\displaystyle \textbf{Question 1: }\text{Find the value of }x \text{ in each of the following figures:} p1
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }BD=DA,\quad \angle ABC=\angle BAD=35^{\circ}.
\displaystyle \text{(Angles opposite equal sides are equal)}
\displaystyle \text{In }\triangle ABD,\quad \angle BDA=180^{\circ}-35^{\circ}-35^{\circ}=110^{\circ}.
\displaystyle \text{Since }D,E,C\text{ are collinear,}
\displaystyle \angle ADE=180^{\circ}-110^{\circ}=70^{\circ}.
\displaystyle \text{Similarly, since }AE=EC,\quad \angle EAC=\angle ECA=28^{\circ}.
\displaystyle \text{In }\triangle AEC,\quad \angle AEC=180^{\circ}-28^{\circ}-28^{\circ}=124^{\circ}.
\displaystyle \text{Since }D,E,C\text{ are collinear,}
\displaystyle \angle AED=180^{\circ}-124^{\circ}=56^{\circ}.
\displaystyle \text{In }\triangle ADE,\quad x+70^{\circ}+56^{\circ}=180^{\circ}.
\displaystyle \therefore x=54^{\circ}.

\displaystyle \text{(ii) Since }OB=OC,\quad \angle OBC=\angle OCB=40^{\circ}.
\displaystyle \text{In }\triangle BOC,\quad \angle BOC=180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ}.
\displaystyle \text{Since }OA=OC,\quad \angle OCA=\angle OAC=30^{\circ}.
\displaystyle \text{In }\triangle AOC,\quad \angle AOC=180^{\circ}-30^{\circ}-30^{\circ}=120^{\circ}.
\displaystyle \angle AOB+\angle BOC+\angle COA=360^{\circ}.
\displaystyle \Rightarrow \angle AOB =360^{\circ}-100^{\circ}-120^{\circ}=140^{\circ}.
\displaystyle \text{Since }OA=OB,\quad \angle ABO=\angle BAO=x.
\displaystyle \text{In }\triangle AOB,\quad x+x+140^{\circ}=180^{\circ}.
\displaystyle \Rightarrow 2x=40^{\circ}.
\displaystyle \therefore x=20^{\circ}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{State, giving reasons, whether it is possible}
\displaystyle \text{to construct a triangle with sides of lengths:}
\displaystyle \text{(i) }3\text{ cm},\ 4\text{ cm},\ 7\text{ cm} \qquad \text{(ii) }9\text{ cm},\ 8\text{ cm},\ 16\text{ cm}
\displaystyle \text{(iii) }7\text{ cm},\ 9.2\text{ cm},\ 6.7\text{ cm} \qquad \text{(iv) }3\text{ cm},\ 6.2\text{ cm},\ 10.8\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \text{The sum of any two sides of a triangle must be greater}
\displaystyle \text{than the third side.}
\displaystyle \text{(i) }3+4=7.
\displaystyle \text{Since }3+4\text{ is not greater than }7,\text{ the triangle cannot}
\displaystyle \text{be constructed.}

\displaystyle \text{(ii) }9+8=17>16,\quad 9+16>8,\quad 8+16>9.
\displaystyle \text{Hence, the triangle can be constructed.}

\displaystyle \text{(iii) }7+6.7=13.7>9.2,\quad 7+9.2>6.7,\quad 9.2+6.7>7.
\displaystyle \text{Hence, the triangle can be constructed.}

\displaystyle \text{(iv) }3+6.2=9.2<10.8.
\displaystyle \text{Hence, the triangle cannot be constructed.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In }\Delta ABC,\ \angle B=60^{\circ}\text{ and }\angle C=45^{\circ}.
\displaystyle \text{Find }\angle A.\text{ Name (i) the largest side, (ii) the smallest side,}
\displaystyle \text{and (iii) write the sides in ascending order of their lengths.}
\displaystyle \text{Answer:}
\displaystyle \angle A=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle \text{Since }75^{\circ}>60^{\circ}>45^{\circ},
\displaystyle \text{(i) the largest side is }BC.

\displaystyle \text{(ii) The smallest side is }AB.

