EXERCISE
1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
2. E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm 3. In Fig., if LM || CB and LN || CD, A B C D M L N prove that ${AM}/{AB}={AN}/{AD}$ 4. In Fig., DE || AC and DF || AE. A C B D E F Prove that ${BF}/{FE}={BE}/{EC}$ 5. In Fig. DE || OQ and DF || OR. P Q R O E F D Show that EF || QR. 6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. P Q R O A B C
Show that BC || QR.
7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle
parallel to another side bisects the third side.
8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is
parallel to the third side.
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
Show that ${AO}/{BO}={CO}/{DO}$
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that
${AO}/{BO}={CO}/{DO}$.Show that ABCD is a trapezium.
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Solutions: 1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution In ∆ABC, DE || BC. Therefore, by theorem , If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. i.e ${AD}/{DB}={AE}/{EC}$ i.e ${1.5}/{3}={1}/{EC}$ i.e ${1.5}/{3}={1}/{EC}$ $EC=2$ cm. In fig (ii), i.e ${AD}/{DB}={AE}/{EC}$ i.e ${AD}/{7.2}={1.8}/{5.4}$ i.e ${AD}={7.2}/{3}$=$2.4$ cm 2. E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm Solution In ∆PQR, if EF || QR , then ${PE}/{EQ}={PF}/{FR}$ So,${PE}/{EQ}=3.9/3 = 1.3$ ${PF}/{FR}=3.6/2.4=3/2=1.5$ Since, ${PE}/{EQ}≠{PF}/{FR}$, therefore EF is not parallel to QR. . (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm Solution In ∆PQR, if EF || QR , then ${PE}/{EQ}={PF}/{FR}$ So,${PE}/{EQ}= 4/4.5=8/9$ and ${PF}/{FR}= 8/9$ Since ${PE}/{EQ}={PF}/{FR}$, EF is parallel to QR . (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm Solution In ∆PQR, if EF || QR , then ${PE}/{EQ}={PF}/{FR}$ So,${PE}/{EQ}= 1.28/2.56=1/2$ and ${PF}/{FR}= 0.18/0.36=1/2$ Since ${PE}/{EQ}={PF}/{FR}$, EF is parallel to QR . 3. In Fig., if LM || CB and LN || CD, prove that ${AM}/{AB}={AN}/{AD}$ Solution: A B C D M L N In ∆ABC, since LM || BC, ${AM}/{MB}={AL}/{LC}$ or ${MB}/{AM}= {LC}/ {AL}$ Adding 1 on both side, ${MB}/{AM}+1={LC}/ {AL} +1$ ${MB+AM}/{AM }={LC+AL}/{AL}$ Since (MB+AM)=AB and LC+AL = AC ${AB}/{AM }={AC}/{AL}$ ..............(1) Similairly in ∆ACD, since LN||CD, ${AN}/{ND}={AL}/{LC}$ or ${ND}/{AN}={LC}/{AL}$ Adding 1 on both sides, ${ND}/{AN}+1={LC}/{AL}+1$ ${AN+ND}/{AN}={AL+LC}/{AL}$ Since AN+ND = AD and AL+LC=AC ${AD}/{AN}={AC}/{AL}$ ...........(2) Therefore from eqn (1) and (2) ${AB}/{AM }={AC}/{AL}= {AD}/{AN}$ i.e ${AB}/{AM }= {AD}/{AN}$ i.e ${AM}/{AB}={AN}/{AD}$.. Hence Proved. 4. In Fig. DE || AC and DF || AE. Prove that ${BF}/{FE}={BE}/{EC}$ Solution: A C B D E F In ∆ABC,since DE|| AC, ${BD}/{DA}={BE}/{EC}$ ...................(1) In ∆ABE, DF||AE, Therefore, ${BD}/{DA}={BF}/{FE}$ ..............(2) From (1) and (2), ${BF}/{FE}={BE}/{EC}$ ..Hence Proved 5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR. Solution: P Q R O E F D P O Q E D P O R F D In ∆OPQ, since DE || OQ, ${PE}/{EQ}={PD}/{DO}$ ............(1) In ∆OPR, since DF || OR, ${PF}/{FR}={PD}/{DO}$ .............(2) Therefore from (1) and (2) , ${PE}/{EQ}= {PF}/{FR}$ Since E & F are the points on PQ & PR of ∆PQR, and if ${PE}/{EQ}= {PF}/{FR}$ then EF must be parallel to QR. Hence Proved.. 6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. Solution: P Q R O A B C Q P O A B R P O A C In ∆OPQ, since AB || PQ, Therefore, ${OA}/{AP}= {OB}/{BQ}$ ...........(1) In ∆OPR, since AC || PR, Therefore, ${OA}/{AP}={OC}/{CR}$ ...........(2) From (1) and (2), ${OB}/{BQ}= {OC}/{CR}$ ..........(3) Since B & C are the points on OQ & OR of ∆OQR, therefore BC must be parallel to QR to satisfy the above relation. 7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. Solution: A B C D E Since according to theorem, If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Therefore in ∆ABC,since DE || BC ${AD}/{DB}={AE}/{EC}$ Since D is mid point of AB, AD= DB ${AD}/{AD}={AE}/{EC}$ ${EC}={AE}$
