SEBA Class 10 Maths Chapter 6 Tringles

SEBA Class 10 Maths Chapter 6 Tringles Solutions English Medium As Per New Syllabus. to each chapter is provided in the list so that you can easily browse through different chapters SEBA Class 10 Maths Chapter 6 Tringles Notes and select need one. SEBA Class 10 Maths Chapter 6 Tringles Question Answers Download PDF. SEBA Class 10 General Mathematics Solutions.

SEBA Class 10 Maths Chapter 6 Tringles

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Tringles

Chapter: 6

Exercise 6.1

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

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

Ans: Similar.

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

Ans: Similar.

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

Ans: Equilateral.

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

(a) their corresponding angles are __________ and 

Ans: Equal.

(b) Their corresponding sides are __________. (equal,

proportional)

Ans: Proportional.

2. Give two different examples of pair of

(i) Similar figures.

Ans: Two equilateral triangles with sides 1 cm and 2 cm

(ii) Non-similar figures

Ans: 

3. State whether the following quadrilaterals are similar or not: 

Ans: Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.

4. Which of the following are definitely similar to each other? 

(A) Any two Rhombuses.

(B) Any two right angled triangles.

(C) Any two regular Pentagons.

(D) Any two isosceles triangles.

Ans: (C) Any two regular Pentagons.

5. Consider the following statements:

Statement (i): If three sides of a triangle are equal to three sides of another triangle, then the triangles are congruent.

Statement (ii): If three angles of a triangle are respectively equal to three angles of another triangle, then the two triangles are congruent.

Choose the correct option:

(A) (i) is correct (ii) is false.

(B) Both (i) and (ii) are false.

(C) Both (i) and (ii) are correct.

(D) (i) is false, (ii) is correct.

Ans: (A) (i) is correct (ii) is false.

Exercise 6.2

1. In figure. 6.19. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Ans: Figure (i):

In ΔABC, we have

∴ DE || BC                                   

∴ AD/DB = AE/EC                             

⇒ 1.5/3 = 1/EC

⇒ 1.5 EC = 3

⇒ EC = 3/1.5

⇒ EC = 2

∴ EC = 2 cm

Figure (ii):

In ΔABC, we have

∴ DE || BC                                  

∴ AD/DB = AE/EC                            

⇒ AD/7.2 = 1.8/5.4

⇒ 5.4 AD = 7.2 × 1.8

⇒ AD = (1.8 × 7.2)/5.4

⇒ AD = 2.4

∴ AD = 2.4 cm

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.

Ans:

Given: In ΔPQR, E and F are points on the sides PQ and PR respectively.

(i) Here,

PE = 3.9 cm

EQ = 3 cm,

PF = 3.6 cm and FR = 2.4 cm

Now,

PE/EQ = 3.9/3 = 1.3 ………. (i)

and PF/FR = 3.6/2.4 = 3/2 = 1.5 ………. (ii)

From equations (i) and (ii), we have

PE/EQ ≠ PF/FR

i.e. EF does not divide the two sides PQ and PR of ΔPQR in the same ratio.

Therefore, EF is not parallel to QR.    

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

Ans: Here,

PE = 4 cm, QE = 4.5 cm,

PF = 8 cm and RF = 9 cm

Now,

PE/EQ = 4/4.5 = 40/45 = 8/9 ………. (i)

and PF/FR = 8/9 ………. (ii)

From equations (i) and (ii), we have

PE/EQ = PF/FR

i.e. EF divides the two sides PQ and PR of ΔPQR in the same ratio.

Therefore, EF || QR.

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

Ans: Here,

PQ = 1.28 cm, PR = 2.56 cm,

PE = 0.18 cm and PF = 0.36 cm

EQ = PQ − PE

= (1.28 − 0.18) cm

= 1.10 cm

FR = PR − PF

= (2.56 − 0.36) cm

= 2.20 cm

Now,

PE/EQ = 0.18/1.10 = 18/110 = 9/55 ………. (i)

and

PF/FR = 0.36/2.20 = 36/220 = 9/55 ………. (ii)

From equations (i) and (ii), we have

PE/EQ = PF/FR

i.e. EF divides the two sides PQ and PR of ΔPQR in the same ratio.

