Class 10: Triangles – Exercise 7.6

$\displaystyle \textbf{Question 1: }\text{The sides of certain triangles are given below. Determine which of} \\ \text{them are right triangles.}$
$\displaystyle \text{(i) }a=7\text{ cm},\ b=24\text{ cm and }c=25\text{ cm}$
$\displaystyle \text{(ii) }a=9\text{ cm},\ b=16\text{ cm and }c=18\text{ cm}$
$\displaystyle \text{(iii) }a=1.6\text{ cm},\ b=3.8\text{ cm and }c=4\text{ cm}$
$\displaystyle \text{(iv) }a=8\text{ cm},\ b=10\text{ cm and }c=6\text{ cm}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{By the converse of Pythagoras Theorem, if the square of the longest side is equal to the sum}$
$\displaystyle \text{of the squares of the other two sides, then the triangle is a right triangle.}$

$\displaystyle \text{(i) Longest side}=25\text{ cm}$
$\displaystyle 25^2=625$
$\displaystyle 7^2+24^2=49+576=625$
$\displaystyle \therefore 25^2=7^2+24^2$
$\displaystyle \therefore \text{The triangle is a right triangle.}$

$\displaystyle \text{(ii) Longest side}=18\text{ cm}$
$\displaystyle 18^2=324$
$\displaystyle 9^2+16^2=81+256=337$
$\displaystyle \therefore 18^2\ne9^2+16^2$
$\displaystyle \therefore \text{The triangle is not a right triangle.}$

$\displaystyle \text{(iii) Longest side}=4\text{ cm}$
$\displaystyle 4^2=16$
$\displaystyle (1.6)^2+(3.8)^2=2.56+14.44=17$
$\displaystyle \therefore 4^2\ne(1.6)^2+(3.8)^2$
$\displaystyle \therefore \text{The triangle is not a right triangle.}$

