Ex 6.2 Maths Triangles Class 10

Ex 6.2

 Maths 

Triangles 

Class 10 

Q. 1. In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

As we all know the Basic Proportionality Theorem (B.P.T) or (Thales Theorem)

Two triangles are similar if :

(i) Their corresponding angles are equal

(ii) Their corresponding sides are in the same ratio (or proportion)

Proof (1):

(i) InΔABC

BC||DE

InΔABC & ΔADE

∠ABC = ∠ADE        

(∵ Correspondinga angles)

∠ACB=∠AED

(∵Corresponding angles)

∠A = ∠A   (Common)

⇒ΔABC ∼ ΔADE

AD/DB=AE/EC

1.5/3 = 1/EC

EC = (3×1)/1.5

EC = 2cm

Proof (2):

(ii) InΔABC

BC||DE

InΔABC & ΔADE

∠ABC = ∠ADE        

(∵ Correspondinga angles)

∠ACB=∠AED

(∵Corresponding angles)

∠A = ∠A   (Common)

⇒ΔABC ∼ ΔADE

AD/DB=AE/EC

AD/7.2 = 1.8/5.4

AD = (1.8 × 7.2)/5.4

AD = 2.4 cm

Q.  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
Solution:

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

(i) Proof (1):

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (by converse of $BPT$)

Steps:

Here,

$$\begin{array}{c}\frac{P\mathrm{E/}}{E\mathrm{Q\; =}}\end{array}$$PF/FR

$$\begin{array}{c}\frac{\mathrm{3.9/}}{\mathrm{3\; =\; 3.6/2.4}}\end{array}$$

$$1.3\text{cm}$$≠ 1.5cm

Hence,

$\begin{array}{c}\frac{\mathrm{PE/EQ\; \ne \; PF/FR}}{}\end{array}$

According to converse of  $BPT,EF$ is not parallel to $QR$.

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

(ii) Proof:

Here

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (by converse of $BPT$)

Here,

$$\begin{array}{c}\frac{P\mathrm{E/}}{E\mathrm{Q\; =}}\end{array}$$PF/FR

$$\begin{array}{c}\frac{\mathrm{4/4.5}}{\mathrm{=\; 8/9}}\end{array}$$

$$\mathrm{8/9 =\; 8/9}$$

$$1.3\text{cm}$$= 1.5cm

Hence,

$\begin{array}{c}\frac{\mathrm{PE/EQ\; = PF/FR}}{}\end{array}$

According to converse of  $BPT,EF$ is parallel to $QR$.

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

(ii) Proof:

Here

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (by converse of $BPT$)

Here,

PQ = 1.28  and PE = 0.18

EQ = PQ – PE = 1.28 – 0.18 = 1.10

PR = 2.56 and PF = 0.36

FR = PR – PF = 2.56 – 0.36 = 2.20

Now,

$$\begin{array}{c}\frac{P\mathrm{E/}}{E\mathrm{Q\; =}}\end{array}$$PF/FR

$$\begin{array}{c}\frac{\mathrm{0.18/1.10}}{\mathrm{=\; 0.36/2.20}}\end{array}$$

$$\mathrm{9/55 =\; 9/55}$$

Hence,

$\begin{array}{c}\frac{\mathrm{PE/EQ\; = PF/FR}}{}\end{array}$

According to converse of  $BPT,EF$ is parallel to $QR$.

3. In the given figure, if LM || CB and LN || CD. Prove that AM/AB=AN/AD∙

Solution:

As we know 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.

Steps:

In ΔABC, LM || CB

$$\begin{array}{c}\\ & \frac{A\mathrm{M/}}{MB}=\frac{A\mathrm{L/}}{LC}\dots (\end{array}$$1) (by BPT)

In $\Delta ACD$L, LN || CD

AN/DN=AL/LC           …(2) (by BPT)

From equations $\left(1\right)$ and $\left(2\right)$

$$\begin{array}{c}\frac{A\mathrm{M/}}{MB}=\frac{A\mathrm{N/DN}}{}\end{array}$$

$$\begin{array}{c}\Rightarrow \frac{M\mathrm{B/AM}}{}=\frac{D\mathrm{N/}}{AN}\end{array}$$

Adding $1$ on both sides

$$\begin{array}{cc}\frac{M\mathrm{B/}}{AM}+1& =\frac{DN}{\mathrm{A/}N}+1\end{array}$$

$$\begin{array}{cc}\frac{(MB+A\mathrm{M)/}}{AM}& =(\frac{DN+A\mathrm{N)/}}{AN}\end{array}$$

$$\begin{array}{c}\frac{}{}\\ \frac{A\mathrm{B/}}{AM}& =\frac{A\mathrm{D/}}{AN}\end{array}$$

$$\begin{array}{c}\frac{}{}\\ \frac{A\mathrm{M/}}{AB}& =\frac{A\mathrm{N/}}{AD}\end{array}$$

Q. 4. In the given figure, DE || AC and DF || AE.
Prove that BF/FE=BE/EC∙

Solution:

As we know 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.

