[5 marks sum]

Question 15 Marks

From the following figure;

prove that:$(i) AB > BD;(ii) A C>C D;(iii) A B+A C>B C$.A

Answer

$(i)\angle ADC + \angle ADB = 180^\circ\dots ...[ \text{BDC}$ is a straight line $]$
$\angle ADC = 90^\circ \dots...[$ Given $]$
$90^\circ + \angle ADB = 180^\circ$
$\angle ADB = 90^\circ\dots ....(i)$
In $\triangle ADB,$
$\angle ADB = 90^\circ\dots ....[$ From $(i) ]$
$\therefore \angle B + \angle BAD = 90^\circ$
Therefore,$ \angle B$ and $\angle BAD$ are both acute, that is less than $90^\circ .$
$\therefore AB > BD \dots….(ii)[$ Side opposite $90^\circ$ angle is greater than the side opposite acute angle $]$
$(ii)$ In $ \triangle ADC,$
$\angle ADB = 90^\circ$
$\therefore \angle C + \angle DAC = 90^\circ$
Therefore,$\angle C$ and $\angle DAC$ are both acute, which is less than $90^\circ .$
$\therefore AC > CD \dots...(iii)[$ Side opposite $90^\circ$ angle is greater than side opposite acute angle $]$
Adding $(ii)$ and $(iii)$
$AB + AC > BD + CD$
$\Rightarrow AB + AC > BC$

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Question 25 Marks

In the following figure $; A C=C D ; \angle B A D=110^{\circ}$ and $\angle A C B=74^{\circ}$.

Prove that: $BC > CD$.

Answer

$\angle ACB = 74^\circ \dots...(i)[$ Given $]$
$\angle ACB + \angle ACD = 180^\circ \dots....[ \text{BCD}$ is a straight line $]$
$\Rightarrow 74^\circ + \angle ACD = 180^\circ $
$\Rightarrow \angle ACD = 106^\circ \dots…..(ii)$
In $\triangle ACD,$
$\angle ACD + \angle ADC+ \angle CAD = 180^\circ $
Given that $AC = CD$
$\Rightarrow \angle ADC= \angle CAD$
$\Rightarrow 106^\circ + \angle CAD + \angle CAD = 180^\circ \dots....[$From $(ii)]$
$\Rightarrow 2\angle CAD = 74^\circ $
$\Rightarrow \angle CAD = 37^\circ = \angle ADC \dots...(iii)$
Now,
$\angle BAD = 110^\circ\dots ....[$Given$]$
$\angle BAC + \angle CAD = 110^\circ $
$\angle BAC + 37^\circ = 110^\circ $
$\angle BAC = 73^\circ \dots….(iv)$
In $\text{ABC},$
$\angle B + \angle BAC + \angle ACB = 180^\circ $
$\angle B + 73^\circ + 74^\circ = 180^\circ \dots...[$From $(i)$ and $(iv)]$
$\angle B + 147^\circ = 180^\circ $
$\angle B = 33^\circ …..(v)$
$\therefore \angle BAC > \angle B ...[$ From $(iv)$ and $(v)]$
$\Rightarrow BC > AC$
But,
$AC = CD\dots ...[ $Given $]$
$\Rightarrow BC > CD$

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Question 35 Marks

In the following figure, $\angle B A C=60^{\circ}$ and $\angle A B C=65^{\circ}$.

