[3 marks sum]

Question 13 Marks

In the following figure, $\text{AB}=\text{EF},\text{BC}=\text{DE}$ and $\angle B=\angle E=90^{\circ}$.

Prove that $\text{AD}=\text{FC}$

Answer

Given that, $\mathrm{BC}=\mathrm{DE}$
$\Rightarrow \text{BC}+\text{CD}=\text{DE}+\text{CD} \ldots .($ Adding $\text{CD}$ on both sides $)$
$\Rightarrow \mathrm{BD}=\mathrm{CE} ....(i)$
Now, in $\triangle \mathrm{ABD}$ and $\triangle \mathrm{FEC}$,
$\mathrm{AB}=\mathrm{EF} ....($given$)$
$\angle \mathrm{ABD}=\angle \mathrm{FEC}$
$\ldots\left(\right.$ Each $\left.90^{\circ}\right)$
$\text{BD}=\text{CE} ....[$ From $(i) ]$
$\Rightarrow \triangle \mathrm{ABD} \cong \triangle \mathrm{FEC} ...($by $\text{SAS}$ congruence criterion$)$
$\Rightarrow \text{AD}=\text{FC} ...($c.p.c.t.$)$

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Question 23 Marks

$\text{AD}$ and $\text{BC}$ are equal perpendiculars to a line segment $\text{AB}$. If $\text{AD}$ and $\text{BC}$ are on different sides of $\text{AB}$ prove that $\text{CD}$ bisects $\text{AB}.$

Answer

In $\triangle AOD$ and $\triangle BOC$,
$\angle \mathrm{AOD}=\angle \mathrm{BOC} \quad \ldots. ($vertically opposite angles$)$
$\angle \mathrm{DAO}=\angle \mathrm{CBO} \quad \ldots . .\left(\right.$ each $\left.90^{\circ}\right)$
$\text{AD}=\text{BC} \quad \quad \ldots. ($given$)$
$\therefore \triangle \mathrm{AOD} \cong \triangle \mathrm{BOC} \quad...($by $\text{AAS}$ congruence criterion$)$
$\Rightarrow \text{AO}=\text{BO} ...($c.p.c.t$.)$
$\Rightarrow O$ is the mid$-$point of $\text{AB}$. Hence, $\text{CD}$ bisects $\text{AB}$.

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Question 33 Marks

If the following pair of the triangle is congruent? state the condition of congruency:In $\triangle ABC$ and $\triangle PQR, \text{AB} = \text{PQ}, \text{AB} = \text{PQ},$ and $\text{BC} = \text{QR}.$

Answer

In $\triangle ABC$ and $\triangle PQR$
$\mathrm{AB}=\mathrm{PQ}[$Given$]$
$ \mathrm{AB}=\mathrm{PQ}[$Given $]$
$ \mathrm{BC}=\mathrm{QR}[$ Given $]$

By Side$-$ Side $-$ Side criterion of congruency, the triangles $\triangle ABC$ and $\triangle PQR$ are congruent to each other.
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}$

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Question 43 Marks

If the following pair of the triangle is congruent? state the condition of congruency:In $\triangle ABC$ and $\triangle QRP, \text{AB} = \text{QR}, \angle B = \angle R$ and $\angle C = P.$

Answer

In $\triangle ABC$ and $\triangle QRP$
$\angle B=\angle R [$ Given $]$
$\angle \mathrm{C}=\angle \mathrm{P} [$ Given $]$
$\mathrm{AB}=\mathrm{QR}[$ Given $]$

By Angel$-$Angel SIde criterion of congruency, the triangles $\triangle ABC$ and $\triangle QRP$ are congruent to each other.
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{QRP}$

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Question 53 Marks

In $\triangle ABC,\text{AB} = \text{AC}.$ Show that the altitude $\text{AD}$ is median also.

Answer

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ADC}$,
$\mathrm{AB}=\mathrm{AC} \quad.... ($ Since is ani isosceles triangle$)$
$\mathrm{AD}=\mathrm{AD} \quad..... ($ common side $)$
$\angle ADB=\angle ADC \quad \ldots .\left(\right.$ Since $\text{AD}$ is the altitude so each is $\left.90^{\circ}\right)$
$\Rightarrow \triangle ADB \cong \triangle ADC \quad..... (\text{RHS}$ congruence criterion $)$
$\mathrm{BD}=\mathrm{DC} \quad.... ($cpct$)$
$\Rightarrow \text{AD}$ is the median.

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Question 63 Marks

If the following pair of the triangle is congruent? state the condition of congruency :In $\triangle ABC$ and $\triangle DEF,\angle B = \angle E = 90^o; \text{AC} =\text{DF}$ and $\text{BC} =\text{EF}.$

Answer

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$
$\angle \mathrm{B}=\angle \mathrm{E}=90^{\circ}$
$HYp. \text{AC}= HYp. \text{DF}$

$\mathrm{BC}=\mathrm{EF}$
By Right Angel$-$ Hypotenuse$-$Side Criterion of congruency, the triangles
$\triangle ABC$ and $\triangle DEF$ are congruent to each other.
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{DEF}$

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Question 73 Marks

If the following pair of the triangle is congruent? state the condition of congruency :In $\triangle ABC$ and $\triangle DEF, \text{AB} = \text{DE}, \text{BC} =\text{ EF}$ and $\angle B = \angle E.$

Answer

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$,
$\mathrm{AB}=\mathrm{DE} \quad \ldots [$ Given $]$
$\angle B=\angle \mathrm{E} \quad...[$ Given$]$
$\mathrm{BC}=\mathrm{EF} \quad \ldots[$ Given $]$

By Side $-$ Angle $-$ Side criterion of congruency, the triangles $\triangle ABC$ and $\triangle DEF$ are congruent to each other.
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{DEF}$

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MATHEMATICS [3 marks sum]

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