\displaystyle \textbf{Exercise - 13(A)}

\displaystyle \textbf{Question 1: }\text{A ladder }13\text{ m long rests against a vertical wall. If the foot of the}
\displaystyle \text{ladder is }5\text{ m from the foot of the wall, find the distance of the other end of}
\displaystyle \text{the ladder from the ground.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the distance of the upper end of the ladder from the ground }=h\text{ m.}
\displaystyle \text{The ladder, wall and ground form a right-angled triangle.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle h^2+5^2=13^2.
\displaystyle \therefore h^2=169-25=144.
\displaystyle \therefore h=\sqrt{144}=12\text{ m.}
\displaystyle {\therefore\ \text{The other end of the ladder is }12\text{ m above the ground.}}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A man goes }40\text{ m due north and then }50\text{ m due west. Find his}
\displaystyle \text{distance from the starting point.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The distances travelled north and west form the perpendicular sides of a}
\displaystyle \text{right-angled triangle.}
\displaystyle \text{Let the distance from the starting point }=d\text{ m.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle d^2=40^2+50^2.
\displaystyle \therefore d^2=1600+2500=4100.
\displaystyle \therefore d=\sqrt{4100}=10\sqrt{41}\text{ m.}
\displaystyle \therefore d\approx64.03\text{ m.}
\displaystyle {\therefore\ \text{His distance from the starting point is }10\sqrt{41}\text{ m,}}
\displaystyle \text{i.e., }64.03\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the figure, }\angle PSQ=90^\circ,\ PQ=10\text{ cm},\ QS=6\text{ cm}
\displaystyle \text{and }RQ=9\text{ cm. Calculate the length of }PR. \displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled }\triangle PQS,
\displaystyle PQ^2=PS^2+QS^2.
\displaystyle \therefore PS^2=PQ^2-QS^2.
\displaystyle \therefore PS^2=10^2-6^2=100-36=64.
\displaystyle \therefore PS=\sqrt{64}=8\text{ cm.}
\displaystyle \text{Since }R,\ Q\text{ and }S\text{ are collinear,}
\displaystyle RS=RQ+QS=9+6=15\text{ cm.}
\displaystyle \text{In right-angled }\triangle RPS,
\displaystyle PR^2=PS^2+RS^2.
\displaystyle \therefore PR^2=8^2+15^2=64+225=289.
\displaystyle \therefore PR=\sqrt{289}=17\text{ cm.}
\displaystyle {\therefore\ \text{The length of }PR\text{ is }17\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The given figure shows a quadrilateral }ABCD\text{ in which }AD=13\text{ cm},
\displaystyle DC=12\text{ cm},\ BC=3\text{ cm and }\angle ABD=\angle BCD=90^\circ. \displaystyle \text{Calculate the length of }AB.
\displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled }\triangle BCD,
\displaystyle BD^2=BC^2+CD^2.
\displaystyle \therefore BD^2=3^2+12^2=9+144=153.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AD^2=AB^2+BD^2.
\displaystyle \therefore AB^2=AD^2-BD^2.
\displaystyle \therefore AB^2=13^2-153=169-153=16.
\displaystyle \therefore AB=\sqrt{16}=4\text{ cm.}
\displaystyle {\therefore\ \text{The length of }AB\text{ is }4\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{Question 5: }AD\text{ is drawn perpendicular to base }BC\text{ of an equilateral triangle }ABC.
\displaystyle \text{Given }BC=10\text{ cm, find the length of }AD,\text{ correct to one place of decimal.}
\displaystyle \textbf{Answer:}\displaystyle \text{Since }ABC\text{ is an equilateral triangle,}
\displaystyle AB=BC=AC=10\text{ cm.}
\displaystyle \text{The perpendicular }AD\text{ bisects }BC.
\displaystyle \therefore BD=\frac{1}{2}BC=\frac{1}{2}\times10=5\text{ cm.}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AB^2=AD^2+BD^2.
\displaystyle \therefore AD^2=AB^2-BD^2.