Therefore, EF || QR.

3. In Fig. 6.20, if LM || CB and LN || CD prove that 

AM/AB = AN/AD 

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

To prove: AM/AB = AN/AD

Proof: ∵ LM || CB

In ΔABC, we have

AM/MB = AL/LC

4. In Fig. 6.21, DE || AC and DF || AE. Prove that

BF/FE = BE/EC

Ans: Given: DE || AC and DF || AE

To prove: BF/FE = BE/EC

Proof: ∵ DE || AC [Given]

∴ In ΔBAC, we have

BD/DA = BE/EC ….(i)

Similarly, ∵ DF || AE

∴ In ΔBAE, we have

BD/DA = BF/FE …(ii)

From equations (i) and (ii), we have

BF/FE = BE/EC

Therefore, BF/FE = BE/EC.

5. In the following figure, DE || OQ and DF || OR, show that EF || QR. 

Ans: Given: DE || OQ and DF || OR

To prove: EF || QR

Proof: ∵ DE || OQ

∴ In ΔPQO, we have

PE/EQ = PD/DO                         

Again,

∵ DF || OR

∴ In ΔPOR, we have

∴ PD/DO = PF/FR                 

From equations (i) and (ii), we have

PE/EQ = PF/FR

i.e. EF divides the two sides PQ and PR of ΔPQR in the same ratio. We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.

Therefore, EF || QR

6. In fig.6.23, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. 

Ans: Given: A, B and C are points on OP, OQ and OR respectively

such that AB || PQ and AC || PR

To prove: BC || QR

Proof:

∵ AB || PQ   [Given]

∴ In ΔOPQ, we have

OA/AP = OB/BQ  [Thales Theorem] …. (i)

Similarly,

∵ AC || PR   [Given]

∴ In ΔOPR, we have

OA/AP = OC/CR   [Thales Theorem] …. (ii)

From equations (i) and (ii), we have

OB/BQ = OC/CR

i.e. BC divides the two sides OQ and OR of ΔOQR in the same ratio.

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

Therefore, BC || QR

7. Using theorem 6.1, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Ans: 

8. Using theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX). 

Ans: 

Let ABC be a triangle in which D and E are the mid-points of AB and AC respectively.

To prove: DE || BC

Proof: Since D and E are the mid-points of AB and AC respectively.

∴ AD = DB …(i)

⇒ AD/DB = 1

and AE = EC …(ii)

⇒ AE/EC = 1

From equations (i) and (ii), we have

AD/DB = AE/EC

i.e. DE divides the two sides AB and AC of ΔABC in the same ratio.

Therefore, DE || BC.

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.

Ans: 

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is trapezium.

Now, we have BA || OE and OE || CD

⇒ AB || CD ⇒ Quadrilateral ABCD is a trapezium.

11. Assertion (A): If in ΔABC, D and E are two points on the sides AB and AC respectively such that DE || BC, then AD/AB = AE/AC

Reason (R): If a line is drawn parallel to one side of a triangle intersecting the two other sides, then it divides the two sides in the same ratio.

(A) Both (A) and (R) are true and (R) is the correct explanation of A

(B) Both (A) and (R) are true and (R) is not the correct explanation of A

(C) A is true but R is false

(D) A is false but R is true.

Ans: (A) Both (A) and (R) are true and (R) is the correct explanation of A

12. Given three statements from the adjacent figure where (Fig.6.24)

(i) PQ/QR = PO/OC

(ii) PO/OB = BC/AB

(iii) PQ/QR = AB/BC

Which one of the following is true?

(A) Only (i) is true

(B) (i) and (ii) are true

(C) Only (iii) is true

(D) (i) and (iii) are true

Ans: (D) (i) and (iii) are true

13. In a triangle ABC D and E are two points on the sides AB and AC respectively such that DE||BC. If AD = x, AE = (x + 2) BD = (x – 2) and CE = (x – 1) , the value of x is

(A) 2

(B) 3

(C) 4

(D) 5

Ans: (C) 4