\displaystyle \text{(iii) The sides in ascending order are}
\displaystyle AB<AC<BC.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In }\Delta XYZ,\ \angle X=53^{\circ}\text{ and }\angle Y=67^{\circ}.
\displaystyle \text{Name (i) the smallest side, (ii) the largest side, and (iii) write}
\displaystyle \text{the sides in ascending order of their lengths.}
\displaystyle \text{Answer:}
\displaystyle \angle Z=180^{\circ}-53^{\circ}-67^{\circ}=60^{\circ}.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle \text{Since }67^{\circ}>60^{\circ}>53^{\circ},
\displaystyle \text{(i) the smallest side is }YZ.

\displaystyle \text{(ii) The largest side is }XZ.

\displaystyle \text{(iii) The sides in ascending order are}
\displaystyle YZ<XY<XZ.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In }\Delta PQR,\ \angle P=\angle R=64^{\circ}.
\displaystyle \text{Name the smallest side and the equal sides.}
\displaystyle \text{Answer:}
\displaystyle \angle Q=180^{\circ}-64^{\circ}-64^{\circ}=52^{\circ}.
\displaystyle \text{Since }\angle Q\text{ is the smallest angle, the side opposite to it}
\displaystyle \text{is the smallest side.}
\displaystyle \therefore\ \text{Smallest side}=PR.
\displaystyle \text{Also, }\angle P=\angle R.
\displaystyle \text{Sides opposite to equal angles of a triangle are equal.}
\displaystyle \therefore\ \text{Equal sides}=QR\text{ and }PQ.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In }\Delta ABC,\ \angle A=\angle B\text{ and }\angle C=100^{\circ}.
\displaystyle \text{Name the largest side and the equal sides.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\angle A=\angle B=x.
\displaystyle x+x+100^{\circ}=180^{\circ}.
\displaystyle 2x=80^{\circ}\quad\Rightarrow\quad x=40^{\circ}.
\displaystyle \therefore\ \angle A=\angle B=40^{\circ}.
\displaystyle \text{Since }\angle C=100^{\circ}\text{ is the largest angle, the side opposite}
\displaystyle \text{to it is the largest side.}
\displaystyle \therefore\ \text{Largest side}=AB.
\displaystyle \text{Since }\angle A=\angle B,\text{ the sides opposite to them are equal.}
\displaystyle \therefore\ \text{Equal sides}=BC\text{ and }AC.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In }\Delta DEF,\ \angle D:\angle E:\angle F=7:3:5.
\displaystyle \text{Name the smallest side and the largest side.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\angle D=7x,\quad\angle E=3x,\quad\angle F=5x.
\displaystyle 7x+3x+5x=180^{\circ}.
\displaystyle 15x=180^{\circ}\quad\Rightarrow\quad x=12^{\circ}.
\displaystyle \therefore\ \angle D=84^{\circ},\quad \angle E=36^{\circ},\quad\angle F=60^{\circ}.
\displaystyle \text{Since }\angle E\text{ is the smallest angle, the side opposite to it}
\displaystyle \text{is the smallest side.}
\displaystyle \therefore\ \text{Smallest side}=DF.
\displaystyle \text{Since }\angle D\text{ is the largest angle, the side opposite to it}
\displaystyle \text{is the largest side.}
\displaystyle \therefore\ \text{Largest side}=EF.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the adjoining figure, }AC=DC,\ \angle CAD=50^{\circ}
\displaystyle \text{and }\angle BAD=23^{\circ}.\text{ Find }\angle ADC,\ \angle ACD,\ \angle ADB
\displaystyle \text{and }\angle ABD.\text{ Also show that }AD>AC,\ AD>BD \text{ and }AB>BC. p8\displaystyle \text{Answer:}
\displaystyle AC=DC.
\displaystyle \therefore\ \angle ADC=\angle CAD=50^{\circ} \quad\text{(angles opposite equal sides)}.
\displaystyle \angle ACD=180^{\circ}-50^{\circ}-50^{\circ} =80^{\circ}.
\displaystyle \text{Since }B,D,C\text{ are collinear,}
\displaystyle \angle ADB=180^{\circ}-\angle ADC =180^{\circ}-50^{\circ}=130^{\circ}.
\displaystyle \text{In }\Delta ABD,
\displaystyle \angle ABD=180^{\circ}-130^{\circ}-23^{\circ} =27^{\circ}.
\displaystyle \therefore\ \angle ADC=50^{\circ},\quad \angle ACD=80^{\circ},\quad\angle ADB=130^{\circ},\quad \angle ABD=27^{\circ}.
\displaystyle \text{In }\Delta ACD,\quad \angle ACD=80^{\circ}>\angle ADC=50^{\circ}.
\displaystyle \therefore\ AD>AC \quad\text{(greater angle has the greater opposite side)}.
\displaystyle \text{In }\Delta ABD,\quad \angle ABD=27^{\circ}>\angle BAD=23^{\circ}.
\displaystyle \therefore\ AD>BD.
\displaystyle \angle BAC=\angle BAD+\angle DAC =23^{\circ}+50^{\circ}=73^{\circ}.