\displaystyle \therefore AD^2=10^2-5^2=100-25=75.
\displaystyle \therefore AD=\sqrt{75}=5\sqrt{3}\text{ cm.}
\displaystyle \therefore AD\approx8.7\text{ cm.}
\displaystyle {\therefore\ \text{The length of }AD\text{ is }8.7\text{ cm, correct to one decimal place.}}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In triangle }ABC,\text{ given below, }AB=8\text{ cm},\ BC=6\text{ cm}
\displaystyle \text{and }AC=3\text{ cm. Calculate the length of }OC. \displaystyle \textbf{Answer:}
\displaystyle \text{Let }OC=x\text{ cm. Then }OB=BC+OC=(6+x)\text{ cm.}
\displaystyle \text{In right-angled }\triangle ACO,
\displaystyle AO^2=AC^2-OC^2=3^2-x^2=9-x^2.
\displaystyle \text{In right-angled }\triangle ABO,
\displaystyle AB^2=AO^2+OB^2.
\displaystyle \therefore 8^2=(9-x^2)+(6+x)^2.
\displaystyle \therefore 64=9-x^2+36+12x+x^2.
\displaystyle \therefore 64=45+12x.
\displaystyle \therefore 12x=19.
\displaystyle \therefore x=\frac{19}{12}\text{ cm.}
\displaystyle {\therefore\ \text{The length of }OC\text{ is }\frac{19}{12}\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In triangle }ABC,\ AB=AC=x,\ BC=10\text{ cm and the area of the}
\displaystyle \text{triangle is }60\text{ cm}^2.\text{ Find }x.
\displaystyle \textbf{Answer:}\displaystyle \text{Let }AD\perp BC.
\displaystyle \text{Since }\triangle ABC\text{ is isosceles, }BD=DC=\frac{10}{2}=5\text{ cm.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times BC\times AD=60.
\displaystyle \therefore \frac{1}{2}\times10\times AD=60.
\displaystyle \therefore AD=12\text{ cm.}
\displaystyle \text{In right-angled }\triangle ADB,
\displaystyle AB^2=AD^2+BD^2.
\displaystyle \therefore x^2=12^2+5^2=144+25=169.
\displaystyle \therefore x=13\text{ cm.}
\displaystyle {\therefore\ \text{The value of }x\text{ is }13\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If the sides of a triangle are in the ratio }1:\sqrt{2}:1,\text{ show that it is a}
\displaystyle \text{right-angled triangle.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the sides of the triangle be }k,\ k\sqrt{2}\text{ and }k,
\displaystyle \text{where }k>0.
\displaystyle \text{Here, }k\sqrt{2}\text{ is the largest side.}
\displaystyle (k\sqrt{2})^2=2k^2.
\displaystyle k^2+k^2=2k^2.
\displaystyle \therefore (k\sqrt{2})^2=k^2+k^2.
\displaystyle \text{Hence, the square of the largest side is equal to the sum of the}
\displaystyle \text{squares of the other two sides.}
\displaystyle \text{By the converse of Pythagoras Theorem, the triangle is right-angled.}
\displaystyle {\therefore\ \text{The given triangle is a right-angled triangle.}}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Two poles of heights }6\text{ m and }11\text{ m stand vertically on a plane}
\displaystyle \text{ground. If the distance between their feet is }12\text{ m, find the distance between}
\displaystyle \text{their tips.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Difference between the heights of the poles}=11-6=5\text{ m.}
\displaystyle \text{The line joining the tips forms the hypotenuse of a right-angled triangle}
\displaystyle \text{whose other two sides are }12\text{ m and }5\text{ m.}
\displaystyle \text{Let the distance between the tips be }d\text{ m.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle d^2=12^2+5^2=144+25=169.
\displaystyle \therefore d=\sqrt{169}=13\text{ m.}
\displaystyle {\therefore\ \text{The distance between the tips of the two poles is }13\text{ m.}}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the given figure, }AB\parallel CD,\ AB=7\text{ cm},\ BD=25\text{ cm}
\displaystyle \text{and }CD=17\text{ cm};\text{ find the length of side }BC. \displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle BD^2=AB^2+AD^2.
