\displaystyle \textbf{Question 1: }\text{In }\Delta ABC,\text{ right-angled at }A, \text{ find }BC\text{ when:}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\Delta ABC\text{ is right-angled at }A,\text{ by the Pythagoras theorem,}
\displaystyle BC^2=AB^2+AC^2. 301
\displaystyle \text{(i) }AB=24\text{ cm and }AC=7\text{ cm}.
\displaystyle BC=\sqrt{24^2+7^2}
\displaystyle =\sqrt{576+49}=\sqrt{625}=25\text{ cm}.

\displaystyle \text{(ii) }AB=28\text{ cm and }AC=45\text{ cm}.
\displaystyle BC=\sqrt{28^2+45^2}
\displaystyle =\sqrt{784+2025}=\sqrt{2809}=53\text{ cm}.

\displaystyle \text{(iii) }AB=1.6\text{ cm and }AC=3\text{ cm}.
\displaystyle BC=\sqrt{(1.6)^2+3^2}
\displaystyle =\sqrt{2.56+9}=\sqrt{11.56}=3.4\text{ cm}.

\displaystyle \text{(iv) }AB=5.6\text{ cm and }AC=4.2\text{ cm}.
\displaystyle BC=\sqrt{(5.6)^2+(4.2)^2}
\displaystyle =\sqrt{31.36+17.64}=\sqrt{49}=7\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In }\Delta PQR,\text{ right-angled at }Q, \text{ find }PQ\text{ when:}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\Delta PQR\text{ is right-angled at }Q,\text{ by the Pythagoras theorem,}
\displaystyle PR^2=PQ^2+QR^2\Rightarrow PQ=\sqrt{PR^2-QR^2}. 302
\displaystyle \text{(i) }PR=29\text{ cm and }QR=21\text{ cm}.
\displaystyle PQ=\sqrt{29^2-21^2}
\displaystyle =\sqrt{841-441}=\sqrt{400}=20\text{ cm}.

\displaystyle \text{(ii) }PR=5\text{ cm and }QR=4.8\text{ cm}.
\displaystyle PQ=\sqrt{5^2-(4.8)^2}
\displaystyle =\sqrt{25-23.04}=\sqrt{1.96}=1.4\text{ cm}.

\displaystyle \text{(iii) }PR=3.4\text{ cm and }QR=3\text{ cm}.
\displaystyle PQ=\sqrt{(3.4)^2-3^2}
\displaystyle =\sqrt{11.56-9}=\sqrt{2.56}=1.6\text{ cm}.

\displaystyle \text{(iv) }PR=5.3\text{ cm and }QR=4.5\text{ cm}.
\displaystyle PQ=\sqrt{(5.3)^2-(4.5)^2}
\displaystyle =\sqrt{28.09-20.25}=\sqrt{7.84}=2.8\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The lengths of the sides of some triangles}
\displaystyle \text{are given below. Which of them are right-angled triangles?}
\displaystyle \text{In each case of a right-angled triangle, find the angle}
\displaystyle \text{which measures }90^{\circ}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }AB=7\text{ cm},\ BC=8\text{ cm and }AC=15\text{ cm}.
\displaystyle AB+BC=7+8=15=AC.
\displaystyle \text{Hence, these sides do not form a triangle.}
\displaystyle \therefore\text{ It cannot be a right-angled triangle.}

\displaystyle \text{(ii) }PQ=24\text{ cm},\ QR=20\text{ cm and }PR=32\text{ cm}.
\displaystyle PQ^2+QR^2=24^2+20^2=576+400=976.
\displaystyle PR^2=32^2=1024.
\displaystyle PQ^2+QR^2\ne PR^2.
\displaystyle \therefore\text{ It is not a right-angled triangle.}

\displaystyle \text{(iii) }XY=24\text{ cm},\ YZ=26\text{ cm and }ZX=10\text{ cm}.
\displaystyle XY^2+ZX^2=24^2+10^2=576+100=676.
\displaystyle YZ^2=26^2=676.
\displaystyle \therefore XY^2+ZX^2=YZ^2.
\displaystyle \text{Hence, by the converse of the Pythagoras theorem,}
\displaystyle \Delta XYZ\text{ is a right-angled triangle.}
\displaystyle \therefore\angle X=90^{\circ}.

