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Learn Exercise 6.2 with Free Lessons & Tips

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Question 1:

In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution 1:

(i)

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

$\frac{AB}{DB}=\frac{AE}{EC}$

$\frac{1.5}{3}=\frac{1}{x}$

$x=2, EC=2cm$

(ii) Let AD = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

$\frac{AD}{DB}= \frac{AE}{EC}$

$\frac{x}{7.2}=\frac{1.8}{5.4}$

$x=2.4 \therefore AD=2.4cm$

Comments

Tasneem

Question 2:

E and F are points on the sides PQ and PR respectively of a triangle 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

Solution 2:

(i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

$\frac{PE}{EQ}=\frac{3.9}{3}=1.3$

$\frac{PF}{FR}=\frac{3.6}{2.4}=1.5$

$\therefore \frac{PE}{EQ} eq \frac{PF}{FR}$

Therefore, EF is not parallel to QR

(ii)

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

$\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}$

$\frac{PF}{FR}=\frac{8}{9}$

therefore, EF is parallel to QR.

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

$\frac{PE}{EQ}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}$

$\frac{PF}{PR}=\frac{9}{64}$

Hence $\frac{PE}{PQ}=\frac{PF}{PR}$

Therefore, EF is parallel to QR.

Comments

Tasneem

Question 3:

In Fig., DE || OQ and DF || OR. Show that EF || QR.

Solution 3:

In Δ POQ, DE || OQ

$\therefore \frac{PE}{EQ}=\frac{PD}{DO}.............(i)$ (basic proportionality theorem)

$In \Delta POR, DF || OR$

$\therefore \frac{PF}{FR}=\frac{PD}{DO}$ $...............(ii) BPT$

From (i) and (ii) we get,

$\frac{PE}{EQ}=\frac{PF}{FR}$

$\therefore EF\parallel QR$ (converse of BPT)

Comments

Tasneem

Question 4:

In Fig., A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution 4:

In Δ POQ, AB || PQ

$\therefore \frac{OA}{AP}=\frac{OB}{BQ}.............(i)$

$In \Delta POR, AC || PR$

$\therefore \frac{OA}{AP}=\frac{OC}{CR}..........(ii)$

From (i) and (ii) we get

$\frac{OB}{BQ}=\frac{OC}{CR}$

$\therefore BC\parallel QR$

Comments

Tasneem

Question 5:

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).

Solution 5:

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 $\frac{AP}{PB}=\frac{1}{1}$ and $\frac{AQ}{QC}=\frac{1}{1}$

$\therefore \frac{AP}{PB}=\frac{AQ}{QC}$

Hence, by using basic proportionality theorem, we obtain

Comments

Tasneem

Question 6:

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\frac{AO}{BO}=\frac{CO}{DO}$

Solution 6:

Draw a line EF through point O, such that $EF\parallel CD$

In ΔADC, $EO\parallel CD$

By using basic proportionality theorem, we obtain

$\frac{AE}{ED}=\frac{AO}{OC}..................(i)$

In $\Delta ABD, OE\parallel AB$

So, by using basic proportionality theorem, we obtain

$\frac{ED}{AE}=\frac{OD}{BO}$

$\Rightarrow \frac{AE}{ED}=\frac{BO}{OD}$

$\Rightarrow \frac{AO}{BO}=\frac{OC}{OD}$

Comments

Tasneem

Question 7:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{BO}=\frac{CO}{DO}$ ⋅ Show that ABCD is a trapezium.

Solution 7:

Consider the following figure

Draw a line OE || AB

In ΔABD, $OE || AB$

By using basic proportionality theorem, we obtain

$\frac{AE}{ED}=\frac{AO}{OC}$

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

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.

Comments

Tasneem

All Exercises - Chapter 6 - Triangles

Exercise 6.2

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Learn Exercise 6.2 with Free Lessons & Tips

All Exercises - Chapter 6 - Triangles

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