Prove that:$(i) CF > AF;(ii)DC > DF$

Answer

In $\triangle BEC,$
$\angle B + \angle BEC + \angle BCE = 180^\circ $
$\angle B = 65^\circ\dots ...[$Given$]$
$\angle BEC = 90^\circ\dots ...[CE$ is perpendicular to $AB]$
$\Rightarrow 65^\circ + 90^\circ + \angle BCE = 180^\circ $
$\Rightarrow \angle BCE = 180^\circ - 155^\circ $
$\Rightarrow \angle BCE = 25^\circ = \angle DCF\dots …(i)$
In $\triangle CDF,$
$\angle DCF + \angle FDC + \angle CFD = 180^\circ $
$\angle DCF = 25^\circ \dots....[$From $(i)]$
$\angle FDC = 90^\circ\dots ...[ AD$ is perpendicular to $BC]$
$\Rightarrow 25^\circ + 90^\circ + \angle CFD = 180^\circ $
$\Rightarrow \angle CFD = 180^\circ - 115^\circ $
$\Rightarrow \angle CFD = 65^\circ \dots…(ii)$
Now, $\angle AFC + \angle CFD = 180^\circ \dots....[\text{AFD}$ is a straight line$]$
$\Rightarrow \angle AFC + 65^\circ = 180^\circ $
$\Rightarrow \angle AFC = 115^\circ \dots…(iii)$
In $\triangle ACE,$
$\angle ACE + \angle CEA + \angle BAC = 180^\circ $
$\angle BAC = 60^\circ\dots ....[$Given$]$
$\Rightarrow \angle CEA = 90^\circ \dots...[CE$ is perpendicular to $AB]$
$\Rightarrow \angle ACE + 90^\circ + 60^\circ = 180^\circ $
$\Rightarrow \angle ACE = 180^\circ - 150^\circ $
$\angle ACE = 30^\circ\dots …(iv)$
In $\triangle AFC,$
$\angle AFC + \angle ACF + \angle FAC = 180^\circ $
$\angle AFC = 115^\circ\dots ....[$From $(iii)]$
$\angle ACF = 30^\circ\dots ...[$From $(iv)]$
$\Rightarrow 115^\circ + 30^\circ + \angle FAC = 180^\circ $
$\Rightarrow \angle FAC = 180^\circ - 145^\circ $
$\Rightarrow \angle FAC = 35^\circ\dots …(v)$
In $\triangle AFC,$
$\Rightarrow \angle FAC = 35^\circ \dots...$[ From $(v) ]$
$\Rightarrow \angle ACF = 30^\circ \dots...$[ From $(iv) ]$
$\therefore \angle FAC > \angle ACF$
$\Rightarrow CF > AF$
In $\triangle CDF,$
$\angle DCF = 25^\circ \dots...[$From $(i)]$
$\angle CFD = 65^\circ\dots ...[$From $(ii)]$
$\therefore \angle CFD > \angle DCF$
$\Rightarrow DC > DF$

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Question 45 Marks

Arrange the sides of $\triangle BOC$ in descending order of their lengths. $BO$ and $C O$ are bisectors of angles $\text{ABC}$ and $\text{ACB}$ respectively.

Answer

$\angle BAC =180^{\circ}-\angle BAD =180^{\circ}-137^{\circ}=43^{\circ}$
$ \angle ABC =180^{\circ}-\angle ABE =180^{\circ}-106^{\circ}=74^{\circ}$
Thus, in $\triangle ABC$,
$\angle ACB =180^{\circ}-\angle BAC -\angle ABC$
$ \Rightarrow \angle ACB =180^{\circ}-43^{\circ}-74^{\circ}=63^{\circ}$
Now, $\angle ABC =\angle OBC +\angle ABO$
$\Rightarrow \angle A B C=2 \angle O B C\ldots( OB$ is biosector of $\angle ABC )$
$\Rightarrow 74^{\circ}=2 \angle OBC$
$ \Rightarrow \angle OBC =37^{\circ}$
Similarly,
$\angle A C B=\angle O C B+\angle A C O$
$\Rightarrow \angle A C B=2 \angle O C B \dots... (OC$ is bisector of $\text{ACB})$
$\Rightarrow 63^{\circ}=2 \angle OCB$
$ \Rightarrow \angle O C B=31.5^{\circ}$
Now, in $\triangle BOC$,
$\angle BOC =180^{\circ}-\angle OBC -\angle OCB$
$ \Rightarrow \angle BOC =180^{\circ}-37^{\circ}-31.5^{\circ}$
$ \Rightarrow \angle BOC =111.5^{\circ}$
Since, $\angle B O C>\angle O B C>\angle O C B$, we have
$BC > OC > OB$

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Question 55 Marks

In the following figure, write $BC , AC$, and $CD$ in ascending order of their lengths.