\displaystyle \text{(iv) }LM=1.6\text{ cm},\ MN=1.4\text{ cm and }LN=1.8\text{ cm}.
\displaystyle LM^2+MN^2=(1.6)^2+(1.4)^2=2.56+1.96=4.52.
\displaystyle LN^2=(1.8)^2=3.24.
\displaystyle LM^2+MN^2\ne LN^2.
\displaystyle \therefore\text{ It is not a right-angled triangle.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the unknown side of each of the}
\displaystyle \text{following figures.}
\displaystyle \text{Answer:} 303
\displaystyle \text{(i) }ZY=\sqrt{32^2+24^2}
\displaystyle =\sqrt{1024+576}=\sqrt{1600}=40\text{ cm}.

\displaystyle \text{(ii) }LN=\sqrt{2^2+(1.5)^2}
\displaystyle =\sqrt{4+2.25}=\sqrt{6.25}=2.5\text{ cm}.

\displaystyle \text{(iii) }RQ=\sqrt{34^2-16^2}
\displaystyle =\sqrt{1156-256}=\sqrt{900}=30\text{ cm}.

\displaystyle \text{(iv) }AC=\sqrt{20^2-12^2}
\displaystyle =\sqrt{400-144}=\sqrt{256}=16\text{ cm}.

\displaystyle \text{(v) }DF=\sqrt{(0.7)^2+(2.4)^2}
\displaystyle =\sqrt{0.49+5.76}=\sqrt{6.25}=2.5\text{ cm}.

\displaystyle \text{(vi) }KL=\sqrt{(2.5)^2+6^2}
\displaystyle =\sqrt{6.25+36}=\sqrt{42.25}=6.5\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the adjoining figure, it is given that}
\displaystyle AD\perp BC,\ AB=25\text{ cm},\ AD=15\text{ cm and }AC=17\text{ cm}.
\displaystyle \text{Find the length of: (i) }BD,\quad\text{(ii) }DC,\quad\text{(iii) }BC.
\displaystyle \text{Answer:} 304
\displaystyle \text{In }\Delta ABD,\quad AB^2=AD^2+BD^2.
\displaystyle BD=\sqrt{AB^2-AD^2}=\sqrt{25^2-15^2}
\displaystyle =\sqrt{625-225}=\sqrt{400}=20\text{ cm}.

\displaystyle \text{In }\Delta ADC,\quad AC^2=AD^2+DC^2.
\displaystyle DC=\sqrt{AC^2-AD^2}=\sqrt{17^2-15^2}
\displaystyle =\sqrt{289-225}=\sqrt{64}=8\text{ cm}.

\displaystyle BC=BD+DC=20+8=28\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The diagonal of a rectangle is }37\text{ cm}
\displaystyle \text{and the length of its shorter side is }12\text{ cm}.
\displaystyle \text{Find: (i) the length of its longer side, (ii) its perimeter,}
\displaystyle \text{and (iii) its area.} 305
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the longer side of the rectangle be }AB.
\displaystyle AB=\sqrt{37^2-12^2}
\displaystyle =\sqrt{1369-144}=\sqrt{1225}=35\text{ cm}.

\displaystyle \text{(ii) Perimeter}=2(35+12)=2\times47=94\text{ cm}.

\displaystyle \text{(iii) Area}=35\times12=420\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The base of an isosceles triangle is }28\text{ cm}
\displaystyle \text{long and }AB=AC=50\text{ cm}. \text{ Let }AD\perp BC.
\displaystyle \text{Find: (i) the length of }AD,\quad\text{(ii) the area of }\Delta ABC.
\displaystyle \text{Answer:} 306
\displaystyle \text{Since }AB=AC,\ AD\text{ bisects }BC.
\displaystyle \therefore BD=DC=\frac{28}{2}=14\text{ cm}.
\displaystyle \text{In }\Delta ABD,\quad AB^2=AD^2+BD^2.
\displaystyle AD=\sqrt{AB^2-BD^2}=\sqrt{50^2-14^2}
\displaystyle =\sqrt{2500-196}=\sqrt{2304}=48\text{ cm}.
\displaystyle \text{(i) Therefore, }AD=48\text{ cm}.

