Question 1: Find the value of $x$ in each of the following figures:

i)    $\angle ABC=\angle BAC=35^{\circ} \ (since \ BD=DA)$

$\angle BDA = 180 - 35 - 35 = 110^{\circ}$

Similarly, $\angle ECA=\angle AEC=28^{\circ} (since EC=EA)$

$\angle AEC=180-28-28=124^{\circ}$

$x+ \angle ADE+ \angle AED=180^o$

$x+(180-110)+(180-124)=180^o$

$\Rightarrow x=54^{\circ}$

ii)   $\angle OBC= \angle OCB=40^{\circ} (Since \ OB=OC)$

$\angle BOC=180-40-40=100^o$

$\angle OCA= \angle OAC=30^{\circ} (Since \ OA=OC)$

$\angle AOC=180-30-30=120^{\circ}$

We know $\angle BOC+ \angle COA+ \angle AOB=360^o$

$\Rightarrow \angle AOB=(360-100-120)=140^{\circ}$

$\Rightarrow 2x+140=180^o$

$\Rightarrow x=20^{\circ}$

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Question 2: State giving reasons, whether it is possible to construct a triangle or not with sides of lengths:

i) $3 \ cm, \ 4 \ cm,\ 7 \ cm$     ii) $9 \ cm, \ 8 \ cm,\ 16 \ cm$

iii) $7 \ cm, \ 9.2 \ cm, \ 6.7 \ cm$     iv) $3 \ cm, \ 6.2 \ cm, \ 10.8 \ cm$

Answer: (Note: The sum of any two sides of a triangle is always greater than the third side)

i) Not Possible to construct the triangle because (3+4 not greater than 7

ii) Possible to construct the triangle as sum of any two sides is greater then the third side.

iii) Possible to construct the triangle as sum of any two sides is greater then the third side.

iv) Not possible to construct a triangle as sum of $4.3+6.2<10.8$

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Question 3: In $\Delta ABC, \angle B=60^{\circ} \ and\ \angle C=45^{\circ}$. Find $\angle A$. Name i) the largest side of $\Delta ABC$, ii) the smallest side of $\Delta ABC$ iii) write the sides of $\Delta ABC$ in ascending order of their lengths.

$\angle A=180-60-45=75^{\circ}$   .   Therefore

i) the largest side of $\Delta ABC=BC$

ii) the smallest side of $\Delta ABC=AB$

iii) $AB < AC < AB$

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Question 4: In $\Delta XYZ, \angle X=53^{\circ} \ and\ \angle Y=67^{\circ}$   .  Name i) the smallest side of $\Delta XYZ$, ii) the largest side of $\Delta XYZ$ iii) write the sides of $\Delta XYZ$ in ascending order of their lengths.

$\angle Z=180-53-67=60^{\circ}$   .   Therefore

i) the largest side of $\Delta XYZ=XZ$

ii) the smallest side of $\Delta XYZ=YZ$

iii) $YZ < XY

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Question 5: In $\Delta PQR, \angle P= \angle R=64^{\circ}$   .  Name the smallest side and the equal sides.

$\angle Q=180-64-64=52^{\circ}$

Smallest side $=PR$

Equal Sides $=QR \ \&\ PQ$

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Question 6: In $\Delta ABC, \angle A= \angle B \ and\ \angle C=100^{\circ}$. Name the largest side and the equal sides.

$x+x+100=180^o (let \ \angle A= \angle B=x)$

$\Rightarrow x=40^{\circ}$

Largest side $=AB$, Equal Sides $=AC \ \&\ BC$

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Question 7: In $\Delta DEF, \angle D \colon \angle E \colon \angle F=7 \colon 3\colon 5$. Name the smallest side and the largest sides.

