Class 10: Triangles – Exercise 7.2

$\displaystyle \textbf{Question 1: }\text{In a }\triangle ABC,\ D\text{ and }E\text{ are points on the sides }AB\text{ and }AC$
$\displaystyle \text{respectively such that }DE\parallel BC.$
$\displaystyle \text{(i) If }AD=6\text{ cm},\ DB=9\text{ cm and }AE=8\text{ cm, find }AC.$
$\displaystyle \text{(ii) If }\frac{AD}{DB}=\frac{3}{4}\text{ and }AC=15\text{ cm, find }AE.$
$\displaystyle \text{(iii) If }AD=2.5\text{ cm},\ BD=3.0\text{ cm and }AE=3.75\text{ cm, find the length of }AC.$
$\displaystyle \hfill\text{[CBSE 2006C]}$
$\displaystyle \text{(iv) If }AD=4,\ AE=8,\ DB=x-4\text{ and }EC=3x-19,\text{ find }x.$
$\displaystyle \text{(v) If }AD=8\text{ cm},\ AB=12\text{ cm and }AE=12\text{ cm, find }CE.$
$\displaystyle \text{(vi) If }AD=2.4\text{ cm},\ AE=3.2\text{ cm},\ DE=2\text{ cm and }BC=5\text{ cm, find }BD\text{ and }CE.$
$\displaystyle \hfill\text{[CBSE 2001C]}$
$\displaystyle \text{(vii) If }AD=x,\ DB=x-2,\ AE=x+2\text{ and }EC=x-1,\text{ find }x.$
$\displaystyle \hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }DE\parallel BC,\text{ by the Basic Proportionality Theorem,}$
$\displaystyle \frac{AD}{DB}=\frac{AE}{EC}.$

$\displaystyle \text{(i) }\frac{6}{9}=\frac{8}{EC}$
$\displaystyle 6EC=72\Rightarrow EC=12\text{ cm}$
$\displaystyle AC=AE+EC=8+12=20\text{ cm}$
$\displaystyle \therefore AC=20\text{ cm.}$

$\displaystyle \text{(ii) }\frac{AD}{DB}=\frac{AE}{EC}=\frac{3}{4}$
$\displaystyle \therefore AE:EC=3:4$
$\displaystyle AC=AE+EC=15\text{ cm}$
$\displaystyle AE=\frac{3}{3+4}\times15=\frac{45}{7}\text{ cm}$
$\displaystyle \therefore AE=\frac{45}{7}\text{ cm.}$

$\displaystyle \text{(iii) }\frac{2.5}{3}=\frac{3.75}{EC}$
$\displaystyle 2.5EC=3\times3.75=11.25$
$\displaystyle EC=\frac{11.25}{2.5}=4.5\text{ cm}$
$\displaystyle AC=AE+EC=3.75+4.5=8.25\text{ cm}$
$\displaystyle \therefore AC=8.25\text{ cm.}$

$\displaystyle \text{(iv) }\frac{4}{x-4}=\frac{8}{3x-19}$
$\displaystyle 4(3x-19)=8(x-4)$
$\displaystyle 12x-76=8x-32$
$\displaystyle 4x=44\Rightarrow x=11$
$\displaystyle \therefore x=11.$

$\displaystyle \text{(v) }DB=AB-AD=12-8=4\text{ cm}$
$\displaystyle \frac{8}{4}=\frac{12}{CE}$
$\displaystyle 8CE=48\Rightarrow CE=6\text{ cm}$
$\displaystyle \therefore CE=6\text{ cm.}$

$\displaystyle \text{(vi) Since }DE\parallel BC,\ \frac{AD}{AB}=\frac{DE}{BC}$
$\displaystyle \frac{2.4}{AB}=\frac{2}{5}\Rightarrow AB=\frac{2.4\times5}{2}=6\text{ cm}$
$\displaystyle BD=AB-AD=6-2.4=3.6\text{ cm}$
$\displaystyle \text{Also, }\frac{AE}{AC}=\frac{DE}{BC}$
$\displaystyle \frac{3.2}{AC}=\frac{2}{5}\Rightarrow AC=\frac{3.2\times5}{2}=8\text{ cm}$
$\displaystyle CE=AC-AE=8-3.2=4.8\text{ cm}$
$\displaystyle \therefore BD=3.6\text{ cm and }CE=4.8\text{ cm.}$