\displaystyle \text{Also, since }B,D,C\text{ are collinear,}\quad \angle ACB=\angle ACD=80^{\circ}.
\displaystyle \text{In }\Delta ABC,\quad \angle ACB=80^{\circ}>\angle BAC=73^{\circ}.
\displaystyle \therefore\ AB>BC.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the adjoining figure, }BA\parallel CD \text{ and }AB=BC.
\displaystyle \text{If }\angle BAC=56^{\circ}\text{ and }\angle COD=85^{\circ}, \text{ find }x,\ y\text{ and }z. p9\displaystyle \text{Answer:}
\displaystyle \text{Since }AB=BC,\quad \angle BAC=\angle ACB=56^{\circ}.
\displaystyle \text{In }\Delta ABC,\quad \angle ABC=180^{\circ}-56^{\circ}-56^{\circ}=68^{\circ}.
\displaystyle \therefore\ x+y=68^{\circ}.\qquad\text{...(i)}
\displaystyle \angle AOB=\angle COD=85^{\circ} \quad\text{(vertically opposite angles)}.
\displaystyle \text{In }\Delta AOB,\quad x+56^{\circ}+85^{\circ}=180^{\circ}.
\displaystyle \therefore\ x=39^{\circ}.
\displaystyle \text{Substituting }x=39^{\circ}\text{ in (i),}
\displaystyle 39^{\circ}+y=68^{\circ} \quad\Rightarrow\quad y=29^{\circ}.
\displaystyle \text{Since }BA\parallel CD\text{ and }BC\text{ is a transversal,}
\displaystyle \angle ABC=\angle DCZ \quad\text{(corresponding angles)}.
\displaystyle \therefore\ z=\angle DCZ=\angle ABC=68^{\circ}.
\displaystyle \therefore\ x=39^{\circ},\quad y=29^{\circ},\quad z=68^{\circ}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the adjoining figure, }BA\parallel CD,\ BA=BC\text{ and }CE=DE.
\displaystyle \text{If }\angle ABC=66^{\circ},\ \angle DCE=48^{\circ} \text{ and }\angle DAC=63^{\circ},
\displaystyle \text{find the values of }x,\ y\text{ and }z. p10\displaystyle \text{Answer:}
\displaystyle \text{Since }BA=BC,\quad \angle BAC=\angle BCA=y.
\displaystyle \text{In }\Delta ABC,\quad 2y+66^{\circ}=180^{\circ}.
\displaystyle \therefore\ 2y=114^{\circ} \quad\Rightarrow\quad y=57^{\circ}.
\displaystyle \text{Since }CE=DE,\quad \angle ECD=\angle CDE=48^{\circ}.
\displaystyle \text{In }\Delta CDE,\quad x+48^{\circ}+48^{\circ} =180^{\circ}.
\displaystyle \therefore\ x=84^{\circ}.
\displaystyle \text{Since }BA\parallel CD\text{ and }AC\text{ is a transversal,}
\displaystyle \angle BAC=\angle ACD=57^{\circ} \quad\text{(alternate interior angles)}.
\displaystyle \text{In }\Delta ACD,\quad 63^{\circ}+57^{\circ}+z=180^{\circ}.
\displaystyle \therefore\ z=60^{\circ}.
\displaystyle \therefore\ x=84^{\circ},\quad y=57^{\circ},\quad z=60^{\circ}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the adjoining figure, }AD=BD=CD \text{ and }\angle CAD=40^{\circ}.
\displaystyle \text{Show that }AC>AD,\ AB>AD\text{ and }AC>AB. p11\displaystyle \text{Answer:}
\displaystyle \text{Since }AD=CD,\quad \angle DAC=\angle ACD=40^{\circ}.
\displaystyle \text{In }\Delta ACD,\quad \angle ADC=180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ}.
\displaystyle \text{Since }B,D,C\text{ are collinear,}
\displaystyle \angle ADB=180^{\circ}-\angle ADC =180^{\circ}-100^{\circ}=80^{\circ}.
\displaystyle \text{Since }AD=BD,\quad \angle DAB=\angle DBA.
\displaystyle \text{In }\Delta ABD,\quad \angle DAB+\angle DBA+80^{\circ}=180^{\circ}.
\displaystyle \therefore\ \angle DAB=\angle DBA=50^{\circ}.
\displaystyle \text{In }\Delta ACD,\quad \angle ADC=100^{\circ}>\angle ACD=40^{\circ}.
\displaystyle \therefore\ AC>AD \quad\text{(greater angle has the greater opposite side)}.
\displaystyle \text{In }\Delta ABD,\quad \angle ADB=80^{\circ}>\angle ABD=50^{\circ}.
\displaystyle \therefore\ AB>AD.
\displaystyle \angle BAC=\angle BAD+\angle DAC =50^{\circ}+40^{\circ}=90^{\circ}.
\displaystyle \angle ABC=\angle ABD=50^{\circ}.
\displaystyle \text{In }\Delta ABC,\quad \angle BAC=90^{\circ}>\angle ABC=50^{\circ}.
\displaystyle \therefore\ BC>AC.
\displaystyle \text{But }BC=BD+DC=2AD.
\displaystyle \text{Also, in }\Delta ABC,\quad \angle ACB=40^{\circ}<\angle ABC=50^{\circ}.
\displaystyle \therefore\ AB>AC.
\displaystyle \text{Hence }AB>AC>AD.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }D\text{ is a point on side }BC \text{ of }\Delta ABC,\text{ prove that:} p12
\displaystyle \text{(i) }AB+BD>AD\qquad \text{(ii) }AC+CD>AD
\displaystyle \text{(iii) }AB+BC+AC>2AD.
\displaystyle \text{Answer:}
\displaystyle \text{(i) In }\Delta ABD,\text{ by the triangle inequality,}
\displaystyle AB+BD>AD.