$7x+3x+5x=180 \Rightarrow x=12$

$\angle D=84^{\circ} , \angle E=36^{\circ} \ and\ \angle F=60^{\circ}$

Largest side $=EF$, Smallest Side $=DF$

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Question 8: In the adjoining figure, $AC=DC, \angle CAD=50^{\circ} \ and\ \angle BAD=23^{\circ}$. Find $\angle ADC, \angle ACD, \angle ADB \ and\ \angle ABD$. Also show that $AD>AC, \ AD>BD \ and\ AB>BC$

$AD=AD \Rightarrow \angle ADC= \angle DAC=50^{\circ}$

$\angle ACD=180-50-50=80^{\circ}$

Therefore $\angle ABD=180-50=130^{\circ}$

$\Rightarrow \angle ABD=180-130-23=27^{\circ}$

Since $\angle ACD> \angle ADC \Rightarrow AD>AC$

Since $\angle ABD> \angle BAD \Rightarrow AD>BD$

Since $\angle ACB> \angle BAC\Rightarrow AB>BC$

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Question 9: In the adjoining figure, $BA \parallel CD \ and\ AB=BC$. If $\angle BAC=56^{\circ} \ and \ \angle COD=85^{\circ}$, find the values of $x, \ y \ and \ z$.

Since $AB=BC, \angle BAC= \angle ACB=56^{\circ}$

$x+y=180-112=68^o ...i)$

$\angle AOB= \angle DOC=85^{\circ}$  (opposite angles)

Therefore $x+56+85=180^o \Rightarrow x=39^o$

Which implies that $y=68-39=29^o$

Becasue $AB \parallel CD, \angle ABC=\angle DCX=x+y=39+29=68^{\circ}$

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Question 10: In the adjoining figure, $BA \parallel CD, BA=BC and CE=DE$. If $\angle ABC=66^{\circ} , \angle DCE=48^{\circ} \ and\ \angle DAC=63^{\circ}$  , find the value of $x, \ y \ and\ z$.

$\angle BAC= \angle BCA=y$

$2y+66=180 \Rightarrow y=57^{\circ}$

$\angle ECD= \angle EDC=48^o$

$x+48+48=180 \Rightarrow x=84^{\circ}$

Since $BA \parallel CD, \angle BAC= \angle ACD=y=57^o$

Therefore $63+57+z=180^o \Rightarrow z=60^{\circ}$

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Question 11: In the adjoining figure, $AD=BD=CD$ and $\angle CAD=40^{\circ}$. Show that $AC>AD, AB>AD, AC>AB$.

Since $AD=CD, \angle DAC= \angle DCA=40^{\circ}$

$\angle ADB=180-(180-40-40)=80^{\circ}$

Since $AD=BD, \angle DAB= \angle DBA$

$\angle ADC=100 \Rightarrow \angle ADB=80^{\circ}$

$\angle DAB= \angle DBA=50^{\circ}$

Since$\angle ADC> \angle ACD \Rightarrow AC>AD$

Since $\angle ADB> \angle ABD \Rightarrow AB>AD$

Since $\angle ABD> \angle ACB\Rightarrow AC>AB$

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Question 12: If $D$ is a point on side $BC of \Delta ABC$, prove that: $i) \ AB+BD>AD \ ii)\ AC+CD>AD \ iii)\ AB+BC+AC>2AD$

In $\Delta ABD$, because it is a valid triangle,  $AB+BD>AD$

In $\Delta ADC$, because it is a valid triangle,  $AC+CD>AD$

In $\Delta ABC$, because it is a valid triangle,

$AB+(BD+CD)+AC>2 AD$

$\Rightarrow AB+BC+AC>2AD$

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Question 13: In the adjoing figure, $ABCD$ is a quadrilateral. Prove that: i)  $AD+DC>AC$     ii) $AB+BC>AC$       iii) $AB+BC+AD+DC>2AC$ iv) $AD+AB+BC>CD$

In $\Delta ADC, AD+DC>AC$
In $\Delta ABC AB+BC>AC$
$AB+BC+AD+DC>2AC$
$AB+BC>AC$ and $AD+AC>CD$. Add these two expressions we get
$AD+AB+BC>CD$