# Chapter 6: Triangles

## Class 10 Math Triangle NCERT Solution

**TEXTBOOK’S EXERCISE 6.1 **

**1. **Fill in the blanks using the correct word given in brackets :

(i) All circles are _______ . (congruent, similar)

(ii) All squares are _______. (similar, congruent)

(iii) All _______ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if

(a) their corresponding angles are _______ and (b) their corresponding sides are _______. (equal, proportional)

**Sol. **(i) All circles are **similar**.

(ii) All squares are **similar.
**

(iii) All **equilateral
**triangles are similar.

(iv) Two polygons of the same number of sides are similar, if

(a) their
corresponding angles are **equal **and

(b) their
corresponding sides are **proportional. **

**2. **Give two different examples of pairs of W

(i) similar figures. (ii) non-similar figures.

Sol. (i) Two different examples of similar figures are :

(a) Two twenty-rupees notes.

(b) Two two-rupees coins.

(ii) Two different examples of non-similar figures are :

(a) One-rupee coin and a five rupees coin.

(b) One-rupee note and ten rupees note.

**3. **State whether the following quadri-laterals are similar or not :

**Sol. **Two quadrilaterals in the given
figure are not similar, because their corresponding angles are not equal.

**TEXTBOOK’S EXERCISE 6.2 **

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

**Sol.
**(i) In ΔABC, DE || BC [Given]

Therefore, AD DB = AE EC

[By Basic Proportionality Theorem]

⇒ 1.5/3 = 1/EC ⇒ EC = 3/1.5

⇒ EC = 3 X 10/ 15 cm × = 2

Hence, EC = 2 cm

(ii) In ΔABC, DE || BC [Given]

Therefore, AD DB = AE EC

[By Basic Proportionality Theorem]

⇒ AD/7.2 = 1.8/ 5.4

⇒ AD = 2.4

Hence, AD = 2.4 cm.

**2. **E and F are points on the sides PQ and
PR respectively of a ΔPQR. 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. **Sol. **ΔPQR, E and F are two points on sides PQ and PR respectively

(i) PE = 3.9 cm, EQ = 3 cm,

PF = 3.6 cm, FR = 2.4 cm [Given]

Therefore, PE/EQ= 3. 9/3=39/30 =13/10= 1.3

And, PF/FR= 3. 6 /2. 4 =36/24= 3/2= 1 5

So, PE /EQ ≠PF/FR

Hence, EF is not parallel to QR.

Hence, EF is not parallel to QR. [By converse of BPT]

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

RF = 9 cm [Given]

Hence, EF is parallel to QR. [By converse of BPT]

(iii) PQ = 1.28 cm, PR = 2.56 cm,

PE = 0.18 cm, PF = 0.36 cm [Given]

Here, EQ = PQ – PE = (1.28 – 0.18) cm

= 1.10 cm

And, FR = PR – PF = (2.56 – 0.36) cm

= 2.20 cm

Hence, EF is parallel to QR. [By converse of BPT]

**3. **In figure, LM || CB and LN || CD, prove that AM/AB = AN/AD

**Sol.
**In ΔABC, we have ML || BC [Given]

⇒ AM/MB =AL/ LC …(i)

[By Basic Proportionality Theorem]

In ΔACD, we have LN || DC [Given]

⇒ AN/ND= AL/LC … (ii)

[By Basic Proportionality Theorem]

From (i) and (ii), we get

AM/MB=AN/ND or MB/AM=ND/AN

[Adding 1 in both sides]

**4. **In figure, DE || AC, and DF || AE, prove that BF/ FE = BE/EC.

**Sol.
**In ΔABC, DE || AC [Given]

Therefore, BD/DA = BE/ EC …(i) [By Basic Proportionality Theorem]

In ΔABE, DF || AE [Given]

Therefore, BD/DA = BF/FE …(ii) [By Basic Proportionality Theorem]

