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NCERT Solutions · Class 10 · Chapter 6: Triangles
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2
Exercise 6.2: Basic Proportionality Theorem and its converse. BPT: a line parallel to one side of a triangle, cutting the other two sides, divides them in the same ratio: \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\). Converse: if a line divides two sides in the same ratio, it is parallel to the third side.
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Exercise 6.2 questions and solutions
Exercise 6.2, Question 1
In \(\triangle ABC\), \(DE \parallel BC\) with D on AB and E on AC.
(i) \(AD = 1.5,\ DB = 3,\ AE = 1\) cm. Find EC.
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BPT: \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\). \[\begin{aligned}&\dfrac{1.5}{3} = \dfrac{1}{EC} \\ \Rightarrow\ &EC = 2\end{aligned}\] Answer: \(EC = 2\) cm
(ii) \(DB = 7.2,\ AE = 1.8,\ EC = 5.4\) cm. Find AD.
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\[\begin{aligned}\dfrac{AD}{7.2} &= \dfrac{1.8}{5.4} \\ &= \dfrac13\end{aligned}\] \(AD = 2.4\). Answer: \(AD = 2.4\) cm
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Exercise 6.2, Question 2
E and F are points on PQ and PR of \(\triangle PQR\). Is \(EF \parallel QR\)?
(i) \(PE = 3.9,\ EQ = 3,\ PF = 3.6,\ FR = 2.4\) cm
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\(\dfrac{PE}{EQ} = \dfrac{3.9}{3} = 1.3\) and \(\dfrac{PF}{FR} = \dfrac{3.6}{2.4} = 1.5\). The ratios differ, so by the converse of BPT the line is not parallel. Answer: No
(ii) \(PE = 4,\ QE = 4.5,\ PF = 8,\ RF = 9\) cm
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\[\begin{aligned}\dfrac{PE}{EQ} &= \dfrac{4}{4.5} \\ &= \dfrac89\end{aligned}\] and \(\dfrac{PF}{FR} = \dfrac89\). Equal ratios, so \(EF \parallel QR\) (converse of BPT). Answer: Yes
(iii) \(PQ = 1.28,\ PR = 2.56,\ PE = 0.18,\ PF = 0.36\) cm
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\[\begin{aligned}\dfrac{PE}{PQ} &= \dfrac{0.18}{1.28} \\ &= \dfrac{9}{64}\end{aligned}\] and \[\begin{aligned}\dfrac{PF}{PR} &= \dfrac{0.36}{2.56} \\ &= \dfrac{9}{64}\end{aligned}\] Equal, so \(EF \parallel QR\). Answer: Yes
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Exercise 6.2, Question 3
In the figure, L is on AC, M on AB and N on AD, with \(LM \parallel CB\) and \(LN \parallel CD\). Prove \(\dfrac{AM}{AB} = \dfrac{AN}{AD}\).
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In \(\triangle ABC\), \(LM \parallel CB\), so by BPT \(\dfrac{AM}{AB} = \dfrac{AL}{AC}\). In \(\triangle ADC\), \(LN \parallel CD\), so \(\dfrac{AN}{AD} = \dfrac{AL}{AC}\). Both equal \(\dfrac{AL}{AC}\), so \(\dfrac{AM}{AB} = \dfrac{AN}{AD}\). Answer: Proved (both ratios equal \(\dfrac{AL}{AC}\)).
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Exercise 6.2, Question 4
In \(\triangle ABC\), D is on AB, E on BC and F on BE, with \(DE \parallel AC\) and \(DF \parallel AE\). Prove \(\dfrac{BF}{FE} = \dfrac{BE}{EC}\).
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In \(\triangle ABC\), \(DE \parallel AC\): \(\dfrac{BD}{DA} = \dfrac{BE}{EC}\). In \(\triangle ABE\), \(DF \parallel AE\): \(\dfrac{BD}{DA} = \dfrac{BF}{FE}\). So \(\dfrac{BF}{FE} = \dfrac{BE}{EC}\). Answer: Proved.
