A and B are two points on the sides PQ and PR respectively such that AB QR is a trapezium

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.

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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