$\displaystyle \text{(vii) }\frac{x}{x-2}=\frac{x+2}{x-1}$
$\displaystyle x(x-1)=(x+2)(x-2)$
$\displaystyle x^2-x=x^2-4$
$\displaystyle x=4$
$\displaystyle \therefore x=4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{In a }\triangle ABC,\ D\text{ and }E\text{ are points on the sides }AB\text{ and }AC\text{ respectively.}$
$\displaystyle \text{For each of the following cases show that }DE\parallel BC:$
$\displaystyle \text{(i) }AB=12\text{ cm},\ AD=8\text{ cm},\ AE=12\text{ cm and }AC=18\text{ cm}.$
$\displaystyle \text{(ii) }AD=5.7\text{ cm},\ BD=9.5\text{ cm},\ AE=3.3\text{ cm and }EC=5.5\text{ cm}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }DB=AB-AD=12-8=4\text{ cm}$
$\displaystyle EC=AC-AE=18-12=6\text{ cm}$
$\displaystyle \frac{AD}{DB}=\frac{8}{4}=2$
$\displaystyle \frac{AE}{EC}=\frac{12}{6}=2$
$\displaystyle \therefore \frac{AD}{DB}=\frac{AE}{EC}$
$\displaystyle \text{Thus, }D\text{ and }E\text{ divide the sides }AB\text{ and }AC\text{ respectively in the same ratio.}$
$\displaystyle \therefore DE\parallel BC\qquad\text{[By the converse of Basic Proportionality Theorem]}$

