $MX$.
So that $MQ _1= Q _1 Q _2= Q _2 Q _3= Q _3 Q _4= Q _4 Q _5$
4. Join $Q _5 N$ and draw a line through $Q _4$. Parallel to $Q _5 N$ to intersect $MN$ at $N ^{\prime}$.
5. Draw a line through $N^{\prime}$ parallel to the line $L N$ to intersect $M L$ at $L^{\prime}$.
$\therefore \Delta L ^{\prime} MN ^{\prime}$ is the required triangle.

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Question 545 Marks

Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{2}{3}$ of the corresponding sides of the triangle PQR (scale factor $\frac{2}{3}<1$ )

Answer

Given $\triangle P Q R$, we are required to construct another triangle whose sides are $\frac{2}{3}$ of the corresponding sides of the $\triangle P Q R$

Steps of construction:
(i) Construct a $\triangle PQR$ with any measurement.
(ii) Draw a ray $QX$ making an acute angle with $QR$ on the side opposite to the vertex $P$.
(iii) Locate 3 points $Q _1, Q _2$ and $Q _3$ on $QX$. So that $QQ _1= Q _1 Q _2= Q _2 Q _3$
(iv) Join $Q_3 R$ and draw a line through $Q_2$ parallel to $Q_3 R$ to intersect $Q R$ at $R^{\prime}$.
(v) Draw a line through $R^{\prime}$ parallel to the line $R P$ to intersect $Q P$ at $P^{\prime}$. Then $\triangle P^{\prime} Q R^{\prime}$ is the required triangle.

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MATHS [5 Mark Questions]