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Exercise 6.2, Question 5
O is a point inside \(\triangle PQR\). D is on OP, E on OQ and F on OR, with \(DE \parallel PQ\) and \(DF \parallel PR\). Show \(EF \parallel QR\).
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In \(\triangle POQ\), \(DE \parallel PQ\): \(\dfrac{OD}{DP} = \dfrac{OE}{EQ}\). In \(\triangle POR\), \(DF \parallel PR\): \(\dfrac{OD}{DP} = \dfrac{OF}{FR}\). So \(\dfrac{OE}{EQ} = \dfrac{OF}{FR}\); by the converse of BPT in \(\triangle OQR\), \(EF \parallel QR\). Answer: Proved: \(EF \parallel QR\).
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Exercise 6.2, Question 6
A, B, C are points on OP, OQ, OR with \(AB \parallel PQ\) and \(AC \parallel PR\). Show \(BC \parallel QR\).
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In \(\triangle OPQ\), \(AB \parallel PQ\): \(\dfrac{OA}{AP} = \dfrac{OB}{BQ}\). In \(\triangle OPR\), \(AC \parallel PR\): \(\dfrac{OA}{AP} = \dfrac{OC}{CR}\). So \(\dfrac{OB}{BQ} = \dfrac{OC}{CR}\), and by the converse of BPT in \(\triangle OQR\), \(BC \parallel QR\). Answer: Proved: \(BC \parallel QR\).
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Exercise 6.2, Question 7
Using BPT, prove that a line through the midpoint of one side of a triangle, parallel to another side, bisects the third side.
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Let D be the midpoint of AB and \(DE \parallel BC\) with E on AC. BPT: \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\). Since \(AD = DB\), the left side is 1. So \(AE = EC\): E is the midpoint of AC. Answer: Proved.
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Exercise 6.2, Question 8
Using the converse of BPT, prove that the line joining the midpoints of two sides of a triangle is parallel to the third side.
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Let D, E be the midpoints of AB and AC. \(\dfrac{AD}{DB} = 1 = \dfrac{AE}{EC}\). Equal ratios, so by the converse of BPT, \(DE \parallel BC\). Answer: Proved.
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Exercise 6.2, Question 9
ABCD is a trapezium with \(AB \parallel DC\); its diagonals meet at O. Show \(\dfrac{AO}{BO} = \dfrac{CO}{DO}\).
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Draw \(EO \parallel AB\) (so also \(\parallel DC\)) with E on AD. In \(\triangle ADC\), \(EO \parallel DC\): \(\dfrac{AE}{ED} = \dfrac{AO}{OC}\). In \(\triangle DAB\), \(EO \parallel AB\): \(\dfrac{AE}{ED} = \dfrac{BO}{OD}\). So \(\dfrac{AO}{OC} = \dfrac{BO}{OD}\), which rearranges to \(\dfrac{AO}{BO} = \dfrac{CO}{DO}\). Answer: Proved.
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Exercise 6.2, Question 10
The diagonals of quadrilateral ABCD meet at O with \(\dfrac{AO}{BO} = \dfrac{CO}{DO}\). Show that ABCD is a trapezium.
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Draw \(OE \parallel AB\) with E on AD. In \(\triangle DAB\), \(\dfrac{AE}{ED} = \dfrac{BO}{OD}\) (BPT). Given \(\dfrac{AO}{BO} = \dfrac{CO}{DO}\), so \[\begin{aligned}\dfrac{AO}{OC} &= \dfrac{BO}{OD} \\ &= \dfrac{AE}{ED}\end{aligned}\] In \(\triangle ADC\) the converse of BPT gives \(EO \parallel DC\). Since also \(EO \parallel AB\), \(AB \parallel DC\): a trapezium. Answer: Proved: \(AB \parallel DC\).
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Triangles in our sample papers: Sample paper 1 (questions 8, 23, 28, 33) · Sample paper 2 (questions 9, 23, 29, 33) · Sample paper 3 (questions 10, 12, 19, 28, 33) · Sample paper 4 (questions 9, 29, 33) · Sample paper 5 (questions 9, 10, 28, 33).
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