$\displaystyle \text{(iv) Longest side}=10\text{ cm}$
$\displaystyle 10^2=100$
$\displaystyle 8^2+6^2=64+36=100$
$\displaystyle \therefore 10^2=8^2+6^2$
$\displaystyle \therefore \text{The triangle is a right triangle.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A ladder }17\text{ m long reaches a window of a building }15\text{ m above}$
$\displaystyle \text{the ground. Find the distance of the foot of the ladder from the building.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance of the foot of the ladder from the building}=x\text{ m.}$
$\displaystyle \text{The ladder, wall and ground form a right triangle with hypotenuse }17\text{ m.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle x^2+15^2=17^2$
$\displaystyle x^2+225=289$
$\displaystyle x^2=64$
$\displaystyle x=8\text{ m}$
$\displaystyle \therefore \text{The foot of the ladder is }8\text{ m from the building.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Two poles of heights }6\text{ m and }11\text{ m stand on a plane ground. If the distance}$
$\displaystyle \text{between their feet is }12\text{ m, find the distance between their tops.}\hfill\text{[CBSE 2002C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference in the heights of the two poles}=11-6=5\text{ m}$
$\displaystyle \text{Distance between their feet}=12\text{ m}$
$\displaystyle \text{Let the distance between their tops}=x\text{ m.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle x^2=12^2+5^2$
$\displaystyle x^2=144+25=169$
$\displaystyle x=13\text{ m}$
$\displaystyle \therefore \text{The distance between the tops of the two poles is }13\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{In an isosceles triangle }ABC,\ AB=AC=25\text{ cm},$
$\displaystyle \ BC=14\text{ cm. Calculate the} \ \text{altitude from }A\text{ on }BC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }AD\perp BC.$
$\displaystyle \text{Since }\triangle ABC\text{ is isosceles, the altitude }AD\text{ bisects }BC.$
$\displaystyle \therefore BD=DC=\frac{14}{2}=7\text{ cm}$
$\displaystyle \text{In right }\triangle ABD,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AD^2+BD^2$
$\displaystyle 25^2=AD^2+7^2$
$\displaystyle AD^2=625-49=576$
$\displaystyle AD=24\text{ cm}$
$\displaystyle \therefore \text{The altitude from }A\text{ on }BC\text{ is }24\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{The foot of a ladder is }6\text{ m away from a wall and its top reaches}$
$\displaystyle \text{a window 8 m above the ground. If the ladder is shifted in such a way that its}$
$\displaystyle \text{foot is 8 m away from} \ \text{the wall, to what height does its tip reach?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Initially, the foot of the ladder is }6\text{ m from the wall and its top is }8\text{ m high.}$
$\displaystyle \text{Let the length of the ladder}=l\text{ m.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle l^2=6^2+8^2$
$\displaystyle l^2=36+64=100$
$\displaystyle l=10\text{ m}$
$\displaystyle \text{After shifting, let the height reached by the ladder}=h\text{ m.}$
$\displaystyle \text{The foot of the ladder is now }8\text{ m from the wall.}$
$\displaystyle 10^2=8^2+h^2$
$\displaystyle h^2=100-64=36$
$\displaystyle h=6\text{ m}$
$\displaystyle \therefore \text{The tip of the ladder reaches a height of }6\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{An aeroplane leaves an airport and flies due north at a speed of }$
$\displaystyle 300\text{ km/hr. At the} \ \text{same time, another aeroplane leaves the same airport and flies}$
$\displaystyle \text{due west at a speed of }400\text{ km/hr.} \ \text{How far apart will be the two planes after }1\frac{1}{2}\text{ hours?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Time}=1\frac{1}{2}\text{ hours}=\frac{3}{2}\text{ hours}$