\displaystyle \therefore AD^2=25^2-7^2=625-49=576.
\displaystyle \therefore AD=\sqrt{576}=24\text{ cm.}
\displaystyle \text{Since }AB\parallel CD,\text{ the horizontal difference between }B\text{ and }C
\displaystyle \text{is }CD-AB=17-7=10\text{ cm.}
\displaystyle \text{Now, in the right-angled triangle formed by }BC,
\displaystyle BC^2=10^2+24^2=100+576=676.
\displaystyle \therefore BC=\sqrt{676}=26\text{ cm.}
\displaystyle {\therefore\ \text{The length of }BC\text{ is }26\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the given figure, }\angle B=90^\circ,\ XY\parallel BC,\ AB=12\text{ cm},
\displaystyle AY=8\text{ cm and }AX:XB=1:2.\text{ Find the lengths of }AC\text{ and }BC. \displaystyle \textbf{Answer:}
\displaystyle \text{Given, }AX:XB=1:2.
\displaystyle \therefore AX:AB=1:(1+2)=1:3.
\displaystyle \therefore \frac{AX}{AB}=\frac{1}{3}.
\displaystyle \text{Since }XY\parallel BC,\ \triangle AXY\sim\triangle ABC.
\displaystyle \therefore \frac{AY}{AC}=\frac{AX}{AB}=\frac{1}{3}.
\displaystyle \therefore \frac{8}{AC}=\frac{1}{3}.
\displaystyle \therefore AC=8\times3=24\text{ cm.}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle AC^2=AB^2+BC^2.
\displaystyle \therefore BC^2=AC^2-AB^2.
\displaystyle \therefore BC^2=24^2-12^2=576-144=432.
\displaystyle \therefore BC=\sqrt{432}=12\sqrt{3}\text{ cm.}
\displaystyle {\therefore\ AC=24\text{ cm and }BC=12\sqrt{3}\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In }\triangle ABC,\ \angle B=90^\circ.\text{ Find the sides of the triangle, if:}
\displaystyle \text{(i) }AB=(x-3)\text{ cm},\ BC=(x+4)\text{ cm and }AC=(x+6)\text{ cm}.
\displaystyle \text{(ii) }AB=x\text{ cm},\ BC=(4x+4)\text{ cm and }AC=(4x+5)\text{ cm}.
\displaystyle \textbf{Answer:}
\displaystyle \textbf{(i) }\text{Since }\angle B=90^\circ,\ AC\text{ is the hypotenuse.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle AB^2+BC^2=AC^2.
\displaystyle \therefore (x-3)^2+(x+4)^2=(x+6)^2.
\displaystyle \therefore x^2-6x+9+x^2+8x+16=x^2+12x+36.
\displaystyle \therefore x^2-10x-11=0.
\displaystyle \therefore (x-11)(x+1)=0.
\displaystyle \therefore x=11\text{ or }x=-1.
\displaystyle \text{Since the lengths of the sides are positive and }x>3,\ x=11.
\displaystyle \therefore AB=x-3=11-3=8\text{ cm.}
\displaystyle BC=x+4=11+4=15\text{ cm.}
\displaystyle AC=x+6=11+6=17\text{ cm.}
\displaystyle {\therefore\ \text{The sides of the triangle are }8\text{ cm},\ 15\text{ cm and }17\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{(ii) }\text{Since }\angle B=90^\circ,\ AC\text{ is the hypotenuse.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle AB^2+BC^2=AC^2.
\displaystyle \therefore x^2+(4x+4)^2=(4x+5)^2.
\displaystyle \therefore x^2+16x^2+32x+16=16x^2+40x+25.
\displaystyle \therefore x^2-8x-9=0.
\displaystyle \therefore (x-9)(x+1)=0.
\displaystyle \therefore x=9\text{ or }x=-1.