\displaystyle \text{(ii) Area of }\Delta ABC=\frac{1}{2}\times BC\times AD
\displaystyle =\frac{1}{2}\times28\times48=672\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A diagonal of a rhombus is }16\text{ cm}
\displaystyle \text{long and each of its sides measures }10\text{ cm}.
\displaystyle \text{Find the length of the other diagonal.}
\displaystyle \text{Answer:}
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \therefore BO=OD=\frac{16}{2}=8\text{ cm}. 307
\displaystyle \text{In right-angled }\Delta AOB,\quad AB^2=AO^2+BO^2.
\displaystyle AO=\sqrt{AB^2-BO^2}=\sqrt{10^2-8^2}
\displaystyle =\sqrt{100-64}=\sqrt{36}=6\text{ cm}.
\displaystyle \text{Since }AO=OC,\quad AC=2AO=2\times6=12\text{ cm}.
\displaystyle \therefore\text{ The other diagonal is }12\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The supporting wire to the top of a vertical}
\displaystyle \text{pole is }13\text{ m long and is fastened to the ground at a point}
\displaystyle 5\text{ m away from the foot of the pole. How high is the pole?}
\displaystyle \text{Answer:} 308
\displaystyle \text{Let }AB\text{ be the height of the pole.}
\displaystyle \text{In right-angled }\Delta ABC,\quad AC^2=AB^2+BC^2.
\displaystyle AB=\sqrt{AC^2-BC^2}=\sqrt{13^2-5^2}
\displaystyle =\sqrt{169-25}=\sqrt{144}=12\text{ m}.
\displaystyle \therefore\text{ The pole is }12\text{ m high}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A ladder }26\text{ m long rests against a vertical}
\displaystyle \text{wall with its foot }10\text{ m away from the wall. How high up the}
\displaystyle \text{wall will the ladder reach?}
\displaystyle \text{Answer:} 309
\displaystyle \text{Let }AB\text{ be the height reached by the ladder on the wall.}
\displaystyle \text{In right-angled }\Delta ABC,\quad AC^2=AB^2+BC^2.
\displaystyle AB=\sqrt{AC^2-BC^2}=\sqrt{26^2-10^2}
\displaystyle =\sqrt{676-100}=\sqrt{576}=24\text{ m}.
\displaystyle \therefore\text{ The ladder reaches }24\text{ m up the wall}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A }15\text{ m ladder placed against a vertical}
\displaystyle \text{wall reaches a window at a height of }12\text{ m from the ground.}
\displaystyle \text{How far is the foot of the ladder from the wall?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the distance of the foot of the ladder from the wall be }d. 3010
\displaystyle \text{By Pythagoras theorem,}\quad d^2+12^2=15^2.
\displaystyle d=\sqrt{15^2-12^2}
\displaystyle =\sqrt{225-144}=\sqrt{81}=9\text{ m}.
\displaystyle \therefore\text{ The foot of the ladder is }9\text{ m from the wall}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The heights of two towers are }34\text{ m and}
\displaystyle 10\text{ m respectively. If the distance between their feet is }32\text{ m},
\displaystyle \text{find the distance between their tops.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }l\text{ be the distance between the tops of the two towers}.
\displaystyle \text{Difference between their heights}=34-10=24\text{ m}. 3011
\displaystyle \text{The horizontal distance between the towers}=32\text{ m}.
\displaystyle \text{By Pythagoras theorem,}\quad l^2=24^2+32^2.
\displaystyle l=\sqrt{24^2+32^2}
\displaystyle =\sqrt{576+1024}=\sqrt{1600}=40\text{ m}.
\displaystyle \therefore\text{ The distance between the tops is }40\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the adjoining figure, }ABC\text{ is a}
\displaystyle \text{triangle in which }\angle B=90^{\circ}.\text{ If }D\text{ is the midpoint}
\displaystyle \text{of }BC,\text{ prove that }AC^2=AD^2+3CD^2.
\displaystyle \text{Answer:}
\displaystyle \text{Since }D\text{ is the midpoint of }BC,\quad BD=CD.
\displaystyle \therefore BC=BD+CD=2CD.
\displaystyle \text{In right-angled }\Delta ABC,\text{ by Pythagoras theorem,}
\displaystyle AC^2=AB^2+BC^2 3012
\displaystyle =AB^2+(2CD)^2
\displaystyle =AB^2+4CD^2.
\displaystyle \text{In right-angled }\Delta ABD,\text{ by Pythagoras theorem,}
\displaystyle AD^2=AB^2+BD^2.
\displaystyle \therefore AB^2=AD^2-BD^2=AD^2-CD^2.
\displaystyle \text{Substituting,}
\displaystyle AC^2=AD^2-CD^2+4CD^2
\displaystyle =AD^2+3CD^2.
\displaystyle \therefore AC^2=AD^2+3CD^2.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the adjoining figure, it is given that}
\displaystyle AB=27\text{ cm},\ CD=12\text{ cm},\ AC=36\text{ cm},
\displaystyle \angle BAC=\angle DCA=90^{\circ}\text{ and }AM=CM.
\displaystyle \text{Find: (i) }BM^2,\quad\text{(ii) }MD^2,\quad\text{(iii) }BD^2.
\displaystyle \text{Answer:} 3013
\displaystyle \text{Since }AM=CM\text{ and }AC=36\text{ cm},
\displaystyle AM=CM=\frac{36}{2}=18\text{ cm}.

