[2 Mark Question Answer]

Question 12 Marks

Given is a triangle with sides $3 \ cm, 5 \ cm$ and $6 \ cm.$ Find the sides of a triangle which is similar to the given triangle and its shortest side is $4.5 \ cm.$

Answer

Since the two triangles are similar, so the ratio of the corresponding sides are equal.
Let $x$ and $y$ be the sides of the triangle, where $y$ is the longest side.
$\frac{3}{5}=\frac{4.5}{x}$
$\Rightarrow x =7.5\ cm$
$\frac{5}{6}=\frac{7.5}{y}$
$\Rightarrow y =9 \ cm$
So, the sides of the triangles are $4.5\ cm , 7.5\ cm$ and $9\ cm$.

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Question 22 Marks

A model of a ship is made to a scale of $1:500.$ Find: The length of the ship, if length of the model is $1.2.$

Answer

Scale $= 1:500$
The length of the model $= 11.2\ m$
The actual length
$= 1.2 \times 500$
$= 600\ m.$

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Question 32 Marks

A model of cargo tuck is made to a scale of $1:40.$ The length of the model is $15\ cm.$ Calculate: The length of the truck

Answer

Scale $=1: 40$
The length of the model $=15\ cm$
The actual length
$=15 \times 40 $
$=600 \ cm $
$=\frac{600}{100} $
$=6\ m .$

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Question 42 Marks

$\triangle ABC$ is enlarged, with a scale factor $5.$ Find $:\text{BC}, \text{B'C'} = 16\ cm$

Answer

$\frac{\text { Image length }}{\text { Actual length }}=\text { Scale factor }$
$\frac{ B ^{\prime} C ^{\prime}}{ BC }=5$
$\Rightarrow \text{BC}=\frac{16}{5}$
$\Rightarrow \text{BC} =3.2\ cm .$

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Question 52 Marks

$\triangle ABC$ is enlarged, with a scale factor $5.$ Find $:\text{ A'B'},$ if $\text{AB} = 4\ cm$

Answer

$\frac{\text { lmage length }}{\text { Actual length }}=\text { Scale factor }$
$\frac{A^{\prime} B^{\prime}}{A B}=5$
$\Rightarrow A^{\prime} B^{\prime}=4 \times 5$
$\Rightarrow A^{\prime} B^{\prime}=20 \ cm .$

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Question 62 Marks

$\triangle ABC$ has been reduced by a scale factor $0.6$ to $\triangle A'B'C'$ Calculate: Length of $\text{AB}$, if $A'B' = 5.4\ cm$

Answer

$\frac{\text { Image length }}{\text { Actual length }}=\text { Scale factor }$
$\frac{A^{\prime} B^{\prime}}{8}=0.6$
$\Rightarrow \text{AB}=\frac{5.4}{0.6}$
$\Rightarrow \text{AB}=9 \ cm$

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Question 72 Marks

$\triangle ABC$ has been reduced by a scale factor $0.6$ to $\triangle A'B'C'/$ Calculate:Length of $\text{B' C'},$ if $\text{BC} = 8\ cm$

Answer

$\frac{\text { lmage length }}{\text { Actual length }}=\text { Scale factor }$
$\frac{ B ^{\prime} C ^{\prime}}{8}=0.6$
$\Rightarrow B ^{\prime} C ^{\prime}=8 \times 0.6$
$\Rightarrow B ^{\prime} C ^{\prime}=4.8 \ cm .$

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Question 82 Marks

Given that $\triangle ABC \sim \triangle PRQ,$ name the corresponding angles and the corresponding sides.

Answer

$\triangle ABC \sim \triangle PRQ$
$A \leftrightarrow P, B \leftrightarrow R, C \leftrightarrow Q$
$\angle A \sim \angle P$
$\angle B \sim \angle R$
$\angle C \sim \angle Q$
$\text{AB} \sim \text{PR}, \text{BC} \sim \text{RQ}, \text{AC}\sim \text{PQ}.$

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Question 92 Marks

In a quadrilateral $\text{PQRS},$ the diagonals $\text{PR}$ and $\text{QS}$ intersect each other at the point $T.$ If $\text{PT}:\text{TR} = \text{QT} :\text{TS} = 1:2,$ show that $\text{TP}:\text{TQ} = \text{TR}:\text{TS}$

Answer

Consider $\triangle PTQ$ and $\triangle RTS$,
$\frac{ PT }{ TR }=\frac{ QT }{ TS }=\frac{1}{2} \ldots ($Given$)$
$\angle PTQ =\angle RTS \ldots ($Vertically Opposite angles$) $
$\Rightarrow \triangle PTQ \sim \triangle RTS \ldots \text{SAS}$ criterion for Similarity$) $
$\Rightarrow \frac{ TP }{ TQ }=\frac{ TR }{ TS } \ldots ($Rearranging the terms$) $

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Question 102 Marks

$\text{PQ}$ is perpendicular to $\text{BA}$ and $\text{BD}$ is perpendicular to $\text{AP} . \text{PQ}$ and $\text{BD}$ intersect at $R$. Prove that $\triangle ABD \sim \triangle APQ$ and $\frac{ AB }{ AP }=\frac{ BD }{ PQ }$.

Answer

In $\triangle A B D $ and $\triangle A P Q$
$\angle B D A\angle P Q A=90^{\circ} $
$\angle A=\angle A$
Therefore, $\triangle ABD \sim \triangle APQ...(AA$ axiom$)$
And hence, $\frac{ AB }{ AP }=\frac{ BD }{ PQ }$.

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