Chapter 06 Triangles Exercise-02

EXERCISE 6.2

1. In Fig. 6.17, (i) and (ii), DEBC. Find EC in (i) and AD in (ii).

Fig. 6.17

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Solution

(i)

Let EC=xcm

It is given that DEBC.

By using basic proportionality theorem, we obtain

ADDB=AEEC1.53=1xx=3×11.5x=2EC=2cm

(ii)

Let AD=xcm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

ADDB=AEEC

x7.2=1.85.4

x=1.8×7.25.4

x=2.4

AD=2.4cm

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=4cm,QE=4.5cm,PF=8cm and RF=9cm

(iii) PQ=1.28cm,PR=2.56cm,PE=0.18cm and PF=0.36cm

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Solution

(i)

Given that, PE=3.9cm,EQ=3cm,PF=3.6cm,FR=2.4cm PEEQ=3.93=1.3 PFFR=3.62.4=1.5

Hence, PEEQPFFR

Therefore, EF is not parallel to QR.

(ii)

PE=4cm,QE=4.5cm,PF=8cm,RF=9cm

PEEQ=44.5=89

PFFR=89

Hence, PEEQ=PFFR

Therefore, EF is parallel to QR.

(iii)

PQ=1.28cm,PR=2.56cm,PE=0.18cm,PF=0.36cm

PEPQ=0.181.28=18128=964

PFPR=0.362.56=964

Hence, PEPQ=PFPR

Therefore, EF is parallel to QR.

3. In Fig. 6.18, if LMCB and LNCD, prove that AMAB=ANAD.

Fig. 6.18

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Solution

In the given figure, LMCB

By using basic proportionality theorem, we obtain AMAB=ALAC

Similarly, LNCD

ANAD=ALAC

From (i) and (ii), we obtain

AMAB=ANAD

4. In Fig. 6.19, DEAC and DFAE. Prove that BFFE=BEEC.

Fig. 6.19

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Solution

In ABC,DEAC

BDDA=BEEC (Basic Proportionality Theorem)

In BAE,DFAE

BDDA=BFFE (Basic Proportionality Theorem)

(ii)

From (i) and (ii), we obtain

BEEC=BFFE

5. In Fig. 6.20, DE OQ and DFOR. Show that EF QR.

Fig. 6.20

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Solution

In POQ,DEOQ

PEEQ=PDDO (Basic proportionality theorem)

(i)

In POR,DFOR

PFFR=PDDO (Basic proportionality theorem)

From (i) and (ii), we obtain

PEEQ=PFFR

EFQR (Converse of basic proportionality theorem)

6. In Fig. 6.21, A, B and C are points on OP,OQ and OR respectively such that ABPQ and ACPR. Show that BCQR.

Fig. 6.21

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Solution

In POQ,ABPQ

OAAP=OBBQ (Basic proportionality theorem)

(i)

In POR,ACPR

OAAP=OCCR (By basic proportionality theorem) (ii)

From (i) and (ii), we obtain

OBBQ=OCCR

BCQR (By the converse of basic proportionality theorem)

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. (Recall that you have proved it in Class IX).

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Solution

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQBC

By using basic proportionality theorem, we obtain

AQQC=APPB

AQQC=11 (P is the mid-point of AB.AP=PB )

AQ=QC

Or, Q 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. (Recall that you have done it in Class IX).

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Solution

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.

i.e., AP=PB and AQ=QC

It can be observed that APPB=11

and AQQC=11

APPB=AQQC

Hence, by using basic proportionality theorem, we obtain

PQBC

9. ABCD is a trapezium in which ABDC and its diagonals intersect each other at the point O. Show that AOBO=CODO.

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Solution

Draw a line EF through point O, such that

EFCD

In ADC,EOCD

By using basic proportionality theorem, we obtain

AEED=AOOC

In ABD

OEAB

So, by using basic proportionality theorem, we obtain

EDAE=ODBO

AEED=BOOD

From equations (1) and (2), we obtain AOOC=BOOD

AOBO=OCOD

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO Show that ABCD is a trapezium.

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Solution

Let us consider the following figure for the given question.

Draw a line OEAB

In ABD,OEAB

By using basic proportionality theorem, we obtain

AEED=BOOD

However, it is given that AOOC=OBOD

From equations (1) and (2), we obtain

AEED=AOOC

EO || DC [By the converse of basic proportionality theorem]

AB||OE||DC

ABCD

ABCD is a trapezium.



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