$\displaystyle \text{(ii) }\frac{AD}{DB}=\frac{5.7}{9.5}=\frac{3}{5}$
$\displaystyle \frac{AE}{EC}=\frac{3.3}{5.5}=\frac{3}{5}$
$\displaystyle \therefore \frac{AD}{DB}=\frac{AE}{EC}$
$\displaystyle \text{Thus, }D\text{ and }E\text{ divide the sides }AB\text{ and }AC\text{ respectively in the same ratio.}$
$\displaystyle \therefore DE\parallel BC\qquad\text{[By the converse of Basic Proportionality Theorem]}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{In adjoining Figure, state if }PQ\parallel EF.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle DEF,\ P\text{ and }Q\text{ lie on }DE\text{ and }DF\text{ respectively.}$
$\displaystyle EP=3\text{ cm},\ PD=3.9\text{ cm},\ FQ=2.4\text{ cm and }QD=3.6\text{ cm}$
$\displaystyle \frac{DP}{PE}=\frac{3.9}{3}=\frac{13}{10}$
$\displaystyle \frac{DQ}{QF}=\frac{3.6}{2.4}=\frac{3}{2}$
$\displaystyle \therefore \frac{DP}{PE}\ne\frac{DQ}{QF}$
$\displaystyle \therefore PQ\text{ is not parallel to }EF\text{, by the converse of Basic Proportionality Theorem.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the value of }x\text{ for which }DE\parallel AB\text{ in adjoining Figure }$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }DE\parallel AB,\text{ by the Basic Proportionality Theorem,}$
$\displaystyle \frac{CD}{DA}=\frac{CE}{EB}$
$\displaystyle \frac{x+3}{3x+19}=\frac{x}{3x+4}$
$\displaystyle (x+3)(3x+4)=x(3x+19)$
$\displaystyle 3x^2+13x+12=3x^2+19x$
$\displaystyle 6x=12$
$\displaystyle x=2$
$\displaystyle \therefore x=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{In adjoining Figure, if }AB\parallel CD,\text{ find the value of }x.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AB\parallel CD,\text{ the diagonals of the trapezium divide each other proportionally.}$
$\displaystyle \therefore \frac{AO}{OC}=\frac{BO}{OD}$
$\displaystyle \frac{3x-1}{5x-3}=\frac{2x+1}{6x-5}$
$\displaystyle (3x-1)(6x-5)=(2x+1)(5x-3)$
$\displaystyle 18x^2-21x+5=10x^2-x-3$
$\displaystyle 8x^2-20x+8=0$
$\displaystyle 2x^2-5x+2=0$
$\displaystyle (2x-1)(x-2)=0$
$\displaystyle x=\frac{1}{2}\text{ or }x=2$
$\displaystyle \text{For }x=\frac{1}{2},\ OC=5x-3=-\frac{1}{2},\text{ which is not possible for a length.}$
$\displaystyle \therefore x=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{In adjoining Figure, }AB\parallel CD.\text{ If }OA=3x-19,\ OB=x-4, \\ OC=x-3\text{ and }OD=4,\text{ find }x.\hfill\text{[CBSE 2000C]}$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AB\parallel CD,\text{ the diagonals of the trapezium divide each other proportionally.}$
$\displaystyle \therefore \frac{OA}{OC}=\frac{OB}{OD}$
$\displaystyle \frac{3x-19}{x-3}=\frac{x-4}{4}$
$\displaystyle 4(3x-19)=(x-4)(x-3)$
$\displaystyle 12x-76=x^2-7x+12$
$\displaystyle x^2-19x+88=0$
$\displaystyle (x-8)(x-11)=0$
$\displaystyle x=8\text{ or }x=11$
$\displaystyle \therefore x=8\text{ or }11.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }D\text{ and }E\text{ are points on sides }AB\text{ and }AC\text{ respectively of a }\triangle ABC$
$\displaystyle \text{such that }DE\parallel BC\text{ and }BD=CE,\text{ prove that }\triangle ABC\text{ is isosceles.}$
$\displaystyle \hfill\text{[CBSE 2007, 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }DE\parallel BC,\text{ by the Basic Proportionality Theorem,}$
$\displaystyle \frac{AD}{DB}=\frac{AE}{EC}$
$\displaystyle \text{Given }BD=CE.$
$\displaystyle \therefore \frac{AD}{BD}=\frac{AE}{BD}$
$\displaystyle \therefore AD=AE$
$\displaystyle AB=AD+DB$
$\displaystyle AC=AE+EC$
$\displaystyle \text{Since }AD=AE\text{ and }DB=EC,$
$\displaystyle AB=AC$
$\displaystyle \therefore \triangle ABC\text{ is isosceles.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{State the converse of Basic Proportionality Theorem. Also, find }\frac{BF}{FC}$
$\displaystyle \text{in adjoining Figure, given that }AB\parallel DC\parallel EF\text{ and }\frac{AE}{ED}=\frac{2}{3}.$
$\displaystyle \text{Also, find the length of }EF\text{ if }AB=10\text{ cm and }DC=15\text{ cm.} \hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Converse of Basic Proportionality Theorem: If a line divides any two sides of a}$
$\displaystyle \text{triangle in the same ratio, then the line is parallel to the third side.}$
$\displaystyle \text{Since }AB\parallel EF\parallel DC,\text{ the parallel lines divide the transversals }AD\text{ and }BC$
$\displaystyle \text{proportionally.}$
$\displaystyle \therefore \frac{BF}{FC}=\frac{AE}{ED}=\frac{2}{3}$
$\displaystyle \therefore \frac{BF}{FC}=\frac{2}{3}.$
$\displaystyle \text{Let }AC\text{ intersect }EF\text{ at }G.$
$\displaystyle \text{In }\triangle ADC,\ EG\parallel DC.$
$\displaystyle \therefore \frac{AE}{AD}=\frac{EG}{DC}$
$\displaystyle \frac{AE}{AD}=\frac{2}{2+3}=\frac{2}{5}$
$\displaystyle \therefore \frac{EG}{15}=\frac{2}{5}\Rightarrow EG=6\text{ cm}$
$\displaystyle \text{Also, in }\triangle ABC,\ GF\parallel AB.$
$\displaystyle \frac{CG}{CA}=\frac{GF}{AB}$
$\displaystyle \frac{CG}{CA}=\frac{3}{2+3}=\frac{3}{5}$
$\displaystyle \therefore \frac{GF}{10}=\frac{3}{5}\Rightarrow GF=6\text{ cm}$
$\displaystyle EF=EG+GF=6+6=12\text{ cm}$
$\displaystyle \therefore EF=12\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{State the Basic Proportionality Theorem. Use the theorem to prove the following:}$
$\displaystyle \text{In }\triangle ABC,\ AD\text{ is the angle bisector of angle }A.\ BA\text{ is produced to }E\text{ such that}$
$\displaystyle CE\parallel AD.\text{ Prove that }\frac{BD}{DC}=\frac{BA}{AC}.\hfill\text{[CBSE 2025]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle}$
$\displaystyle \text{to intersect the other two sides, then it divides those two sides in the same ratio.}$
$\displaystyle \text{In }\triangle BCE,\ A\text{ lies on }BE,\ D\text{ lies on }BC\text{ and }AD\parallel CE.$
$\displaystyle \therefore \frac{BD}{DC}=\frac{BA}{AE}\qquad\text{[By Basic Proportionality Theorem]}$
$\displaystyle \text{Since }AD\text{ bisects }\angle BAC,$
$\displaystyle \angle BAD=\angle DAC$
$\displaystyle \text{Also, }CE\parallel AD\text{ and }B,A,E\text{ are collinear.}$
$\displaystyle \therefore \angle AEC=\angle BAD\text{ and }\angle ACE=\angle DAC$
$\displaystyle \therefore \angle AEC=\angle ACE$
$\displaystyle \therefore AE=AC\qquad\text{[Sides opposite to equal angles are equal]}$
$\displaystyle \text{Substituting }AE=AC\text{ in }\frac{BD}{DC}=\frac{BA}{AE},$
$\displaystyle \therefore \frac{BD}{DC}=\frac{BA}{AC}.$
$\displaystyle \\$