Since E is a point on AC and AE=EC, so E is the mid point of AC. 8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. Solution: A B C D E In ∆ABC, let D be the mid point on AB and E be the mid point on AC. Therfore $AD=DB$ and $AE=EC$ Ratio of ${AD}/{DB}=1$ and ${AE}/{EC}=1$ Therefore ${AD}/{DB}= {AE}/{EC}$ This relationship is only fulfilled be DE || BC. Hence Proved. 9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that ${AO}/{BO}={CO}/{DO}$ Solution: A B C D O E F Let draw a line passing through O, parallel to AB and intersecting at E on AD and F on BC. Therefore in ∆ACD, ${AE}/{ED}={AO}/{OC}$ ............(1) ∆BCD, ${BF}/{FC}={BO}/{OD}$ .................(2) ∆ABD, ${AE}/{ED}={BO}/{OD}$ .................(3) ∆ABC, ${AO}/{OC}={BF}/{FC}$ ..................(4) Therefore from (1),(2), (3) and (4) ${AO}/{BO}={CO}/{DO}$ Hence Proved .
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that
${AO}/{BO}={CO}/{DO}$.Show that ABCD is a trapezium.
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ii E and F are points on the sides PQ and PR respectively of a Δ PQR. For the following case, state whether EF || QR. ii PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
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Solution : Let ladder be AB, B be the window and CB be the wall. Then, ABC is a right triangle, angled at C. \(\therefore\) \({AB}^2\) = \({AC}^2\) + \({BC}^2\) So, \({10}^2\) = \({AC}^2\) + \(8^2\) or \({AC}^2\) = 100 – 64 \(\implies\) \({AC}^2\) = 36 \(\implies\) AC = 6 m
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Solution:
We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.
Consider the trapezium ABCD as shown below.
In trapezium ABCD,
AB || CD
Also, AC and BD intersect at ‘O’
Construct XY parallel to AB and CD (XY || AB, XY || CD) through ‘O’
In ΔABC
OY || AB (construction)
According to theorem 6.1 (Basic Proportionality Theorem)
BY/CY = AO/OC................. (1)
In ΔBCD
OY || CD (construction)
According to theorem 6.1 (Basic Proportionality Theorem)
BY/CY = OB/OD................. (2)
From equations (1) and (2)
OA/OC = OB/OD
⇒ OA/OB = OC/OD
Hence proved.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 6
Video Solution:
Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 9
Summary:
Hence it is proved that AO/BO = CO/DO if ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
☛ Related Questions:
- In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)
- E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
- In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
- In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
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E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.(ii) PE = 4 cm, QE = 4.5 cm. PF = 8 cm and RF = 9 cm.(iii) PQ = 1.28 cm. PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
(i) We have,
From (i) and (ii), we have
Therefore, EF is not parallel to QR [By using converse of Basic proportionality theorem]
(ii) We have,
From (i) and (ii), we have
Therefore,
[Using converse of Basic proportionality theorem]
(iii) We have,
From (i) and (ii), we have
Therefore,
[Using converse of Basic proportionality theorem]