Answer

In $\triangle ABC$
$A B=A C$
$\Rightarrow \angle ABC =\angle ACB\dots..($angles opposite to equal sides are equal$)$
$\Rightarrow \angle ABC =\angle ACB =67^{\circ}$
$\Rightarrow \angle BAC =180^{\circ}-\angle ABC -\angle ACB\dots..($angle sum property$)$
$\Rightarrow \angle BAC =180^{\circ}-67^{\circ}-67^{\circ}=46^{\circ}$
Since $\angle B A C<\angle A B C$, we have
$B C$
$\Rightarrow \angle ACD =180^{\circ}-67^{\circ}=113^{\circ}$
Thus, in $\triangle ACD$,
$\angle C A D=180^{\circ}-\angle A C D-\angle A D C$
$ \Rightarrow \angle C A D=180^{\circ}-113^{\circ}-33^{\circ}=34^{\circ}$
Since $\angle A D C<\angle C A D$, we have
$A C$ From $(1)$ and $(2),$ we have
$BC < AC < CD \text {. }$

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Question 65 Marks

From the following figure, prove that: $A B>C D$.

Answer

In $\triangle ABC,$
$AB = AC \dots...[$ Given $]$
$\therefore \angle ACB = \angle B \dots...[$ angles opposite to equal sides are equal $]$
$\angle B = 70^\circ \dots...[$ Given $]$
$\Rightarrow \angle ACB = 70^\circ \dots...(i)$
Now,
$\angle ACB +\angle ACD = 180^\circ\dots ...[ \text{BCD}$ is a straight line$]$
$\Rightarrow 70^\circ + \angle ACD = 180^\circ $
$\Rightarrow \angle ACD = 110^\circ \dots...(ii)$
In $\triangle ACD,$
$\angle CAD + \angle ACD + \angle D = 180^\circ $
$\Rightarrow \angle CAD + 110^\circ + \angle D = 180^\circ \dots...[$ From $(ii) ]$
$\Rightarrow \angle CAD + \angle D = 70^\circ $
But $\angle D = 40^\circ \dots...[$ Given $]$
$\Rightarrow \angle CAD + 40^\circ = 70^\circ $
$\Rightarrow \angle CAD = 30^\circ \dots…(iii)$
In $\triangle ACD,$
$\angle ACD = 110^\circ \dots...$[ From $(ii) ]$
$\angle CAD = 30^\circ\dots ..$.[ From $(iii) ]$
$\angle D = 40^\circ\dots ...[$ Given $]$
$\therefore D > \angle CAD$
$\Rightarrow AC > CD\dots ....[$Greater angle has greater side opposite to it$]$
Also,
$AB = AC\dots ...[$ Given $]$
Therefore,$ AB > CD.$

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Question 75 Marks

In an isosceles $\triangle ABC$, sides $AB$ and $AC$ are equal. If point $D$ lies in base $BC$ and point $E$ lies on $BC$ produced $(BC$ being produced through vertex $C)$, prove that:$(i) AC > AD,(ii) AE > AC,(iii) AE > AD$

Answer

We know that the bisector of the angle at the vertex of an isosceles triangle bisects the base at the right angle.
Using Pythagoras theorem in $\text{AFB},$
$A B^2=A F^2+B F^2\ldots (i)$
In $\text{AFD},$
$A D^2=A F^2+D F^2\dots ...(ii)$
We know $A B C$ is isosceles triangle and $A B=A C$
$A C^2=A F^2+B F^2 \dots..(iii)[$ From$ (i)]$
Subtracting $(ii)$ from $(iii)$
$A C^2-A D^2=A F^2+B F^2-A F^2-D F^2$
$ A C^2-A D^2=B F^2-D F^2$
Let $2 D F=B F$
$ A C^2-A D^2=(2 D F)^2-D F^2$
$ A C^2-A D^2=4 D F^2-D F^2$
$ A C^2=A D^2+3 D F^2$
$ \Rightarrow A C^2>A D^2$
$ \Rightarrow A C>A D$
Similarly, $A E>A C$ and $A E>A D$.