$\displaystyle \text{Distance travelled by the first aeroplane}=300\times\frac{3}{2}=450\text{ km}$
$\displaystyle \text{Distance travelled by the second aeroplane}=400\times\frac{3}{2}=600\text{ km}$
$\displaystyle \text{Since north and west directions are perpendicular, let the distance between the planes}=d\text{ km.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle d^2=450^2+600^2$
$\displaystyle d^2=202500+360000=562500$
$\displaystyle d=\sqrt{562500}=750\text{ km}$
$\displaystyle \therefore \text{The two aeroplanes will be }750\text{ km apart.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{A triangle has sides }5\text{ cm},\ 12\text{ cm and }13\text{ cm. Find the length, to one decimal}$
$\displaystyle \text{place, of the perpendicular from the opposite vertex to the side whose length is }13\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle 13^2=169\text{ and }5^2+12^2=25+144=169$
$\displaystyle \therefore 13^2=5^2+12^2$
$\displaystyle \therefore \text{The triangle is right-angled, with hypotenuse }13\text{ cm.}$
$\displaystyle \text{Let the perpendicular to the side of length }13\text{ cm be }h.$
$\displaystyle \text{Area of triangle}=\frac{1}{2}\times5\times12=30\text{ cm}^2$
$\displaystyle \text{Also, Area of triangle}=\frac{1}{2}\times13\times h$
$\displaystyle \therefore \frac{1}{2}\times13\times h=30$
$\displaystyle 13h=60$
$\displaystyle h=\frac{60}{13}=4.615\ldots\text{ cm}$
$\displaystyle \therefore h=4.6\text{ cm, to one decimal place.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }ABCD\text{ is a square. }F\text{ is the mid-point of }AB.\ BE\text{ is one third of }BC.$
$\displaystyle \text{If the area of }\triangle FBE=108\text{ cm}^2,\text{ find the length of }AC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the side of the square }ABCD=x\text{ cm.}$
$\displaystyle \therefore AB=BC=x\text{ cm}$
$\displaystyle \text{Since }F\text{ is the mid-point of }AB,\ FB=\frac{x}{2}\text{ cm}$
$\displaystyle \text{Also, }BE=\frac{1}{3}BC=\frac{x}{3}\text{ cm}$
$\displaystyle \text{Since }AB\perp BC,\ \angle FBE=90^\circ.$
$\displaystyle \text{Area }(\triangle FBE)=\frac{1}{2}\times FB\times BE$
$\displaystyle 108=\frac{1}{2}\times\frac{x}{2}\times\frac{x}{3}$
$\displaystyle 108=\frac{x^2}{12}$
$\displaystyle x^2=1296$
$\displaystyle x=36\text{ cm}$
$\displaystyle \text{In right }\triangle ABC,\text{ by Pythagoras Theorem,}$
$\displaystyle AC^2=AB^2+BC^2$
$\displaystyle AC^2=36^2+36^2=2(36)^2$
$\displaystyle AC=36\sqrt{2}\text{ cm}$
$\displaystyle \therefore AC=36\sqrt{2}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{In an isosceles triangle }ABC,\text{ if }AB=AC=13\text{ cm and the altitude from }A$
$\displaystyle \text{on }BC\text{ is }5\text{ cm, find }BC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }AD\perp BC.$
$\displaystyle \text{Since }\triangle ABC\text{ is isosceles, the altitude }AD\text{ bisects }BC.$
$\displaystyle \therefore BD=DC$
$\displaystyle \text{In right }\triangle ABD,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AD^2+BD^2$
$\displaystyle 13^2=5^2+BD^2$
$\displaystyle BD^2=169-25=144$
$\displaystyle BD=12\text{ cm}$
$\displaystyle BC=BD+DC=12+12=24\text{ cm}$
$\displaystyle \therefore BC=24\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{A guy wire attached to a vertical pole of height }18\text{ m is }24\text{ m long and has a}$
$\displaystyle \text{stake attached to the other end. How far from the base of the pole should the stake be}$
$\displaystyle \text{driven so that the wire will be taut?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance of the stake from the base of the pole}=x\text{ m.}$
$\displaystyle \text{The pole, ground and guy wire form a right triangle with hypotenuse }24\text{ m.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle x^2+18^2=24^2$
$\displaystyle x^2=576-324=252$
