[2 Mark Question Answer]

Question 12 Marks

Use the information in the given figure to prove:$(i) \mathrm{AB}=\mathrm{FE}\ (ii) \ \mathrm{BD}=\mathrm{CF}$

Answer

ln $\triangle ABC$ and $\triangle EFD,$
$AB II EF \Rightarrow \angle ABC = \angle EFD...($ alternate angles $)$
$AC = ED...($ given $)$
$\angle ACB = \angle EDF...($ given $)$
$\therefore \triangle ABC ≅\triangle EFD....( AAS$ congruence criterion $)$
$\Rightarrow AB = FE...($ cpct $)$
and $BC= DF....($ cpct $)$
$\Rightarrow BD + DC = CF + DC.....( B-D-C-F )$
$\Rightarrow BD = CF$

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Question 22 Marks

In the following figure$, A B=A C$ and $A D$ is perpendicular to $B C$. $B E$ bisects angle $B$ and $E F$ is perpendicular to $A B$. Prove that$: \mathrm{ED}=\mathrm{EF}$

Answer

In $\triangle EFB $ and $\triangle EDB,$
$\angle EFB = \angle EDB ($ both are $90^\circ )$
$EB = EB ($ common side $)$
$\angle FBE = \angle DBE ( $ given$ )$
$\triangle EFB ≅ \triangle EDB(AAS$ congruence criterion$)$
$\Rightarrow EF = ED ($cpct$ )$
that is $, Ed = EF.$

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Question 32 Marks

In the following figure $, A B=A C$ and $A D$ is perpendicular to $B C$. $B E$ bisects angle $B$ and $E F$ is perpendicular to $A B.$ Prove that $: \mathrm{BD}=\mathrm{CD}$

Answer

In $\triangle ADB$ and $\triangle ADC,$
$\angle ADB = \angle ADC....($Since $AD$ is perpendicular to $BC)$
$AB = AC....($given$)$
$AD = AD....($common side$)$
$\therefore \triangle ADB ≅ \triangle ADC....(RHS$ congruence criterion$)$
$\Rightarrow BD = CD....($cpct$)$

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Question 42 Marks

In the following figure $, \mathrm{BL}=\mathrm{CM}.$

Prove that $AD$ is a median of triangle $ABC$.

Answer

In $\triangle DLB$ and $\triangle DMC,$
$BL = CM...($ given $)$
$\angle DLB = \angle DMC...($ Both are $90^\circ)$
$\angle BDL = \angle CDM....($ vertically opposite angels $)$
$\therefore \triangle DLB ≅ \triangle DMC....( AAS$ congruence criterion $)$
$BD = CD....($ cpct $)$
Hence $, AD$ is the median of $\triangle ABC.$

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MATHEMATICS [2 Mark Question Answer]

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