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Question 85 Marks

In the following diagram; $A D=A B$ and $A E$ bisect angle $A$.

Prove that:$(i) BE = DE;(ii)\angle A B D>\angle C$

Answer

Const: Join $ED.$
In $\triangle AOB$ and $\triangle AOD$,
$A B=A D \dots...[$ Given $]$
$A O=A O \dots....[$ Common $]$
$\angle BAO =\angle DAO\dots ....[AO$ is bisector of $A ]$
$\therefore \triangle AOB \cong \triangle AOD ....[ SAS$ criterion $]$
Hence,
$B O=O D\dots ...(i) [ \text{c.p.c.t.} ]$
$\angle AOB =\angle AOD\dots ...(ii)[ \text{c.p.c.t.} ]$
$\angle ABO =\angle ADO \Rightarrow \angle ABD =\angle ADB \dots...(iii)[ \text{c.p.c.t.} ]$
Now,
$\angle AOB =\angle DOE \dots...[$Vertically opposite angles$]$
$\angle AOD =\angle BOE\dots ...[$Vertically opposite angles$]$
$\angle BOE =\angle DOE \dots...(iv)[$ From $(ii) ]$
$(i)$ In $\triangle BOE$ and $\triangle DOE$,
$B O=C D\dots ...[$ From $(i) ]$
$OE = OE\dots ...[$ Common $]$
$\angle BOE =\angle DOE\dots ... [ $From $(iv) ][ \text{SAS}$ criterion $]$
Hence, $BE = DE\dots ...[ \text{c.p.c.t. }]$
$(ii)$ In $BCD$
$\angle ADB =\angle C+\angle C B D$ $\ldots[$ Ext. angle $=$ sum of opp. int. angles $]$
$\Rightarrow \angle A D B>\angle C$
$\Rightarrow \angle ABD >\angle C\dots ...[$ From $(iii) ]$

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Question 95 Marks

In a quadrilateral $\text{ABCD}$; prove that$:(i) AB+ BC + CD > DA,(ii) AB + BC + CD + DA > 2AC,(iii) AB + BC + CD + DA > 2BD$

Answer

Const: Join $AC$ and $BD$.
$(i)$ In $\triangle ABC$
$A B+B C>A C ....(i)[$ Sum of two sides is greater than the third side $]$
In $\triangle A C D$,
$AC + CD > DA\dots ....(ii)[$ Sum of two sides is greater than the third side $]$
Adding $(i)$ and $(ii)$
$AB + BC + AC + CD > AC + DA$
$ AB + BC + CD > AC + DA - AC$
$AB + BC + CD > DA\dots ....(iii)$
$(ii)$ In $\triangle A C D$,
$CD + DA > AC\dots....(iv) [$Sum of two sides is greater than the third side$]$
Adding $(i)$ and$ (iv)$
$A B+B C+C D+D A>A C+A C$
$ A B+B C+C D+D A>2 A C$
$(iii)$ In $\triangle ABD$,
$AB + DA > BD\dots ....(v)[$Sum of two sides is greater than the third side$]$
In $\triangle B C D$,
$B C+C D>B D\dots....(vi)[$Sum of two sides is greater than the third side$]$
Adding $(v)$ and $(vi)$
$ A B+D A+B C+C D>B D+B D$
$ A B+D A+B C+C D>2 B D$

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MATHEMATICS [5 marks sum]

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