\displaystyle \text{(ii) In }\Delta ACD,\text{ by the triangle inequality,}
\displaystyle AC+CD>AD.

\displaystyle \text{(iii) Adding the inequalities obtained in (i) and (ii),}
\displaystyle AB+BD+AC+CD>2AD.
\displaystyle AB+AC+(BD+CD)>2AD.
\displaystyle \text{Since }D\text{ lies on }BC,\quad BD+CD=BC.
\displaystyle \therefore\ AB+BC+AC>2AD.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the adjoining figure, }ABCD \text{ is a quadrilateral. Prove that:} p13\displaystyle \text{(i) }AD+DC>AC\qquad \text{(ii) }AB+BC>AC
\displaystyle \text{(iii) }AB+BC+AD+DC>2AC\qquad \text{(iv) }AD+AB+BC>CD.
\displaystyle \text{Answer:}
\displaystyle \text{(i) In }\Delta ADC,\text{ by the triangle inequality,}
\displaystyle AD+DC>AC.

\displaystyle \text{(ii) In }\Delta ABC,\text{ by the triangle inequality,}
\displaystyle AB+BC>AC.

\displaystyle \text{(iii) Adding the inequalities obtained in (i) and (ii),}
\displaystyle AD+DC+AB+BC>AC+AC.
\displaystyle \therefore\ AB+BC+AD+DC>2AC.

\displaystyle \text{(iv) In }\Delta ABC,\text{ by the triangle inequality,}
\displaystyle AB+BC>AC.
\displaystyle \text{In }\Delta ADC,\text{ by the triangle inequality,}
\displaystyle AD+AC>DC.
\displaystyle \text{Adding these two inequalities,}
\displaystyle AB+BC+AD+AC>AC+DC.
\displaystyle \therefore\ AD+AB+BC>DC.
\displaystyle \\

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