4 Marks Questions

Question 14 Marks

Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.

Answer

We are given two triangles $\mathrm{ABC}$ and $\mathrm{DEF}$ in which:
and
$
\begin{aligned}
\angle \mathrm{B} & =\angle \mathrm{E}, \angle \mathrm{C}=\angle \mathrm{F} \\
\mathrm{BC} & =\mathrm{EF} \\
\triangle \mathrm{ABC} & \cong \triangle \mathrm{DEF}
\end{aligned}
$
We need to prove that
For proving the congruence of the two triangles see that three cases arise.
Case (i) : Let $A B=D E$ (see Fig. 7.12).
Now what do you observe? You may observe that
$\begin{aligned}
\mathrm{AB} =\mathrm{DE} \quad (Assumed) \\
\angle \mathrm{B} =\angle \mathrm{E} \quad (Given)\\
\mathrm{BC} =\mathrm{EF} \quad (Given) \\
So, \quad \triangle \mathrm{ABC} \cong \triangle \mathrm{DEF}\quad (By SAS rule)
\end{aligned}$

Case (ii): Let if possible AB > DE. So, we can take a point P on AB such that PB = DE. Now consider A PBC and A DEF (see Fig. 7.13).

Observe that in $\triangle \mathrm{PBC}$ and $\triangle \mathrm{DEF}$,
$
\begin{aligned}
\mathrm{PB} & =\mathrm{DE} \quad \text{(By construction)} \\
\angle \mathrm{B} & =\angle \mathrm{E} \quad (Given) \\
\mathrm{BC} & =\mathrm{EF} \quad (Given)
\end{aligned}
$
So, we can conclude that:
$\Delta \mathrm{PBC} \cong \triangle \mathrm{DEF}$, by the SAS axiom for congruence.
Since the triangles are congruent, their corresponding parts will be equal.
So,$\angle \mathrm{PCB}=\angle \mathrm{DFE}$
But, we are given that
$
\begin{aligned}
& \angle \mathrm{ACB}=\angle \mathrm{DFE} \\
& So, \quad \angle \mathrm{ACB}=\angle \mathrm{PCB}
\end{aligned}
$
Is this possible?
This is possible only if $\mathrm{P}$ coincides with $\mathrm{A}$.
$
\begin{aligned}
or, \quad \mathrm{BA} & =\mathrm{ED} \\
So, \quad \triangle \mathrm{ABC} & \cong \triangle \mathrm{DEF}
\end{aligned}
$
(by SAS axiom)
Case (iii) : If $A B

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Question 24 Marks

$\triangle$ABC and $\triangle$DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that :

$\triangle$ABD $\cong $ $\triangle$ACD
$\triangle$ABP $\cong $ $\triangle$ACP
AP bisects $\angle$A as well as $\angle$D
AP is the perpendicular bisector of BC.

Answer

It is given in the question that:
$\triangle$ ABC and $\triangle$DBC are two isosceles triangles

$\triangle$ ABC and $\triangle$DBC
AD = AD (Common)
AB = AC (Triangle ABC is isosceles)
BD = CD (Triangle DBC is isosceles)
By SSS axiom,
$\triangle A B D \cong \triangle A C D$
In $\triangle$ ABP and $\triangle$ACP,
AP = AP (Common)
$\angle$PAB = $\angle$PAC (By c.p.c.t)
AB = AC (Triangle DBC is isoceles)
By SAS axiom,
$\triangle A B P \cong \triangle A C P$
$\angle$PAB = $\angle$PAC { c.p.c.t }
AP bisects $\angle$A (i)
Also,
In $\triangle$ BPD and $\triangle$CPD,
PE = PD (Common)
BD = CD (Triangle DBC is isosceles)
BP = CP ($\triangle A B P \cong \triangle A C P$ so by c.p.c.t)
By SSS axiom,
$\triangle B PD \cong \triangle C PD$
Thus,
$\angle$BDP = $\angle$CDP { c.p.c.t } (ii)
By (i) and (ii), we can say that $\angle$AP bisects $\angle$A as well as D
$\angle$BPD = $\angle$CPD (By c.p.c.t)
And,
BP = CP (i)
Also,
$\angle$BPD = $\angle$CPD = $180^\circ$ (BC is a straight line)
2$\angle$BPD = $180^\circ$
$\angle$BPD = $90^\circ$ (ii)
From (i) and (ii), we get
AP is the perpendicular bisector of BC

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Question 34 Marks

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Answer

Given,
BE and CF are altitudes.
AC = AB To show,
BE = CF
In $\triangle\text{AEB}$ and $\triangle\text{AFC}$
$\angle \text{A}=\angle\text{A}$ (Common)
$\angle\text{AEB}=\angle \text{AFC}$ (Right angles)
AB = AC (Given)
Therefore, $\triangle\text{AEB}\cong \angle\text{AFC}$ by AAS congruence condition.
Thus, BE = CF (by CPCT).

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Question 44 Marks

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. (See figure)

Show that:

$\triangle$AMC $ \cong$ $\triangle$BMD
$\angle $DBC is a right angle.
$\triangle$DBC $ \cong$ $\triangle$ACB
CM = $\frac{1}{2}$ AB

Answer

In $\triangle$AMC and $\triangle$and BMD,
AM = BM [M is the mid-point of AB]
$\angle $AMC =$\angle $ BMD [Vertically opposite angles]
CM = DM [Given]
$\therefore $ $\triangle$AMC $ \cong$ $\triangle$BMD [By SAS congruency]
$\therefore $ $\angle $ ACM =$\angle $ BDM ...(i)
For two lines AC and DB and transversal DC, we have,
$\angle $ACD = $\angle $BDC [Alternate angles]
$\therefore $ AC $\parallel $ DB
Now for parallel lines AC and DB and for transversal BC.
$\angle $ DBC = $\angle $ ACB [Alternate angles] ...(ii)
But$ \Delta $ ABC is a right angled triangle, right angled at C.
$\therefore $ $\angle $ACB = 90° (iii)
Therefore $\angle $ DBC = 90° [Using eq. (ii) and (iii)]
$ \Rightarrow $$\angle $ DBC is a right angle.
Now in $\triangle$DBC and $\triangle$ ABC,
DB = AC [Proved in part (i)]
$\angle $ DBC = $\angle $ACB = 90° [Proved in part (ii)]
BC = BC [Common]
$\therefore $ $\triangle$DBC $ \cong$ $\triangle$ACB [By SAS congruency]
Since $\triangle$DBC $ \cong$ $\triangle$ACB [Proved above]
$\therefore $ DC = AB
$ \Rightarrow $ AM + CM = AB
$ \Rightarrow $ CM + CM = AB [$\because $ DM = CM]
$ \Rightarrow $ 2CM = AB
$ \Rightarrow $ CM = $\frac{1}{2}$ AB

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