$\displaystyle x=\sqrt{252}=6\sqrt{7}\text{ m}$
$\displaystyle \therefore \text{The stake should be driven }6\sqrt{7}\text{ m from the base of the pole.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{The lengths of the diagonals of a rhombus are }24\text{ cm and }10\text{ cm.}$
$\displaystyle \text{Find each side of the rhombus.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}$
$\displaystyle \therefore \text{Half of the diagonals are }\frac{24}{2}=12\text{ cm and }\frac{10}{2}=5\text{ cm.}$
$\displaystyle \text{Let each side of the rhombus be }x\text{ cm.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle x^2=12^2+5^2$
$\displaystyle x^2=144+25=169$
$\displaystyle x=13\text{ cm}$
$\displaystyle \therefore \text{Each side of the rhombus is }13\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Each side of a rhombus is }10\text{ cm. If one of its diagonals is }16\text{ cm,}$
$\displaystyle \text{find the length of the other diagonal.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}$
$\displaystyle \text{Half of the given diagonal}=\frac{16}{2}=8\text{ cm.}$
$\displaystyle \text{Let half of the other diagonal be }x\text{ cm.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle 10^2=8^2+x^2$
$\displaystyle x^2=100-64=36$
$\displaystyle x=6\text{ cm}$
$\displaystyle \therefore \text{Length of the other diagonal}=2x=12\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Calculate the height of an equilateral triangle each of whose sides} \\ \text{measures }12\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }ABC\text{ be an equilateral triangle with }AB=BC=CA=12\text{ cm and }AD\perp BC.$
$\displaystyle \text{The altitude of an equilateral triangle bisects the opposite side.}$
$\displaystyle \therefore BD=DC=\frac{12}{2}=6\text{ cm.}$
$\displaystyle \text{In right }\triangle ABD,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AD^2+BD^2$
$\displaystyle 12^2=AD^2+6^2$
$\displaystyle AD^2=144-36=108$
$\displaystyle AD=\sqrt{108}=6\sqrt{3}\text{ cm}$
$\displaystyle \therefore \text{Height of the equilateral triangle}=6\sqrt{3}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Determine whether the triangle having sides }(a-1)\text{ cm, }2\sqrt{a}\text{ cm and }$
$\displaystyle (a+1)\text{ cm is a right angled} \ \text{triangle.}\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The largest side is }(a+1)\text{ cm.}$
$\displaystyle (a-1)^2+(2\sqrt{a})^2$
$\displaystyle =a^2-2a+1+4a$
$\displaystyle =a^2+2a+1$
$\displaystyle =(a+1)^2$
$\displaystyle \therefore (a-1)^2+(2\sqrt{a})^2=(a+1)^2$
$\displaystyle \text{Hence, by the converse of Pythagoras Theorem, the given triangle is right angled.}$
$\displaystyle \text{The side }(a+1)\text{ cm is the hypotenuse.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{In an equilateral }\triangle ABC,\ AD\perp BC,\text{ prove that }AD^2=3BD^2.$
$\displaystyle \hfill\text{[CBSE 2002C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }ABC\text{ is an equilateral triangle and }AD\perp BC,\text{ therefore }AD\text{ bisects }BC.$
$\displaystyle \therefore BD=DC=\frac{BC}{2}$
$\displaystyle \therefore BC=2BD$
$\displaystyle \text{Also, }AB=BC=2BD.$
$\displaystyle \text{In right }\triangle ADB,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AD^2+BD^2$
$\displaystyle (2BD)^2=AD^2+BD^2$
$\displaystyle 4BD^2=AD^2+BD^2$
$\displaystyle \therefore AD^2=3BD^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\triangle ABD\text{ is a right triangle right-angled at }A\text{ and }AC\perp BD.\text{ Show that}$
$\displaystyle \text{(i) }AB^2=BC\cdot BD\qquad\text{(ii) }AC^2=BC\cdot DC$
$\displaystyle \text{(iii) }AD^2=BD\cdot CD\qquad\text{(iv) }\frac{AB^2}{AC^2}=\frac{BD}{DC}\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AC\perp BD,\ \angle ACB=\angle ACD=90^\circ.$

