CBSE 10th Maths - Triangles - Exercise 6.2

Solution:

(i) Given, in △ ABC, DE∥BC

∴ $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ [Using Basic proportionality theorem]

$\dfrac{1.5}{3} = \dfrac{1}{EC}$

EC = $\dfrac{3}{1.5}$

EC = $\dfrac{3 × 10}{15}$ = 2 cm

EC = 2cm

(ii) Given, in △ ABC, DE∥BC

∴ $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ [Using Basic proportionality theorem]

$\dfrac{AD}{7.2} = \dfrac{1.8}{5.4}$

AD = $\dfrac{1.8 × 7.2}{5.4} = (\dfrac{18}{10})×(\dfrac{72}{10})×(\dfrac{10}{54}) = \dfrac{24}{10}$

AD = 2.4

Hence, AD = 2.4 cm.

Solution:

Given, in ΔPQR, E and F are two points on side PQ and PR, respectively. See the figure below

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

by using Basic proportionality theorem

find $\dfrac{PE}{EQ}$ and $\dfrac{PF}{FR}$

$\dfrac{PE}{EQ} = \dfrac{3.9}{3} = \dfrac{39}{30} = \dfrac{13}{10} = 1.3$

$\dfrac{PF}{FR} = \dfrac{3.6}{2.4} = \dfrac{36}{24} = \dfrac{3}{2} = 1.5$

$\dfrac{PE}{EQ} \neq \dfrac{PF}{FR}$ , EF not Parallel QR

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

by using Basic proportionality theorem

find $ \dfrac{PE}{EQ}$ and $\dfrac{PF}{FR}$

$\dfrac{PE}{QE} = \dfrac{4}{4.5} = \dfrac{40}{45} = \dfrac{8}{9}$

$\dfrac{ PF}{RF} = \dfrac{8}{9}$

$\dfrac{PE}{QE} = \dfrac{PF}{RF}$ , EF is Parallel to QR.

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

by using Basic proportionality theorem

find $ \dfrac{PE}{EQ}$ and $\dfrac{PF}{FR}$

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

FR = PR – PF = 2.56 – 0.36 = 2.20 cm

So, $\dfrac{PE}{EQ} = \dfrac{0.18}{1.10} = \dfrac{18}{110} = \dfrac{9}{55}$…………. (i)

$\dfrac{PE}{FR} = \dfrac{0.36}{2.20} = \dfrac{36}{220} = \dfrac{9}{55}$………… (ii)

$\dfrac{PE}{EQ} = \dfrac{PF}{FR},$ EF is parallel to QR.

Solution:

In the given figure, we can see, LM || CB,

By using basic proportionality theorem, we get,

$\dfrac{AM}{AB} = \dfrac{AL}{AC}$……………………..(i)

Similarly, given, LN || CD and using basic proportionality theorem,

∴$\dfrac{AN}{AD} = \dfrac{AL}{AC}$……………………………(ii)

From equation (i) and (ii), we get,

$\dfrac{AM}{AB} = \dfrac{AN}{AD}$

Hence, proved.

Solution:

In ΔABC , DE || AC

$\dfrac{BD}{AD} = \dfrac{BE}{EC}$ .........(i) [By using Basic Proportionality Theorem]

In ΔABE, DF || AE

$\dfrac{BD}{AD} = \dfrac{BF}{FE}$ ........(ii) [By using Basic Proportionality Theorem]

From (i) and (ii)

$\dfrac{BD}{AD} = \dfrac{BE}{EC} = \dfrac{BF}{FE}$

Thus, $\dfrac{BE}{EC} = \dfrac{BF}{FE}$

Solution:

Given, In ΔPQO, DE || OQ

By using Basic Proportionality Theorem,

$\dfrac{PD}{DO} = \dfrac{PE}{EQ}$……………… ..(i)

Again given, in ΔPOR, DF || OR,

Using Basic Proportionality Theorem,

$\dfrac{PD}{DO} = \dfrac{PF}{FR}$………………… (ii)

From equation (i) and (ii), we get,

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

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in ΔPQR.

Solution:

Given,

In ΔOPQ, AB || PQ

By using Basic Proportionality Theorem,

$\dfrac{OA}{AP} = \dfrac{OB}{BQ}$…………….(i)

Also given,

In ΔOPR, AC || PR

By using Basic Proportionality Theorem

∴ $\dfrac{OA}{AP} = \dfrac{OC}{CR}$……………(ii)

From equation (i) and (ii), we get,

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

Therefore, by converse of Basic Proportionality Theorem,

In ΔOQR, BC || QR.

Solution:

Given, in ΔABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

We have to prove that E is the mid point of AC.

Since, D is the mid-point of AB.

∴ AD=DB

$\dfrac{AD}{DB}$ = 1 …………………………. (i)

In ΔABC, DE || BC,

By using Basic Proportionality Theorem,

Therefore,$\dfrac{AD}{DB} = \dfrac{AE}{EC}$

From equation (i), we can write,

1 = $\dfrac{AE}{EC}$

∴ AE = EC

Hence, proved, E is the midpoint of AC.

Solution:

Given, in ΔABC, D and E are the mid points of AB and AC respectively, such that, AD=BD and AE=EC.

We have to prove that: DE || BC.

Since, D is the midpoint of AB

∴ AD = DB

$\dfrac{AD}{BD}$ = 1……………………………….. (i)

Also given, E is the mid-point of AC.

∴ AE = EC

$\dfrac{AE}{EC}$ = 1……………………………….. (ii)

From equation (i) and (ii), we get,

$\dfrac{AD}{BD} = \dfrac{AE}{EC}$

By converse of Basic Proportionality Theorem,

DE || BC

Hence, proved.

Solution:

Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, AO/BO = CO/DO

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In ΔADC, we have OE || DC

Therefore, By using Basic Proportionality Theorem

$\dfrac{AE}{ED} = \dfrac{AO}{CO}$ ……………..(i)

Now, In ΔABD, OE || AB

Therefore, By using Basic Proportionality Theorem

$\dfrac{DE}{EA} = \dfrac{DO}{BO}$…………….(ii)

From equation (i) and (ii), we get,

$\dfrac{AO}{CO} = \dfrac{BO}{DO}$

$\dfrac{AO}{BO} = \dfrac{CO}{DO}$

Hence proved.

Solution:

Given,

Quadrilateral ABCD where AC and BD intersects each other at O such that, $\dfrac{AO}{BO} = \dfrac{CO}{DO}

We have to prove here, ABCD is a trapezium

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In ΔDAB, EO || AB

Therefore, By using Basic Proportionality Theorem

$\dfrac{DE}{EA} = \dfrac{DO}{OB}$……………………(i)

Also, given,

$\dfrac{AO}{BO} = \dfrac{CO}{DO}$

$\dfrac{AO}{CO} = \dfrac{BO}{DO}$

$\dfrac{CO}{AO} = \dfrac{DO}{BO}$

$\dfrac{DO}{OB} = \dfrac{CO}{AO}$ …………………………..(ii)

From equation (i) and (ii), we get

$\dfrac{DE}{EA} = \dfrac{CO}{AO}$

Therefore, By using converse of Basic Proportionality Theorem,

EO || DC also EO || AB

AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.