Steps:

In $\Delta A\mathrm{BC}$

$$\begin{array}{cc}& DE||AC\end{array}$$

$$\begin{array}{c}\frac{B\mathrm{D/}}{AD}=\frac{B\mathrm{E/}}{E\mathrm{C.}}\dots \text{(i) (by BPT)}\end{array}$$

In$\Delta ABE$

$$\begin{array}{cc}& DF||AE\end{array}$$

$$\begin{array}{c}\\ & \frac{B\mathrm{D/}}{AD}=\frac{B\mathrm{F/}}{F\mathrm{E.}}\dots \text{(ii) (by BPT)}\end{array}$$

From $(i$ and $(i$

$$\begin{array}{cc}& \frac{B\mathrm{D/}}{AD}=\frac{B\mathrm{E/}}{EC}=\frac{B\mathrm{F/}}{FE}\end{array}$$

$$\begin{array}{c}\frac{}{}\\ & \frac{B\mathrm{E/}}{EC}=\frac{B\mathrm{F/}}{FE}\end{array}$$

Q. 5. In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Solution:

As we know if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Steps:

In $\Delta POQ$

$$\begin{array}{cc}DE& ||OQ\text{(Given)}\end{array}$$

$$\begin{array}{c}\\ \frac{P\mathrm{E/}}{EQ}& =\frac{P\mathrm{D/}}{OD}\end{array}$$…(1) 

In $\Delta POR$

$$\begin{array}{cc}& DF||OR\left(\text{Given}\right)\end{array}$$

$$\begin{array}{c}\\ & \frac{P\mathrm{F/}}{FR}=\frac{P\mathrm{D/}}{D\mathrm{O.}}\dots \left(2\right)\end{array}$$

From $\left(1\right)$ & $\left(2\right)$

$$\begin{array}{rl}\frac{P\mathrm{E/}}{EQ}& =\frac{P\mathrm{F/}}{FR}=\frac{P\mathrm{D/}}{DO}\end{array}$$

In $\Delta PQR$

$$\begin{array}{c}\frac{P\mathrm{E/}}{EQ}=\frac{P\mathrm{F/}}{FR}\end{array}$$

$\therefore QR||EF$  (Converse of $BPT$)

Q 6. In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:

As we know if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

In $\Delta OPQ$

$$\begin{array}{cc}AB& ||PQ\text{(Given)}\end{array}$$

$$\begin{array}{c}\\ \frac{O\mathrm{A/}}{AP}& =\frac{O\mathrm{B/}}{B\mathrm{Q.}}\dots \left(i\right)\end{array}$$

  Thales Theorem (BPT)]

In $\Delta OPR$

$$\begin{array}{cc}AC& ||PQ\text{(Given)}\end{array}$$

$$\begin{array}{c}\\ \frac{O\mathrm{A/}}{AP}& =\frac{O\mathrm{C/}}{CR}\dots \left(ii\right)\end{array}$$

[ Thales Theorem ($BPT$)]

From $(i$ & $\left(ii\right)$

$$\begin{array}{c}\frac{O\mathrm{A/}}{AP}=\frac{O\mathrm{B/}}{BR}=\frac{O\mathrm{C/}}{CR}\end{array}$$

$$\begin{array}{c}\frac{O\mathrm{B/}}{BQ}=\frac{O\mathrm{C/}}{CR}\end{array}$$

Now In $\Delta OQR$

$$\begin{array}{c}\frac{O\mathrm{B/}}{BQ}=\frac{O\mathrm{C/}}{CR}\end{array}$$

$BC||QR[\because \text{Converse of BPT}]$

Q.7 Using B.P.T., 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 your have proved it in class IX)
Solution:

We know that theorem $6.1$ states that “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 ($BPT$)”.