From (i) and (ii), we have BE/EC = BF/FE . **Proved.**

**5. In figure DE || OQ and DF || OR. Show that EF || QR.**

**Sol.
Given : **DE || OQ and DF || OR

**To
prove : **FE || QR.

**Proof
: **In ΔPQO, DE || OQ [Given]

Therefore, PD/DO = PE/EQ = …(i) [By Basic Proportionality Theorem]

In ΔPOR, DF || OR

Therefore, PD/DO = PF/ FR = …(ii) [By Basic Proportionality Theorem]

From (i) and (ii), we have PE/EQ = PF/FR

In ΔPQR, by converse of Basic Proportionality Theorem.

EF is parallel to
QR. **Proved. **

**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.

**Sol. Given: **ΔPQR, A, B and C are points on OP, OQ and OR respectively such that

AB || PQ and AC || PR.

**To prove **BC || QR.

**Proof
: **In ΔOPQ, we have

AB || PQ [Given]

Therefore, OA /AP= OB/ BQ …(i) [By Basic Proportionality Theorem]

In ΔOPR, we have

AC || PR [Given]

Therefore, OA /AP= OC/ CR = …(ii) [By Basic Proportionality Theorem]

From (i) and (ii), we have

OB /BQ = OC /CR

Hence, in ΔOQR, BC || QR.

[By converse of Basic Proportionality Theorem

**Proved. **

**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.

**Sol. Given: **In ΔABC, D is the mid-point of AB, i.e., AD = DB.

A line DE parallel to BC intersects AC at E as shown in the figure, i.e., DE || BC.

**To prove: **E is the mid-point of AC.

**Proof
: **D is the mid-point of AB.

Therefore, AD = DB

⇒ AD BD =1 …(i)

In ΔABC, DE || BC

Therefore, AD /BD = AE/ EC

[By Basic Proportionality Theorem]

⇒ 1= AE /EC [From (i)]

Therefore, AE = EC

Hence, E is the
mid-point of AC. **Proved. **

**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.

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

**To
prove : **DE || BC

**Proof : **D is the mid-point of AB [Given]

Therefore, AD = BD

⇒ AD /BD = 1 …(i)

Also, E is the mid-point of AC [Given]

Therefore, AE = EC

AE /EC = 1 …(ii)

From (i) and (ii), we have

AD /BD =AE/ EC

Hence, DE || BC.

[By converse of Basic Proportionality Theorem]

**Proved.
**

**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.

**Sol.
**

**Given: **ABCD is a trapezium in which AB || DC, diagonals AC and BD intersect each other O.

**To
prove : **AO/ BO= CO /DO

**Construction
: **Through O, draw EO || DC || AB

**Proof: **In ΔADC, we have

EO || DC [By construction]

Therefore, AO /CO= AE/ DE …(i) [By Basic Proportionality Theorem]

In ΔDBA, we have

EO || AB [By construction]

AE /DE= BO/ DO = …(ii) [By Basic Proportionality Theorem]

From (i) and (ii), we get AO /CO= BO/ DO

⇒ AO/ BO= CO /DO . **Proved.**

**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.

**Sol. Given: **Quadrilateral ABCD, diagonals AC and BD intersect each other at O such that AO/ BO = CO /DO.

**To
prove : **ABCD is a trapezium.

**Construction: **Through O, draw the line EO || AB which meets AD at E.

**Proof : **In ΔDAB, we have EO || AB Therefore, DE/ EA =DO /OB = …(i) [By Basic Proportionality Theorem]

Also, AO /BO =CO /DO [Given]

or, AO/ CO= BO /DO or, CO/ AO= DO/ BO

⇒ DO/ OB= CO /AO = …(ii)

From (i) and (ii), we get DE /EA =CO/ AO

Therefore, By converse of Basic Proportionality Theorem,

EO || DC also EO || AB ⇒ AB || DC. Hence, quadrilateral ABCD is a trapezium with AB || CD. **Proved.**