$\displaystyle \text{(i) In }\triangle ABC\text{ and }\triangle ABD,$
$\displaystyle \angle ACB=\angle BAD=90^\circ$
$\displaystyle \angle ABC=\angle ABD$
$\displaystyle \therefore \triangle ABC\sim\triangle DBA\qquad\text{[By AA criterion]}$
$\displaystyle \therefore \frac{AB}{BD}=\frac{BC}{AB}$
$\displaystyle \therefore AB^2=BC\cdot BD.$

$\displaystyle \text{(ii) In }\triangle ABC\text{ and }\triangle ACD,$
$\displaystyle \angle ACB=\angle ACD=90^\circ$
$\displaystyle \angle BAC=\angle ADC$
$\displaystyle \therefore \triangle ABC\sim\triangle DAC\qquad\text{[By AA criterion]}$
$\displaystyle \therefore \frac{BC}{AC}=\frac{AC}{DC}$
$\displaystyle \therefore AC^2=BC\cdot DC.$

$\displaystyle \text{(iii) In }\triangle ACD\text{ and }\triangle ABD,$
$\displaystyle \angle ACD=\angle BAD=90^\circ$
$\displaystyle \angle ADC=\angle ADB$
$\displaystyle \therefore \triangle ACD\sim\triangle BAD\qquad\text{[By AA criterion]}$
$\displaystyle \therefore \frac{AD}{BD}=\frac{CD}{AD}$
$\displaystyle \therefore AD^2=BD\cdot CD.$

$\displaystyle \text{(iv) From (i) and (ii),}$
$\displaystyle AB^2=BC\cdot BD\text{ and }AC^2=BC\cdot DC$
$\displaystyle \therefore \frac{AB^2}{AC^2}=\frac{BC\cdot BD}{BC\cdot DC}$
$\displaystyle \therefore \frac{AB^2}{AC^2}=\frac{BD}{DC}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{In the adjoining figure, }\angle B<90^\circ\text{ and segment }AD\perp BC,\text{ show that}$
$\displaystyle \text{(i) }b^2=h^2+a^2+x^2-2ax\qquad\text{(ii) }b^2=a^2+c^2-2ax$ $\displaystyle \text{Answer:}$
$\displaystyle \text{From the figure, }AD=h,\ BD=x,\ BC=a\text{ and }DC=a-x.$

$\displaystyle \text{(i) In right }\triangle ADC,\text{ by Pythagoras Theorem,}$
$\displaystyle AC^2=AD^2+DC^2$
$\displaystyle b^2=h^2+(a-x)^2$
$\displaystyle b^2=h^2+a^2-2ax+x^2$
$\displaystyle \therefore b^2=h^2+a^2+x^2-2ax.$

$\displaystyle \text{(ii) In right }\triangle ABD,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AD^2+BD^2$
$\displaystyle c^2=h^2+x^2$
$\displaystyle \therefore h^2=c^2-x^2.$
$\displaystyle \text{Substituting }h^2=c^2-x^2\text{ in (i),}$
$\displaystyle b^2=c^2-x^2+a^2+x^2-2ax$
$\displaystyle \therefore b^2=a^2+c^2-2ax.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{In the adjoining figure, }D\text{ is the mid-point of side }BC\text{ and }AE\perp BC.$
$\displaystyle \text{ If }BC=a, \ AC=b,\ AB=c,\ ED=x,\ AD=p\text{ and }AE=h,\text{ prove that:}$
$\displaystyle \text{(i) }b^2=p^2+ax+\frac{a^2}{4}\qquad\text{(ii) }c^2=p^2-ax+\frac{a^2}{4}$
$\displaystyle \text{(iii) }b^2+c^2=2p^2+\frac{a^2}{2}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }D\text{ is the mid-point of }BC,$
$\displaystyle BD=DC=\frac{a}{2}$
$\displaystyle \text{Also, }ED=x.$
$\displaystyle \therefore EC=ED+DC=x+\frac{a}{2}$
$\displaystyle BE=BD-ED=\frac{a}{2}-x$
$\displaystyle \text{In right }\triangle AED,\text{ by Pythagoras Theorem,}$
$\displaystyle AD^2=AE^2+ED^2$
$\displaystyle p^2=h^2+x^2$
$\displaystyle \therefore h^2=p^2-x^2\qquad\text{...(1)}$

