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).

10th Maths Exercise 6.2
MP Board Solutions

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

Solution: In ∆ABC, DE || BC,

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

\[\frac{1.5}{3}=\frac{1}{EC}\]

\[EC=\frac{3}{1.5}=2\ cm\]

10th Maths Ex 6.2 Question i
MP Board 10th Maths Triangles Ex 6.2

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

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

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

AD=2.4 cm

NCERT Class 10th Maths ex 6.2 ii
MP Board Solutions

(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

NCERT 10th Maths 6.2 Question ii
MP Board Solutions

\[\frac{PE}{EQ}=\frac{4}{4.5}=0.89\]

\[\frac{PF}{FR}=\frac{8}{9}=0.89\]

\[\frac{PE}{EQ}=\frac{PF}{FR}\]

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

\[\frac{AM}{AB}=\frac{AN}{AD}\]

MP Board 10th Maths Ex 6.2 CB and LN CD Proved that
MP Board 10th Maths Solutions

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

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

Proof: ∆ABC, LM || CB

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

IN ∆ADC, LN || CD

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

From (i) and (ii) we get,

\[\frac{AM}{BM}=\ \frac{AN}{DN}\]

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

Adding 1 on both side

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

\[⇒\ \frac{BM+AM}{AM}=\ \frac{DN+AN}{AN}\]

\[⇒\ \frac{AB}{AM}=\ \frac{AD}{AN}\]

NCERT 10th Maths Chapter 6 Ex 6.2 Question 3
MP Board Solutions

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

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

\[\frac{BF}{FE}=\ \frac{BF}{EC}\]

10th Math Exercise 6.2 AC and DF, AE Prove that
MP Board 10th Maths Solutions

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.

10th Maths Exercise 6.2 Question 5
MP Board Solutions

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

MP Board Class 10th Math Chapter 6
MP Board Solutions

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.

NCERT 10th Maths Ex 6.2 Question 6
NCERT 10th Math Solution

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

MP Board 10th Maths Triangles Ex 6.2

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.

MP Board 10th Math Triangles Ex 6.2 AD=DB Proved
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∵ 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
∴ AD=DB

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

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.

MP Board 10th Math Triangles Ex 6.2 AD=DB Proved
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∴ 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.

10th Maths exercise 6.2 ABCED is a trapezium in which AB, DC
MP Board Solutions

Let us draw OE parallel to either AB or DC.
In ∆ADC
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

MP Board 10th Maths Exercise 6.2 Question 9 Solution
MP Board Solutions

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]

10th Maths Chapter 6 triangles Let us draw OE such that OE
MP Board 10th math Triangles Ex 6.2

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

10th maths chapter 6 Triangle Ex 6.1 introduction
MP Board Solutions

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