Given:

In ΔABC ,D is the midpoint of AB and DE||BC

To Prove:

AE = EC

Proof:

In $\Delta ABC$$,D$ is the midpoint of $A\mathrm{B\; and\; DE||BC}$

Therefore,

$$AD=BD$$

$$\frac{A\mathrm{D/}}{BD}=1$$

Now,

$$\begin{array}{cc}DE& ||BC\end{array}$$

$$\begin{array}{c}\\ \Rightarrow \frac{A\mathrm{E/EC}}{}& =\frac{A\mathrm{D/}}{BD}\text{[Theorem}6.1]\end{array}$$

$$\begin{array}{c}\\ \Rightarrow \frac{A\mathrm{E/EC}}{\mathrm{=1}}\end{array}$$

$$\Rightarrow \frac{A\mathrm{E\; =\; EC}}{}$$

Hence $,$

Point E is the midpoint of $\mathrm{AC.}$

Q 8. Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.  (Recall that your have done it in class IX)
Solution:

We know that theorem $6.2$ tells us if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. (Converse of $BPT$)

In $\Delta ABC$

$D$ is the midpoint of $AB$

$$\begin{array}{cc}AD& =BD\end{array}$$

$$\begin{array}{c}\\ \frac{A\mathrm{D/}}{BD}& =1\dots \left(i\right)\end{array}$$

$E$ is the midpoint of $AC$

$$\begin{array}{cc}AE& =CE\end{array}$$

$$\begin{array}{c}\\ \frac{A\mathrm{E/}}{BE}& =1\dots \left(ii\right)\end{array}$$

From (i) and (ii)

$$\begin{array}{cc}\frac{A\mathrm{D/}}{BD}& =\frac{A\mathrm{E/}}{BE}=1\end{array}$$

$$\begin{array}{c}AD/BD=\frac{A\mathrm{E/}}{BE}\end{array}$$

According to theorem $6.2,$ (Converse of $BPT$)

$$DE||BC$$

Q 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:

As we know if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given: ABCD trapezium in which AB||CD. AC and BD intersect at ‘O′.

To Prove: OA/OB=OC/OD

Construct : EO parallel to $AB$ and $C$ $(EO\parallel AB(EO\parallel CD)\; t$hrough $‘O^{\prime .}$

In $\begin{array}{c}\Delta ABC\end{array}$

$$\mathrm{OE}||AB(\because \text{Construction})$$

According to theorem $6.1$ $\left(BPT\right)$

$$\begin{array}{c}\frac{\mathrm{BE}/}{\mathrm{CE}}=\frac{O\mathrm{A/}}{O\mathrm{C.}}\dots \left(i\right)\end{array}$$

In $\Delta BCD$

$\mathrm{OE}||CD(\because Construction$)

According to theorem $6.1$ ($BPT$)

$$\begin{array}{c}\frac{\mathrm{BE}/}{\mathrm{CE}}=\frac{O\mathrm{B/OD}}{}\dots \left(\text{ii}\right)\end{array}$$

From $(i$ and $\left(ii\right)$

$$\begin{array}{cc}\frac{O\mathrm{A/}}{OC}& =\frac{O\mathrm{B/}}{OD}\end{array}$$

$$\begin{array}{c}\frac{}{}\\ \Rightarrow \frac{O\mathrm{A/OB}}{}& =\frac{O\mathrm{C/}}{OD}\end{array}$$

Q. 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.
Solution:

As we know if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.

Given: A  quadrilateral ABCD

with Diagonals AC, BD intersect at ‘O′.

Draw $\mathrm{EF}||AB$

In $\Delta ABC$

$OE||AB$

$$\Rightarrow \frac{O\mathrm{A/OC}}{}=\frac{B\mathrm{E/}}{CE}\left(BPT\right)\cdots \left(1\right)$$

But $\frac{O\mathrm{A/}}{OB}=\frac{O\mathrm{C/}}{OD}\left(Given\right)$

$$\begin{array}{c}\Rightarrow \frac{O\mathrm{A/OC}}{}=\frac{O\mathrm{B/}}{OD}\dots \left(2\right)\end{array}$$

From $(1$ and $\left(2\right)$

$$\begin{array}{c}\frac{O\mathrm{B/}}{OD}=\frac{B\mathrm{E/}}{CE}\end{array}$$

In $\Delta BCD$

$$\begin{array}{cc}\frac{O\mathrm{B/}}{OD}& =\frac{B\mathrm{E/}}{CE}\end{array}$$

$$\begin{array}{c}\frac{}{}\\ OE& ||CD\end{array}$$

(buy Converse of BPT)

$$\begin{array}{c}\\ OE& ||AB\end{array}$$and OE||CD

$$\begin{array}{c}\\ \Rightarrow AB& ||CD\end{array}$$

$$\begin{array}{c}\\ & \Rightarrow ABC\mathrm{D\; is}\text{a trapezium}\end{array}$$