$\displaystyle \text{(i) In right }\triangle AEC,$
$\displaystyle AC^2=AE^2+EC^2$
$\displaystyle b^2=h^2+\left(x+\frac{a}{2}\right)^2$
$\displaystyle =p^2-x^2+x^2+ax+\frac{a^2}{4}\qquad\text{[Using (1)]}$
$\displaystyle \therefore b^2=p^2+ax+\frac{a^2}{4}.$

$\displaystyle \text{(ii) In right }\triangle AEB,$
$\displaystyle AB^2=AE^2+BE^2$
$\displaystyle c^2=h^2+\left(\frac{a}{2}-x\right)^2$
$\displaystyle =p^2-x^2+\frac{a^2}{4}-ax+x^2\qquad\text{[Using (1)]}$
$\displaystyle \therefore c^2=p^2-ax+\frac{a^2}{4}.$

$\displaystyle \text{(iii) Adding (i) and (ii),}$
$\displaystyle b^2+c^2=p^2+ax+\frac{a^2}{4}+p^2-ax+\frac{a^2}{4}$
$\displaystyle \therefore b^2+c^2=2p^2+\frac{a^2}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{In right-angled triangle }ABC\text{ in which }\angle C=90^\circ,\text{ if }D\text{ is the mid-point of }$
$\displaystyle BC, \ \text{prove that }AB^2=4AD^2-3AC^2.\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }D\text{ is the mid-point of }BC,$
$\displaystyle CD=\frac{BC}{2}\Rightarrow BC=2CD.$
$\displaystyle \text{In right }\triangle ADC,\text{ by Pythagoras Theorem,}$
$\displaystyle AD^2=AC^2+CD^2\qquad\text{...(1)}$
$\displaystyle \text{In right }\triangle ABC,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AC^2+BC^2$
$\displaystyle =AC^2+(2CD)^2$
$\displaystyle =AC^2+4CD^2\qquad\text{...(2)}$
$\displaystyle \text{From (1), }CD^2=AD^2-AC^2.$
$\displaystyle \text{Substituting in (2),}$
$\displaystyle AB^2=AC^2+4(AD^2-AC^2)$
$\displaystyle =AC^2+4AD^2-4AC^2$
$\displaystyle \therefore AB^2=4AD^2-3AC^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Using Pythagoras theorem determine the length of }AD\text{ in terms of }$
$\displaystyle b\text{ and }c \text{ shown in the adjoinng figure.}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{In right }\triangle ABC,\ AB=c,\ AC=b\text{ and }\angle A=90^\circ.$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle BC^2=AB^2+AC^2=c^2+b^2$
$\displaystyle \therefore BC=\sqrt{b^2+c^2}$
$\displaystyle \text{Since }AD\perp BC,\text{ using the area of }\triangle ABC,$
$\displaystyle \frac{1}{2}\times AB\times AC=\frac{1}{2}\times BC\times AD$
$\displaystyle \frac{1}{2}\times c\times b=\frac{1}{2}\times\sqrt{b^2+c^2}\times AD$
$\displaystyle bc=\sqrt{b^2+c^2}\,AD$
$\displaystyle \therefore AD=\frac{bc}{\sqrt{b^2+c^2}}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{There is a staircase as shown in Fig. 7.238, connecting points }A\text{ and }B.$
$\displaystyle \text{Measurements of} \ \text{steps are marked in the figure. Find the straight line distance between }A\text{ and }B.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Total horizontal distance}=2+2+2+2=8\text{ units.}$
$\displaystyle \text{Total vertical distance}=1+1.6+1.6+1.8=6\text{ units.}$
$\displaystyle \text{The horizontal and vertical distances form the perpendicular sides of a right triangle.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle AB^2=8^2+6^2$
$\displaystyle =64+36=100$
$\displaystyle \therefore AB=\sqrt{100}=10\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{In a quadrilateral }ABCD,\ \angle B=90^\circ.$
$\displaystyle \text{ If }AD^2=AB^2+BC^2+CD^2, \ \text{then prove that }\angle ACD=90^\circ.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In right }\triangle ABC,\text{ by Pythagoras Theorem,}$
$\displaystyle AC^2=AB^2+BC^2$
$\displaystyle \text{Given, }AD^2=AB^2+BC^2+CD^2$
$\displaystyle \therefore AD^2=AC^2+CD^2$
$\displaystyle \text{Thus, in }\triangle ACD,\ AD^2=AC^2+CD^2.$
$\displaystyle \therefore \angle ACD=90^\circ\qquad\text{[By the converse of Pythagoras Theorem]}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{In }\triangle ABC,\text{ ray }AD\text{ bisects }\angle A\text{ and intersects }$
$\displaystyle BC\text{ in }D.\text{ If }BC=a, \ AC=b\text{ and }AB=c,\text{ prove that}$
$\displaystyle \text{(i) }BD=\frac{ac}{b+c}\qquad\text{(ii) }DC=\frac{ab}{b+c}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AD\text{ bisects }\angle A,\text{ by the Angle Bisector Theorem,}$
$\displaystyle \frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}$
$\displaystyle \text{Also, }BD+DC=BC=a.$

