3 Marks Question

Question 13 Marks

In the given figure, line $l$ is the bisector of an angle $\angle\text{A}$ and $B$ is any point on $l$. If $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle\text{A},$ Show that:
$i. \triangle\text{APB}\cong\triangle\text{AQB}$
$ii. BP = BQ,$ i.e., $B$ is equidistant from the arms of $\angle\text{A}.$

Answer

$i.$ In $\triangle\text{APB}$ and $\triangle\text{AQB},$
$\angle\text{APB}=\angle\text{AQC} ($Each $90^\circ )$
$\angle\text{BAP}=\angle\text{BAQ}$ $\big($line $l$ is the bisector of $\angle\text{A}\big)$
$\text{AB = AB} ($common$)$
$\therefore\triangle\text{APB}\cong\triangle\text{AQB}( $by $\text{AAS}$ congruence criterion$)$
$ii.$ Since $\triangle\text{APB}\cong\triangle\text{AQB},$
$\text{BP = BQ} (\text{C.P.C.T.})$

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Question 23 Marks

In the given figure, $AD$ and $BC$ are equal perpendiculars to a line segment $AB$. Show that $CD$ bisects $AB.$

Answer

In $\triangle\text{AOD}$ and $\triangle\text{BOC},$
$\angle\text{AOD}=\angle\text{BOC}$ (vertically opposite angle)
$\angle\text{DAO}=\angle\text{CBO} ($Each $90^\circ )$ $\text{AD = BC}$ (given)
$\therefore\triangle\text{AOD}\cong\triangle\text{BOC} ($by $AAS$ congruence criterion$)$
$\Rightarrow\text{AO = BO} (C.P.C.T.)$
$\Rightarrow\text{CD bisects AB}.$

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Question 33 Marks

$AD$ is an altitude of an isosceles $\triangle\text{ABC}$ in which $AB = AC.$ Show that:
$i. AD$ bisects $BC,$
$ii. AD$ bisects $\angle\text{A}.$

Answer

$i.$ In $\triangle\text{BAD}$ and $\triangle\text{CAD}$
$\angle\text{ADB}=\angle\text{ADC} ($Each $90^\circ $ as $AD$ is an altitude$)$
$\text{AB = AC} ($given$)$
$\text{AD = AD} ($common$)$
$\therefore\triangle\text{BAD}\cong\triangle\text{CAD} ($by $\text{RHS}$ Congruence criterion$)$
$\Rightarrow\text{BD = CD}(\text{C.P.C.T.})$
Hence $AD$ bisects $BC.$
$ii.$ Also, $\angle\text{BAD}=\angle\text{CAD}(\text{C.P.C.T.})$
Hence, $AD$ bisects $\angle\text{A}.$

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Question 43 Marks

The bisectors of $\angle\text{B}$ and $\angle\text{C}$ of an isosceles triangle with $AB = AC$ intersect each other at a point $O. BO$ is produced to meet $AC$ at a point $M.$ Prove that $\angle\text{MOC}=\angle\text{ABC}.$

Answer

In $\triangle\text{ABC},\text{ AB = AC}$
$\Rightarrow\angle\text{ABC}=\angle\text{ACB}$
$\Rightarrow\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ACB}$
$\angle\text{OBC}=\angle\text{OCB}...\text{(i)}$
Now, by exterior angle property,
$\angle\text{MOC}=\angle\text{OBC}+\angle\text{OCB}$
$\Rightarrow\angle\text{MOC}=2\angle\text{OBC} [$from $(i)]$
$\Rightarrow\angle\text{MOC}=\angle\text{ABC}$
$\big(OB$ is the bisector of $\angle\text{ABC}\big)$

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Question 53 Marks

In the given figure, $PQ > PR$ and $QS$ and $RS$ are the bisectors of $\angle\text{Q}$ and $\angle\text{R}$ respectively. Show that $SQ > SR.$

Answer

Since the angle opposite to the longer side is greater, we have: $\text{PQ > PR}$
$\Rightarrow\angle\text{R}>\angle\text{Q}$
$\Rightarrow\frac{1}{2}\angle\text{R}>\angle\text{Q}$
$\Rightarrow\angle\text{SRQ}>\angle\text{RQS}$
$\Rightarrow\text{QS > SR}$
$\therefore\text{SQ > SR}$

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Question 63 Marks

$ABCD$ is a quadrilateral such that diagonal AC bisects the angles $\angle\text{A}$ and $\angle\text{C}.$ Prove that $AB = AD$ and $CB = CD.$

