# MP Board Class 10th Math Solutions Chapter 6 Triangles Ex 6.2

In this article we will share latest MP Board Class 10th Mathematics Solutions Chapter 6 Triangles Ex 6.2 with pdf.

Question 1. In Fig. below, (i) and (ii) DE || BC. Find EC in (1) and AD in (ii).

(i) Given: DE || BC
To Find: EC

Solution: In ∆ABC, DE || BC,

$\frac{AD}{DB}=\frac{AE}{EC}\ \left(By\ BPT\right)$

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

$EC=\frac{3}{1.5}=2\ cm$

(ii) Given: DE || BC
Solutions: In ∆ABC

$\frac{AD}{BD}=\frac{AE}{CE}\ \left(By\ BPT\right)$

$\frac{AD}{7.2}=\frac{1.8}{5.4}\times7.2$

(i)

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

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

$As\ \frac{PF}{EQ}\ne\frac{PF}{FR}$

EF not parallel to QR

Question 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 = 9cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution:
(i) We have, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) In ∆PQR

$\frac{PE}{EQ}=\frac{4}{4.5}=0.89$

$\frac{PF}{FR}=\frac{8}{9}=0.89$

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

EF || QR (By converse of BPT)

(iii) In ∆PQR
EQ=PQ-PE = 1.28-0.18
EQ=1.10 cm
and FR= PR-PF
= 2.56-0.36
= 2.2 cm

Now,

$\frac{PE}{EQ}=\frac{0.18}{1.10}=0.16$

$and\ \frac{PF}{FR}=\frac{0.36}{2.2}=0.16$

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

EF || QR (By converse of BPT)

Question 3.
In the figure, if LM || CB and LN || CD, prove that

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

Solution:
Given: LM || CB and LN || CD

$To\ \Pr ove:\ \frac{AM}{AB}=\frac{AN}{AD}$

Proof: ∆ABC, LM || CB

$\frac{AM}{BM}=\frac{AL}{LC}\ ...\left(i\right)\ \left(By\ BPT\right)$

$\frac{AN}{DN}=\ \frac{AL}{CL}\ \ ..\left(ii\right)\ \left(BY\ BPT\right)$

From (i) and (ii) we get,

$\frac{AM}{BM}=\ \frac{AN}{DN}$

$⇒\frac{BM}{AM}=\ \frac{DN}{AN}\ \left[Reciprocal\right]$

$⇒\ \frac{BM}{AM}+1=\ \frac{DN}{AN}+1\$

$⇒\ \frac{BM+AM}{AM}=\ \frac{DN+AN}{AN}$

$⇒\ \frac{AB}{AM}=\ \frac{AD}{AN}$

$⇒\ \frac{AM}{AB}=\ \frac{AN}{AD}\ \left(Again\ reciprocal\right)$

Question 4.
In the figure, DE || ACand DF || AE.
Prove that.

$\frac{BF}{FE}=\ \frac{BF}{EC}$

Solution:
In ∆ABC
∵ DE || AC [given]
Using the basic proportionality theorem, we have

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

Solution:
In ∆PQO,
∵ DE || OQ [given]
∴ Using the Basic proportionality theorem, we have

$\frac{PE}{EQ}=\frac{PD}{Do}\ \ ......\left(1\right)$

Again, in ∆POR, DF || OR [given]
∴ Using the Basic proportionality theorem, we have

Now, in ∆PQR,
∵ E and F are two distinct points on PQ and PR respectively and

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

i.e., E and F divide the two sides PQ and PR of ∆PQR in the same ratio.
∴ By converse of Basic proportionality theorem, EF || QR.

Question 6.
In the 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:
In ∆PQR, O is a point and OP, OQ and OR are joined. We have points A, B and C on OP, OQ and OR respectively such that AB || PQ and AC || PR.
Now, in ∆OPQ,
∵ AB || PQ [Given]
Using the Basic proportionality theorem, we have

i.e., B and C divide the sides OQ and OR of ∆OQR in the same ratio.
By converse of Basic proportionality theorem, BC || QR.

Question 7.
Using Basic proportionality theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. A
Solution:
We have ∆ABC, in which D is the midpoint of AB and E is a point on AC such that DE || BC.

∵ DE || BC [given]
∴ Using the Basic proportionality theorem, we get

$\frac{AD}{DB}=\frac{AE}{EC}\ \ \ ......\left(1\right)$

But D is the mid-point of AB

$⇒\ \frac{AD}{DB}=1\ \ ......\left(2\right)$

From (1) and (2),

$1=\ \frac{AE}{EC}\ ⇒EC=AC$

⇒ E is the mid point of AC. Hence, it is proved that a line through the midpoint of one side of a triangle parallel to another side bisects the third side.

Question 8.
Using converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:
We have ∆ABC, in which D and E are the mid-points of sides AB and AC respectively.

∴ AD = DB ……………. (1)
and AE = EC ………….. (2)
From (1) and (2), we have

$\frac{AD}{DB}=1\ and\ \frac{AE}{EC}=1$

$⇒\frac{AD}{DB}=\ \frac{AE}{EC}$

⇒ DE || BC (By converse proportionality theorem).

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AOBO=CODO.
Solution:
We have, a trapezium ABCD such that AB || DC. The diagonals AC and BD intersect each other at O.

Let us draw OE parallel to either AB or DC.
OE || DC [By construction]
∴ Using the Basic proportionality theorem, we get

$\frac{AE}{ED}=\frac{AO}{CO}......\left(1\right)$

In ∆ABD
OE || AB [By construction]
∴ Using the Basic proportionality theorem, we get

Question 10. 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:

$It\ is\ given\ that\ \frac{AO}{BO}=\frac{CO}{DO}$

$From\ \frac{AO}{BO}=\frac{CO}{DO},\ we\ have\ \frac{AO}{CO}=\frac{BO}{DO}$

Let us draw OE such that OE || BA
In ∆ADB, OE || AB [By construction]

i.e., the points O and E divide the sides AC and AD of ∆ADC respectively in the same ratio.
∴ Using the converse of Basic proportionality theorem, we get OE || DC and OE || AB
⇒ AB || DC
⇒ ABCD is a trapezium.

i.e., the points O and E divide the sides AC and AD of ∆ADC respectively in the same ratio.
∴ Using the converse of Basic proportionality theorem, we get OE || DC and OE || AB
⇒ AB || DC
⇒ ABCD is a trapezium.

## Triangles Introduction

You are already aware with the term “congruent”. It means two figures or objects having same shape and size. For e.g. two coins of one rupee. In this chapter we will study about similar triangles, two figures or objects are said to be similar if they have same shape but not necessarily same size. Note: All congruent figures are similar but all similar figures need not be congruent.

Two polygons of the same number of sides are similar, if (1) their corresponding angles are equal (ii) their corresponding sides are in the same ratio (proportion).

If quadrilateral A’B’C’D’ is similar to ABCD then

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