\displaystyle \text{(i) In right-angled }\Delta ABM,
\displaystyle BM^2=AB^2+AM^2
\displaystyle =27^2+18^2=729+324=1053\text{ cm}^2.

\displaystyle \text{(ii) In right-angled }\Delta MCD,
\displaystyle MD^2=MC^2+CD^2
\displaystyle =18^2+12^2=324+144=468\text{ cm}^2.

\displaystyle \text{(iii) Draw }DE\perp AB,\text{ where }E\text{ lies on }AB.
\displaystyle \text{Since }AB\perp AC\text{ and }CD\perp AC,\quad AB\parallel CD.
\displaystyle \therefore AE=CD=12\text{ cm and }ED=AC=36\text{ cm}.
\displaystyle BE=AB-AE=27-12=15\text{ cm}.
\displaystyle \text{In right-angled }\Delta BED,
\displaystyle BD^2=BE^2+ED^2
\displaystyle =15^2+36^2=225+1296=1521\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the following figure, it is given that}
\displaystyle \angle PQR=90^{\circ},\ \angle MNR=90^{\circ},\ PM=RM,
\displaystyle PQ=6\text{ cm},\ QR=8\text{ cm and }MN=12\text{ cm}.
\displaystyle \text{Find the perimeter of: (i) }\Delta PMR,\quad \text{(ii) }\Delta MNR, 3014
\displaystyle \text{(iii) quadrilateral }PQRM.
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\Delta PQR,
\displaystyle PR=\sqrt{PQ^2+QR^2}=\sqrt{6^2+8^2}
\displaystyle =\sqrt{36+64}=\sqrt{100}=10\text{ cm}.
\displaystyle \text{Since }PM=RM,\text{ and }MN\perp PR,\ MN\text{ bisects }PR.
\displaystyle \therefore PN=NR=\frac{PR}{2}=5\text{ cm}.
\displaystyle \text{In right-angled }\Delta MNR,
\displaystyle MR=\sqrt{MN^2+NR^2}=\sqrt{12^2+5^2}
\displaystyle =\sqrt{144+25}=\sqrt{169}=13\text{ cm}.
\displaystyle \therefore PM=MR=13\text{ cm}.

\displaystyle \text{(i) Perimeter of }\Delta PMR=PM+MR+PR
\displaystyle =13+13+10=36\text{ cm}.

\displaystyle \text{(ii) Perimeter of }\Delta MNR=MN+NR+MR
\displaystyle =12+5+13=30\text{ cm}.