\displaystyle \text{Since the lengths of the sides are positive, }x=9.
\displaystyle \therefore AB=x=9\text{ cm.}
\displaystyle BC=4x+4=4(9)+4=40\text{ cm.}
\displaystyle AC=4x+5=4(9)+5=41\text{ cm.}
\displaystyle {\therefore\ \text{The sides of the triangle are }9\text{ cm},\ 40\text{ cm and }41\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{Exercise - 13(B)}

\displaystyle \textbf{Question 1: }\text{In the figure given below, }AD\perp BC.\text{ Prove that }
\displaystyle c^2=a^2+b^2-2ax. \displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled }\triangle ACD,
\displaystyle b^2=h^2+x^2.
\displaystyle \therefore h^2=b^2-x^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle c^2=h^2+(a-x)^2.
\displaystyle \text{Substituting the value of }h^2\text{ from (1),}
\displaystyle c^2=(b^2-x^2)+(a-x)^2.
\displaystyle =b^2-x^2+a^2-2ax+x^2.
\displaystyle =a^2+b^2-2ax.
\displaystyle {\therefore\ c^2=a^2+b^2-2ax.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In equilateral }\triangle ABC,\ AD\perp BC\text{ and }BC=x\text{ cm. Find, in terms}
\displaystyle \text{of }x,\text{ the length of }AD.
\displaystyle \textbf{Answer:}\displaystyle \text{Since }\triangle ABC\text{ is equilateral,}
\displaystyle AB=BC=AC=x\text{ cm.}
\displaystyle \text{The perpendicular }AD\text{ bisects }BC.
\displaystyle \therefore BD=\frac{1}{2}BC=\frac{x}{2}\text{ cm.}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AB^2=AD^2+BD^2.
\displaystyle \therefore AD^2=AB^2-BD^2.
\displaystyle \therefore AD^2=x^2-\left(\frac{x}{2}\right)^2.
\displaystyle \therefore AD^2=x^2-\frac{x^2}{4}=\frac{3x^2}{4}.
\displaystyle \therefore AD=\sqrt{\frac{3x^2}{4}}=\frac{\sqrt{3}}{2}x\text{ cm.}
\displaystyle {\therefore\ \text{The length of }AD\text{ is }\frac{\sqrt{3}}{2}x\text{ cm.}}
\displaystyle \\

\displaystyle \textbf{Question 3: }ABC\text{ is a triangle, right-angled at }B.\ M\text{ is a point on }BC.\text{ Prove that:}
\displaystyle AM^2+BC^2=AC^2+BM^2.
\displaystyle \textbf{Answer:}\displaystyle \text{Since }\angle ABC=90^\circ\text{ and }M\text{ lies on }BC,
\displaystyle \angle ABM=90^\circ.
\displaystyle \text{In right-angled }\triangle ABM,
\displaystyle AM^2=AB^2+BM^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle AC^2=AB^2+BC^2.
\displaystyle \therefore AB^2=AC^2-BC^2.\qquad\cdots(2)
\displaystyle \text{Substituting the value of }AB^2\text{ from (2) in (1),}
\displaystyle AM^2=AC^2-BC^2+BM^2.
\displaystyle \therefore AM^2+BC^2=AC^2+BM^2.
\displaystyle {\therefore\ AM^2+BC^2=AC^2+BM^2.}
\displaystyle \\

\displaystyle \textbf{Question 4: }M\text{ and }N\text{ are the mid-points of the sides }QR\text{ and }PQ\text{ respectively}
\displaystyle \text{of a }\triangle PQR,\text{ right-angled at }Q.\text{ Prove that:}
\displaystyle \text{(i) }PM^2+RN^2=5MN^2
\displaystyle \text{(ii) }4PM^2=4PQ^2+QR^2
\displaystyle \text{(iii) }4RN^2=PQ^2+4QR^2
\displaystyle \text{(iv) }4(PM^2+RN^2)=5PR^2.