$\displaystyle \text{(i) Let }BD=x.\text{ Then }DC=a-x.$
$\displaystyle \frac{x}{a-x}=\frac{c}{b}$
$\displaystyle bx=c(a-x)$
$\displaystyle bx=ac-cx$
$\displaystyle x(b+c)=ac$
$\displaystyle x=\frac{ac}{b+c}$
$\displaystyle \therefore BD=\frac{ac}{b+c}.$

$\displaystyle \text{(ii) }DC=a-BD$
$\displaystyle =a-\frac{ac}{b+c}$
$\displaystyle =\frac{a(b+c)-ac}{b+c}$
$\displaystyle =\frac{ab}{b+c}$
$\displaystyle \therefore DC=\frac{ab}{b+c}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{In }\triangle ABC,\ \angle A=60^\circ. \\ \text{ Prove that }BC^2=AB^2+AC^2-AB\times AC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Draw }CD\perp AB.$
$\displaystyle \text{In right }\triangle ACD,\ \angle CAD=60^\circ.$
$\displaystyle \cos60^\circ=\frac{AD}{AC}$
$\displaystyle \frac{1}{2}=\frac{AD}{AC}\Rightarrow AD=\frac{AC}{2}$
$\displaystyle \text{By Pythagoras Theorem in }\triangle ACD,$
$\displaystyle AC^2=AD^2+CD^2$
$\displaystyle CD^2=AC^2-\frac{AC^2}{4}=\frac{3AC^2}{4}$
$\displaystyle \text{Also, }BD=AB-AD=AB-\frac{AC}{2}$
$\displaystyle \text{In right }\triangle BCD,\text{ by Pythagoras Theorem,}$
$\displaystyle BC^2=BD^2+CD^2$
$\displaystyle =\left(AB-\frac{AC}{2}\right)^2+\frac{3AC^2}{4}$
$\displaystyle =AB^2-AB\times AC+\frac{AC^2}{4}+\frac{3AC^2}{4}$
$\displaystyle \therefore BC^2=AB^2+AC^2-AB\times AC.$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{In }\triangle ABC,\ \angle C\text{ is an obtuse angle. }AD\perp BC$
$\displaystyle \text{ and }AB^2=AC^2+3BC^2. \ \text{Prove that }BC=CD.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }\angle C\text{ is obtuse, }D\text{ lies on }BC\text{ produced beyond }C.$
$\displaystyle \therefore BD=BC+CD$
$\displaystyle \text{In right }\triangle ABD,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AD^2+BD^2$
$\displaystyle =AD^2+(BC+CD)^2\qquad\text{...(1)}$
$\displaystyle \text{In right }\triangle ACD,\text{ by Pythagoras Theorem,}$
$\displaystyle AC^2=AD^2+CD^2\qquad\text{...(2)}$
$\displaystyle \text{Subtracting (2) from (1),}$
$\displaystyle AB^2-AC^2=(BC+CD)^2-CD^2$
$\displaystyle =BC^2+2BC\cdot CD$
$\displaystyle \text{Given, }AB^2-AC^2=3BC^2.$
$\displaystyle \therefore 3BC^2=BC^2+2BC\cdot CD$
$\displaystyle 2BC^2=2BC\cdot CD$
$\displaystyle \therefore BC=CD.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{A point }D\text{ is on the side }BC\text{ of an equilateral triangle }$
$\displaystyle ABC\text{ such that }DC=\frac{1}{4}BC. \ \text{Prove that }AD^2=13CD^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }CD=x.$
$\displaystyle \text{Since }DC=\frac{1}{4}BC,\ BC=4x.$
$\displaystyle \therefore BD=BC-CD=4x-x=3x.$
$\displaystyle \text{Since }\triangle ABC\text{ is equilateral, }AB=BC=4x.$
$\displaystyle \text{Draw }AE\perp BC.$
$\displaystyle \text{In an equilateral triangle, the altitude bisects the opposite side.}$
$\displaystyle \therefore BE=EC=\frac{BC}{2}=2x.$
$\displaystyle ED=BD-BE=3x-2x=x.$
$\displaystyle \text{In right }\triangle ABE,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AE^2+BE^2$
$\displaystyle (4x)^2=AE^2+(2x)^2$
$\displaystyle AE^2=16x^2-4x^2=12x^2$
$\displaystyle \text{In right }\triangle AED,\text{ by Pythagoras Theorem,}$
$\displaystyle AD^2=AE^2+ED^2$
$\displaystyle =12x^2+x^2=13x^2$
$\displaystyle \text{Since }CD=x,$
$\displaystyle \therefore AD^2=13CD^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{In }\triangle ABC,\text{ if }BD\perp AC\text{ and }BC^2=2AC\times CD,\text{ then prove that } \\ AB=AC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }BD\perp AC,\ \triangle BDC\text{ and }\triangle BDA\text{ are right triangles.}$
$\displaystyle \text{In right }\triangle BDC,\text{ by Pythagoras Theorem,}$
$\displaystyle BC^2=BD^2+CD^2\qquad\text{...(1)}$
$\displaystyle \text{Given, }BC^2=2AC\times CD.$
$\displaystyle \therefore BD^2+CD^2=2AC\times CD\qquad\text{[Using (1)]}$
$\displaystyle BD^2=2AC\cdot CD-CD^2\qquad\text{...(2)}$
$\displaystyle \text{Since }D\text{ lies on }AC,\ AD=AC-CD.$
$\displaystyle \text{In right }\triangle ABD,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AD^2+BD^2$
$\displaystyle =(AC-CD)^2+2AC\cdot CD-CD^2\qquad\text{[Using (2)]}$
$\displaystyle =AC^2-2AC\cdot CD+CD^2+2AC\cdot CD-CD^2$
$\displaystyle =AC^2$
$\displaystyle \therefore AB^2=AC^2$
$\displaystyle \therefore AB=AC.$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{In }\triangle ABC,\text{ given that }AB=AC\text{ and }BD\perp AC.\text{ Prove that } \\ BC^2=2AC\times CD.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In right }\triangle ABD,\text{ by Pythagoras Theorem,}$