\displaystyle \text{(iii) Perimeter of quadrilateral }PQRM=PQ+QR+RM+MP
\displaystyle =6+8+13+13=40\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoining figure, } \angle PQR=\angle QRS=90^{\circ}. 3015
\displaystyle \text{Prove that }PR^2-PQ^2=QR^2-SR^2.
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\Delta PQR,\text{ by Pythagoras theorem,}
\displaystyle PR^2=PQ^2+QR^2
\displaystyle \Rightarrow PR^2-PQ^2=QR^2. \qquad\text{...(i)}
\displaystyle \text{In right-angled }\Delta QRS,\text{ by Pythagoras theorem,}
\displaystyle QS^2=QR^2+SR^2
\displaystyle \Rightarrow QR^2-SR^2=QS^2-2SR^2.
\displaystyle \text{Thus, the given statement cannot be proved from the figure.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, } \angle ABC=90^{\circ}.
\displaystyle \text{Prove that }AC^2+PQ^2=AQ^2+PC^2.
\displaystyle \text{Answer:}
\displaystyle \text{Since }AB\perp BC,\text{ the triangles used below are right-angled.}
\displaystyle AC^2=AB^2+BC^2. 3016
\displaystyle AQ^2=AB^2+BQ^2\Rightarrow AB^2=AQ^2-BQ^2.
\displaystyle PC^2=PB^2+BC^2\Rightarrow BC^2=PC^2-PB^2.
\displaystyle \therefore AC^2=(AQ^2-BQ^2)+(PC^2-PB^2)
\displaystyle =AQ^2+PC^2-(BQ^2+PB^2).
\displaystyle \text{Since }\angle PBQ=90^{\circ},
\displaystyle PQ^2=PB^2+BQ^2.
\displaystyle \therefore AC^2=AQ^2+PC^2-PQ^2.
\displaystyle \Rightarrow AC^2+PQ^2=AQ^2+PC^2.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In }\Delta ABC,\text{ if }D\text{ and }E \text{ are the midpoints of }AB\text{ and }AC,
\displaystyle \text{respectively, and }\angle BAC=90^{\circ},\text{ prove that } BE^2+CD^2=5DE^2.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\angle BAC=90^{\circ},\text{ by Pythagoras theorem in } \Delta ABE,
\displaystyle BE^2=AB^2+AE^2. \qquad\text{...(i)} 3017
\displaystyle \text{Similarly, by Pythagoras theorem in }\Delta ACD,
\displaystyle CD^2=AC^2+AD^2. \qquad\text{...(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle BE^2+CD^2=AB^2+AC^2+AE^2+AD^2.
\displaystyle \text{Since }D\text{ and }E\text{ are the midpoints of }AB \text{ and }AC\text{ respectively,}
\displaystyle AB=2AD\quad\text{and}\quad AC=2AE.
\displaystyle \therefore BE^2+CD^2=(2AD)^2+(2AE)^2+AE^2+AD^2
\displaystyle =5AD^2+5AE^2=5(AD^2+AE^2).
\displaystyle \text{In right-angled }\Delta ADE,\quad DE^2=AD^2+AE^2.
\displaystyle \therefore BE^2+CD^2=5DE^2.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In quadrilateral }ABCD,\ \angle B=\angle D=90^{\circ}.
\displaystyle \text{Prove that }AB^2-AD^2=CD^2-CB^2.
\displaystyle \text{Answer:} 3018
\displaystyle \text{In right-angled }\Delta ABC,\text{ by Pythagoras theorem,}
\displaystyle AC^2=AB^2+BC^2
\displaystyle \Rightarrow AB^2=AC^2-BC^2. \qquad\text{...(i)}
\displaystyle \text{In right-angled }\Delta ADC,\text{ by Pythagoras theorem,}
\displaystyle AC^2=AD^2+CD^2
\displaystyle \Rightarrow AD^2=AC^2-CD^2. \qquad\text{...(ii)}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle AB^2-AD^2=(AC^2-BC^2)-(AC^2-CD^2)
\displaystyle =CD^2-BC^2.
\displaystyle \therefore AB^2-AD^2=CD^2-CB^2.
\displaystyle \\

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