\displaystyle \textbf{Answer:}\displaystyle \text{Since }M\text{ and }N\text{ are the mid-points of }QR\text{ and }PQ\text{ respectively,}
\displaystyle QM=MR=\frac{1}{2}QR\text{ and }QN=NP=\frac{1}{2}PQ.
\displaystyle \text{Also, by the Mid-point Theorem,}
\displaystyle MN=\frac{1}{2}PR.
\displaystyle \text{Since }\triangle PQR\text{ is right-angled at }Q,
\displaystyle PR^2=PQ^2+QR^2.\qquad\cdots(1)
\displaystyle \textbf{(ii) }\text{In right-angled }\triangle PQM,
\displaystyle PM^2=PQ^2+QM^2.
\displaystyle \therefore PM^2=PQ^2+\left(\frac{QR}{2}\right)^2.
\displaystyle \therefore PM^2=PQ^2+\frac{QR^2}{4}.
\displaystyle \therefore 4PM^2=4PQ^2+QR^2.
\displaystyle {\therefore\ 4PM^2=4PQ^2+QR^2.}
\displaystyle \textbf{(iii) }\text{In right-angled }\triangle RQN,
\displaystyle RN^2=RQ^2+QN^2.
\displaystyle \therefore RN^2=QR^2+\left(\frac{PQ}{2}\right)^2.
\displaystyle \therefore RN^2=QR^2+\frac{PQ^2}{4}.
\displaystyle \therefore 4RN^2=4QR^2+PQ^2.
\displaystyle {\therefore\ 4RN^2=PQ^2+4QR^2.}
\displaystyle \textbf{(i) }\text{Adding the results obtained in (ii) and (iii),}
\displaystyle 4PM^2+4RN^2=5PQ^2+5QR^2.
\displaystyle \therefore 4(PM^2+RN^2)=5(PQ^2+QR^2).
\displaystyle \text{Using (1),}
\displaystyle 4(PM^2+RN^2)=5PR^2.
\displaystyle \therefore PM^2+RN^2=\frac{5}{4}PR^2.
\displaystyle \text{But }MN=\frac{1}{2}PR.
\displaystyle \therefore MN^2=\frac{1}{4}PR^2.
\displaystyle \therefore 5MN^2=\frac{5}{4}PR^2.
\displaystyle {\therefore\ PM^2+RN^2=5MN^2.}
\displaystyle \textbf{(iv) }\text{From the result obtained above,}
\displaystyle 4(PM^2+RN^2)=5(PQ^2+QR^2).
\displaystyle \text{Using }PR^2=PQ^2+QR^2,
\displaystyle {\therefore\ 4(PM^2+RN^2)=5PR^2.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In triangle }ABC,\ \angle B=90^\circ\text{ and }D\text{ is the mid-point of }BC.
\displaystyle \text{Prove that: }AC^2=AD^2+3CD^2.
\displaystyle \textbf{Answer:}\displaystyle \text{Since }D\text{ is the mid-point of }BC,
\displaystyle BD=CD\text{ and }BC=BD+CD=2CD.
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle AC^2=AB^2+BC^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AD^2=AB^2+BD^2.\qquad\cdots(2)
\displaystyle \text{Subtracting (2) from (1),}
\displaystyle AC^2-AD^2=BC^2-BD^2.
\displaystyle \therefore AC^2-AD^2=(2CD)^2-CD^2.
\displaystyle \therefore AC^2-AD^2=4CD^2-CD^2=3CD^2.
\displaystyle {\therefore\ AC^2=AD^2+3CD^2.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In a rectangle }ABCD,\text{ prove that:}
\displaystyle AC^2+BD^2=AB^2+BC^2+CD^2+DA^2.
\displaystyle \textbf{Answer:}\displaystyle \text{Since }ABCD\text{ is a rectangle,}
\displaystyle AB=CD,\quad BC=DA\quad\text{and}\quad AC=BD.
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle AC^2=AB^2+BC^2.
\displaystyle \text{Since }AC=BD,
\displaystyle BD^2=AB^2+BC^2.
