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# 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]

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

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.

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]

⇒ 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

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

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.