$\displaystyle AB^2=AD^2+BD^2$
$\displaystyle \text{Since }AB=AC,$
$\displaystyle AC^2=AD^2+BD^2$
$\displaystyle \therefore BD^2=AC^2-AD^2$
$\displaystyle =(AC-AD)(AC+AD)$
$\displaystyle =CD(AC+AD)\qquad[\because AC-AD=CD]$
$\displaystyle \text{In right }\triangle BCD,\text{ by Pythagoras Theorem,}$
$\displaystyle BC^2=BD^2+CD^2$
$\displaystyle =CD(AC+AD)+CD^2$
$\displaystyle =CD(AC+AD+CD)$
$\displaystyle =CD(AC+AC)\qquad[\because AD+CD=AC]$
$\displaystyle \therefore BC^2=2AC\times CD.$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }ABCD\text{ is a rectangle. Points }M\text{ and }N\text{ are on }BD\text{ such that }$
$\displaystyle AM\perp BD \ \text{and }CN\perp BD.\text{ Prove that }BM^2+BN^2=DM^2+DN^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In right }\triangle ABD,\ AM\perp BD.$
$\displaystyle \therefore AB^2=BM\cdot BD\text{ and }AD^2=DM\cdot BD.$
$\displaystyle \text{In right }\triangle CBD,\ CN\perp BD.$
$\displaystyle \therefore BC^2=BN\cdot BD\text{ and }CD^2=DN\cdot BD.$
$\displaystyle \text{Since }ABCD\text{ is a rectangle, }AB=CD\text{ and }AD=BC.$
$\displaystyle \therefore AB^2=CD^2$
$\displaystyle BM\cdot BD=DN\cdot BD$
$\displaystyle \therefore BM=DN\qquad\text{...(i)}$
$\displaystyle \text{Also, }AD^2=BC^2$
$\displaystyle DM\cdot BD=BN\cdot BD$
$\displaystyle \therefore DM=BN\qquad\text{...(ii)}$
$\displaystyle \text{Squaring (i) and (ii), we get }BM^2=DN^2\text{ and }DM^2=BN^2.$
$\displaystyle \therefore BM^2+BN^2=DN^2+DM^2$
$\displaystyle \therefore BM^2+BN^2=DM^2+DN^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{In }\triangle ABC,\ \angle ABC=135^\circ. \\ \text{ Prove that }AC^2=AB^2+BC^2+4\,ar(\triangle ABC).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Draw }AD\perp BC\text{ produced beyond }B.$
$\displaystyle \therefore \angle ABD=180^\circ-\angle ABC=180^\circ-135^\circ=45^\circ.$
$\displaystyle \text{In right }\triangle ABD,\ \angle ADB=90^\circ\text{ and }\angle ABD=45^\circ.$
$\displaystyle \therefore \angle BAD=45^\circ$
$\displaystyle \therefore AD=BD.$
$\displaystyle \text{By Pythagoras Theorem in }\triangle ABD,$
$\displaystyle AB^2=AD^2+BD^2=2AD^2\qquad\text{...(i)}$
$\displaystyle \text{In right }\triangle ACD,\text{ by Pythagoras Theorem,}$
$\displaystyle AC^2=AD^2+CD^2$
$\displaystyle =AD^2+(BC+BD)^2$
$\displaystyle =AD^2+BC^2+BD^2+2BC\cdot BD$
$\displaystyle =AB^2+BC^2+2BC\cdot AD\qquad\text{[Using (i) and }BD=AD\text{]}$
$\displaystyle \text{Now, }ar(\triangle ABC)=\frac{1}{2}\times BC\times AD$
$\displaystyle \therefore 2BC\cdot AD=4\,ar(\triangle ABC).$
$\displaystyle \therefore AC^2=AB^2+BC^2+4\,ar(\triangle ABC).$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{Nazima is fly fishing in a stream. The tip of her fishing rod is }1.8\text{ m above the surface}$
$\displaystyle \text{of the water and the fly at the end of the string rests on the water }3.6\text{ m away and }2.4\text{ m}$
$\displaystyle \text{from a point directly under the tip of the rod. Assuming that her string (from the tip of her}$
$\displaystyle \text{rod to the fly) is taut, how much string does she have out? If she pulls the string at the rate}$
$\displaystyle \text{of } 5\text{ cm per second,} \ \text{what will be the horizontal distance of the fly from her after }12\text{ seconds?}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Let }B\text{ be the tip of the rod, }C\text{ the initial position of the fly and }D\text{ the point directly below }B.$
$\displaystyle BD=1.8\text{ m and }CD=2.4\text{ m}$
$\displaystyle \text{In right }\triangle BCD,\text{ by Pythagoras Theorem,}$
$\displaystyle BC^2=BD^2+CD^2$
$\displaystyle =(1.8)^2+(2.4)^2$
$\displaystyle =3.24+5.76=9$
$\displaystyle BC=3\text{ m}$
$\displaystyle \therefore \text{Nazima has }3\text{ m of string out.}$
$\displaystyle \text{In }12\text{ seconds, length of string pulled}=5\times12=60\text{ cm}=0.6\text{ m}$
$\displaystyle \therefore \text{Length of string remaining}=3-0.6=2.4\text{ m}$
$\displaystyle \text{Let the new horizontal distance of the fly from }D\text{ be }x\text{ m.}$
$\displaystyle \text{By Pythagoras Theorem,}$
$\displaystyle x^2+(1.8)^2=(2.4)^2$
$\displaystyle x^2=5.76-3.24=2.52$
$\displaystyle x=\sqrt{2.52}\approx1.59\text{ m}$
$\displaystyle \text{From the figure, the horizontal distance of }D\text{ from Nazima}=3.6-2.4=1.2\text{ m}$
$\displaystyle \therefore \text{Horizontal distance of the fly from Nazima}=1.59+1.2=2.79\text{ m (approx.).}$
$\displaystyle \\$