\displaystyle \therefore AC^2+BD^2=2AB^2+2BC^2.\qquad\cdots(1)
\displaystyle \text{Also, since }CD=AB\text{ and }DA=BC,
\displaystyle AB^2+BC^2+CD^2+DA^2
\displaystyle =AB^2+BC^2+AB^2+BC^2
\displaystyle =2AB^2+2BC^2.\qquad\cdots(2)
\displaystyle \text{From (1) and (2),}
\displaystyle {\therefore\ AC^2+BD^2=AB^2+BC^2+CD^2+DA^2.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In a quadrilateral }ABCD,\ \angle B=90^\circ\text{ and }\angle D=90^\circ.
\displaystyle \text{Prove that: }2AC^2-AB^2=BC^2+CD^2+DA^2.
\displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle AC^2=AB^2+BC^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle ADC,
\displaystyle AC^2=AD^2+CD^2.\qquad\cdots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle 2AC^2=AB^2+BC^2+AD^2+CD^2.
\displaystyle \therefore 2AC^2-AB^2=BC^2+AD^2+CD^2.
\displaystyle {\therefore\ 2AC^2-AB^2=BC^2+CD^2+DA^2.}
\displaystyle \\

\displaystyle \textbf{Question 8: }O\text{ is any point inside a rectangle }ABCD.\text{ Prove that:}
\displaystyle OB^2+OD^2=OC^2+OA^2.
\displaystyle \textbf{Answer:}\displaystyle \text{Draw }OM\perp AB\text{ and }ON\perp BC.
\displaystyle \text{Since }ABCD\text{ is a rectangle, }OM\parallel BC\text{ and }ON\parallel AB.
\displaystyle \text{Let }AM=x,\ MB=AB-x,\ BN=y\text{ and }NC=BC-y.
\displaystyle \text{In right-angled }\triangle AOM,
\displaystyle OA^2=AM^2+OM^2=x^2+y^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle BOM,
\displaystyle OB^2=BM^2+OM^2=(AB-x)^2+y^2.\qquad\cdots(2)
\displaystyle \text{In right-angled }\triangle CON,
\displaystyle OC^2=CN^2+ON^2=(BC-y)^2+(AB-x)^2.\qquad\cdots(3)
\displaystyle \text{In right-angled }\triangle DON,
\displaystyle OD^2=DN^2+ON^2=x^2+(BC-y)^2.\qquad\cdots(4)
\displaystyle \text{Adding (2) and (4),}
\displaystyle OB^2+OD^2=(AB-x)^2+y^2+x^2+(BC-y)^2.
\displaystyle \text{Adding (1) and (3),}
\displaystyle OA^2+OC^2=x^2+y^2+(AB-x)^2+(BC-y)^2.
\displaystyle \therefore OB^2+OD^2=OA^2+OC^2.
\displaystyle {\therefore\ OB^2+OD^2=OC^2+OA^2.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the following figure, }OP,\ OQ\text{ and }OR\text{ are drawn perpendicular}
\displaystyle \text{to the sides }BC,\ CA\text{ and }AB\text{ respectively of }\triangle ABC.\text{ Prove that:}
\displaystyle AR^2+BP^2+CQ^2=AQ^2+CP^2+BR^2. \displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled }\triangle AOR,
\displaystyle AO^2=AR^2+OR^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle AOQ,
\displaystyle AO^2=AQ^2+OQ^2.\qquad\cdots(2)
\displaystyle \text{From (1) and (2),}
\displaystyle AR^2-AQ^2=OQ^2-OR^2.\qquad\cdots(3)
\displaystyle \text{In right-angled }\triangle BOP,
\displaystyle BO^2=BP^2+OP^2.\qquad\cdots(4)
\displaystyle \text{In right-angled }\triangle BOR,
\displaystyle BO^2=BR^2+OR^2.\qquad\cdots(5)
\displaystyle \text{From (4) and (5),}
\displaystyle BP^2-BR^2=OR^2-OP^2.\qquad\cdots(6)
\displaystyle \text{In right-angled }\triangle COQ,
\displaystyle CO^2=CQ^2+OQ^2.\qquad\cdots(7)
\displaystyle \text{In right-angled }\triangle COP,
\displaystyle CO^2=CP^2+OP^2.\qquad\cdots(8)
\displaystyle \text{From (7) and (8),}
\displaystyle CQ^2-CP^2=OP^2-OQ^2.\qquad\cdots(9)
\displaystyle \text{Adding (3), (6) and (9),}
\displaystyle AR^2-AQ^2+BP^2-BR^2+CQ^2-CP^2=0.