Answer

In $\triangle\text{ABC}$ and $\triangle\text{ADC},$
$\angle\text{BAC}=\angle\text{DAC}$
$\big($AC bisects $\angle\text{A}\big)$
$\text{AC = AC}$ (common)
$\angle\text{BCA}=\angle\text{DCA}$
$\big($AC bisects $\angle\text{C}\big)$
$\therefore\triangle\text{ABC}\cong\triangle\text{ADC} ($by $ASA$ congruence criterion$)$
$\Rightarrow\text{AB = AD}$ and $\text{CB = CD} (C.P.C.T.)$

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Question 73 Marks

In the given figure, $BE$ and $CF$ are two equal altitudes of $\triangle\text{ABC}.$
Show that:
$i. \triangle\text{ABE}\cong\triangle\text{ACF},$
$ii. AB = AC.$

Answer

$i.$ In $\triangle\text{ABE}$ and $\triangle\text{ACF},$
$\angle\text{AEB}=\angle\text{AFC} ($Each $90^\circ)$
$\text{BE = CF}$ (given)
$\angle\text{BAE}=\angle\text{CAF}$
$\big($common$\angle\text{A}\big)$
$\therefore\triangle\text{ABE}\cong\triangle\text{ACF} ($by $\text{ASA}$ congruence criterion$)$
$ii.$ Since $\triangle\text{ABE}\cong\triangle\text{ACF},$
$\text{AB = AC} (\text{C.P.C.T.})$

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Question 83 Marks

In the given figure, two parallel line $l$ and $m$ are intersected by two parallel lines $p$ and $q.$
Show that $\triangle\text{ABC}\cong\triangle\text{CDA}.$

Answer

In $\triangle\text{ABC}$ and $\triangle\text{CDA}$
$\angle\text{BAC}=\angle\text{DCA}$
$($alternate interior angles for $p || q)$
$\text{AC = CA}$ (common) $\angle\text{BCA}=\angle\text{DAC}$
$($Alternate interior angles for $l || m)$
$\therefore\triangle\text{ABC}\cong\triangle\text{CDA}$
$($by $ASA$ congruence rule$)$

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Question 93 Marks

In $\triangle\text{ABC},\angle\text{A}=50^{\circ}$ and $\angle\text{B}=60^{\circ}$ Determine the longest and shortest sides of the triangle.

Answer

Given: $\triangle\text{ABC},\angle\text{A}=50^{\circ}$ and $\angle\text{B}=60^{\circ}$ In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ (Angle sum property of a triangle)
$\Rightarrow50^{\circ}+60^{\circ}+\angle\text{C}=180^{\circ}$
$\Rightarrow110^{\circ}+\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{C}=180^{\circ}-110^{\circ}$
$\Rightarrow\angle\text{C}=70^{\circ}$
Hence, the longest side will be opposite to the largest angle $\big(\angle\text{C}=70^{\circ}\big)$
i.e. $AB$. And,
the shortest side will be opposite to the smallest angle $\big(\angle\text{A}=50^{\circ}\big)$ i.e. $BC.$

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Question 103 Marks

In $\triangle\text{ABC, }\angle\text{B}=35^{\circ},\angle\text{C}=65^{\circ}$ and the bisector of $\angle\text{BAC}$ meets $BC$ in $X.$ Arrange $AX, BX$ and $CX$ in descending order.

Answer

Given: in $\triangle\text{ABC, }\angle\text{B}=35^{\circ},\angle\text{C}=65^{\circ}$ and the bisector of $\angle\text{BAC}$ meets $BC$ in $x$ In $\triangle\text{ABX},$
$\because\angle\text{BAX}>\angle\text{ABX}$
$\therefore\text{BX}>\text{AX}...(\text{i})$
$\therefore\text{AX}>\text{CX}...(\text{ii})$ From $(i)$ and $(ii),$ we get $\text{BX > AX > CX}$

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Question 113 Marks

In the given figure, $AB \| CD$ and $O$ is the modpoint of $AD.$
Show that:
$i. \triangle\text{AOB}\cong\triangle\text{DOC}.$
$ii. O$ is the midpoint of $BC.$

Answer

$i.$ In $\triangle\text{AOB}$ and $\triangle\text{DOC},$
$\angle\text{BAO}=\angle\text{CDO} (AB \| CD, $ alternate angle$)$
$\text{AO = DO}$
$(O$ is the mid$-$point of $AD)$
$\angle\text{AOB}=\angle\text{DOC} ($vertically opposite angle$)$
$\therefore\triangle\text{AOB}\cong\triangle\text{DOC} ($by $\text{ASA}$ congruence criterion$)$
$ii.$ Since $\triangle\text{AOB}\cong\triangle\text{DOC},$
$\text{BO = CO} (\text{C.P.C.T.})$
$\Rightarrow O$ is the mid$-$point of $BC.$

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