\displaystyle \therefore AR^2+BP^2+CQ^2=AQ^2+CP^2+BR^2.
\displaystyle {\therefore\ AR^2+BP^2+CQ^2=AQ^2+CP^2+BR^2.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Diagonals of rhombus }ABCD\text{ intersect each other at point }O.
\displaystyle \text{Prove that: }OA^2+OC^2=2AD^2-\frac{BD^2}{2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \therefore OA=OC\text{ and }OD=\frac{1}{2}BD.
\displaystyle \text{In right-angled }\triangle AOD,
\displaystyle AD^2=OA^2+OD^2.
\displaystyle \therefore OA^2=AD^2-OD^2.
\displaystyle \therefore OA^2=AD^2-\left(\frac{BD}{2}\right)^2.
\displaystyle \therefore OA^2=AD^2-\frac{BD^2}{4}.\qquad\cdots(1)
\displaystyle \text{Since }OA=OC,
\displaystyle OA^2+OC^2=2OA^2.
\displaystyle \text{Using (1),}
\displaystyle OA^2+OC^2=2\left(AD^2-\frac{BD^2}{4}\right).
\displaystyle \therefore OA^2+OC^2=2AD^2-\frac{BD^2}{2}.
\displaystyle {\therefore\ OA^2+OC^2=2AD^2-\frac{BD^2}{2}.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the figure, }AB=BC\text{ and }AD\text{ is perpendicular to }CD.
\displaystyle \text{Prove that: }AC^2=2\cdot BC\cdot DC. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }D,\ B\text{ and }C\text{ are collinear and }AD\perp CD,
\displaystyle \angle ADB=\angle ADC=90^\circ.
\displaystyle \text{In right-angled }\triangle ADB,
\displaystyle AB^2=AD^2+DB^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle ADC,
\displaystyle AC^2=AD^2+DC^2.\qquad\cdots(2)
\displaystyle \text{Subtracting (1) from (2),}
\displaystyle AC^2-AB^2=DC^2-DB^2.
\displaystyle \therefore AC^2-AB^2=(DC-DB)(DC+DB).
\displaystyle \text{Since }DC-DB=BC,
\displaystyle AC^2-AB^2=BC(DC+DB).
\displaystyle \text{Also, }DB=DC-BC.
\displaystyle \therefore AC^2-AB^2=BC(DC+DC-BC).
\displaystyle \therefore AC^2-AB^2=2BC\cdot DC-BC^2.
\displaystyle \text{Since }AB=BC,\ AB^2=BC^2.
\displaystyle \therefore AC^2-BC^2=2BC\cdot DC-BC^2.
\displaystyle {\therefore\ AC^2=2\cdot BC\cdot DC.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In an isosceles triangle }ABC,\ AB=AC\text{ and }D\text{ is a point on}
\displaystyle BC\text{ produced. Prove that: }AD^2=AC^2+BD\cdot CD.
\displaystyle \textbf{Answer:}\displaystyle \text{Draw }AE\perp BC.
\displaystyle \text{Since }\triangle ABC\text{ is isosceles and }AB=AC,
\displaystyle BE=EC.
\displaystyle \text{In right-angled }\triangle AED,
\displaystyle AD^2=AE^2+ED^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle AEC,
\displaystyle AC^2=AE^2+EC^2.\qquad\cdots(2)
\displaystyle \text{Subtracting (2) from (1),}
\displaystyle AD^2-AC^2=ED^2-EC^2.
\displaystyle \therefore AD^2-AC^2=(ED-EC)(ED+EC).
\displaystyle \text{Since }ED-EC=CD,
\displaystyle \text{and }ED+EC=(EC+CD)+EC=2EC+CD.
\displaystyle \text{But }BE=EC,\text{ therefore }BC=BE+EC=2EC.
\displaystyle \therefore ED+EC=BC+CD=BD.
\displaystyle \therefore AD^2-AC^2=CD\cdot BD.
\displaystyle {\therefore\ AD^2=AC^2+BD\cdot CD.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In triangle }ABC,\ \angle A=90^\circ,\ CA=AB\text{ and }D\text{ is a point}
\displaystyle \text{on }AB\text{ produced. Prove that: }DC^2-BD^2=2AB\cdot AD. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }D\text{ lies on }AB\text{ produced and }\angle A=90^\circ,
\displaystyle \angle CAD=90^\circ.
\displaystyle \text{In right-angled }\triangle ACD,
\displaystyle DC^2=AC^2+AD^2.
\displaystyle \therefore DC^2-BD^2=AC^2+AD^2-BD^2.
\displaystyle \text{Since }AC=AB,
\displaystyle DC^2-BD^2=AB^2+(AD^2-BD^2).
\displaystyle =AB^2+(AD-BD)(AD+BD).
\displaystyle \text{Since }AD-BD=AB,
\displaystyle DC^2-BD^2=AB^2+AB(AD+BD).
\displaystyle =AB(AB+AD+BD).
\displaystyle \text{But }AB+BD=AD.
\displaystyle \therefore DC^2-BD^2=AB(AD+AD).
\displaystyle {\therefore\ DC^2-BD^2=2AB\cdot AD.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In triangle }ABC,\ AB=AC\text{ and }BD\text{ is perpendicular to }AC.
\displaystyle \text{Prove that: }BD^2-CD^2=2CD\cdot AD.
\displaystyle \textbf{Answer:}\displaystyle \text{Since }BD\perp AC,\ \triangle ABD\text{ is right-angled at }D.
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle AB^2=AD^2+BD^2.
\displaystyle \therefore BD^2=AB^2-AD^2.
\displaystyle \text{Since }AB=AC,
\displaystyle BD^2=AC^2-AD^2.
\displaystyle \text{Also, }AC=AD+CD.
\displaystyle \therefore BD^2=(AD+CD)^2-AD^2.
\displaystyle \therefore BD^2=AD^2+2AD\cdot CD+CD^2-AD^2.
\displaystyle \therefore BD^2=2AD\cdot CD+CD^2.
\displaystyle {\therefore\ BD^2-CD^2=2CD\cdot AD.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the following figure, }AD\text{ is perpendicular to }BC\text{ and }D
\displaystyle \text{divides }BC\text{ in the ratio }1:3.\text{ Prove that:}
\displaystyle 2AC^2=2AB^2+BC^2. \displaystyle \textbf{Answer:}
\displaystyle \text{Given, }BD:DC=1:3.
\displaystyle \text{Let }BD=x\text{ and }DC=3x.
\displaystyle \therefore BC=BD+DC=x+3x=4x.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AB^2=AD^2+BD^2=AD^2+x^2.\qquad\cdots(1)
\displaystyle \text{In right-angled }\triangle ACD,
\displaystyle AC^2=AD^2+DC^2=AD^2+9x^2.\qquad\cdots(2)
\displaystyle \text{Subtracting (1) from (2),}
\displaystyle AC^2-AB^2=8x^2.
\displaystyle \therefore 2AC^2-2AB^2=16x^2.
\displaystyle \text{But }BC=4x.
\displaystyle \therefore BC^2=16x^2.
\displaystyle \therefore 2AC^2-2AB^2=BC^2.
\displaystyle {\therefore\ 2AC^2=2AB^2+BC^2.}
\displaystyle \\

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