SEBA Class 10 Mathematics Chapter 6 Tringles

SEBA Class 10 Mathematics Chapter 6 Tringles Solutions, SEBA Class 10 Maths Textbook Notes in English Medium, SEBA Class 10 Mathematics Chapter 6 Tringles Notes in English to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 10 Mathematics Chapter 6 Tringles Notes and select needs one.

SEBA Class 10 Mathematics Chapter 6 Tringles

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 10 Mathematics Chapter 6 Tringles Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 10 Mathematics Chapter 6 Tringles Solutions for All Subject, You can practice these here.

Tringles

Chapter – 6

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: 

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

1. In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (i).

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

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

3.

Ans: 

In the given figure, LM || CB By using basic proportionality theorem, we obtain

4.

Ans: 

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

Ans: 

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

Ans:

7. Using Basic proportionality theorem, 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 Converse of basic proportionality theorem, 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: 

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively. i.e., AP = PB and AQ = QC It can be observed that

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.

Ans: Let us consider the following figure for the given question. 

1. State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Ans: (i) ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, ∆ABC ∼ ∆PQR [By AAA similarity criterion] 

(ii)

Ans: Do yourself.

Ans: The given triangles are not similar as the corresponding sides are not proportional.

Ans: The given triangles are not similar as the corresponding sides are not proportional.

(v) 

Ans: The given triangles are not similar as the corresponding sides are not proportional.

(vi) 

In ∆DEF,

∠D +∠E +∠F = 180º (Sum of the measures of the angles of a 

triangle is 180º.)

70º + 80º +∠F = 180º

∠F = 30º

Similarly, in ∆PQR,

∠P +∠Q +∠R = 180º

(Sum of the measures of the angles of a triangle is 180º.)

∠P + 80º +30º = 180º

∠P = 70º

In ∆DEF and ∆PQR,

∠D = ∠P (Each 70°)

∠E = ∠Q (Each 80°)

∠F = ∠R (Each 30°)

∴ ∆DEF ∼ ∆PQR [By AAA similarity criterion]

2. In the following figure, ∆ODC ∼ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB

Ans: Do yourself

3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangle.show that AO/OC = OB/OD  

Ans: 

In ∆DOC and ∆BOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠ DOC = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles] 

∴ ∆DOC ∼ ∆BOA [AAA similarity criterion]

4.

Ans: 

5. S and T are point on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ∼ ∆RTS. 

Ans: 

In ∆RPQ and ∆RST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ∆RPQ ∼ ∆RTS (By AA similarity criterion)

 6. In the following figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ∼ ∆ABC.

Ans: 

It is given that ∆ABE ≅ ∆ACD.

∴ AB = AC [By CPCT] …………………(1)

And, AD = AE [By CPCT] …………….(2)

In ∆ADE and ∆ABC,

7. In the following figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that: 

(i) ∆AEP – ∆CDP 

Ans: 

In ∆AEP and ∆CDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion,

∆AEP – ∆CDP 

(ii) ∆ABD – ∆CBE 

Ans:

In ∆ABD and ∆CBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion,

∆ABD – ∆CBE

(iii) ∆AEP – ∆ADB

Ans: 

In ∆PDC and ∆BEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

∆PDC – ∆BEC

(iv) ∆PDC – ∆BEC

Ans: 

In ∆PDC and ∆BEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

∆PDC –  ∆BEC

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ∼ ∆CFB

In ∆ABE and ∆CFB,

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴ ∆ABE –  ∆CFB (By AA similarity criterion) 

9. In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

In ∆ABC and ∆AMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ∆ABC –  ∆AMP (By AA similarity criterion)

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC – ∆FEG, Show that:

Ans: 

11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD – ∆ECF

Ans:  It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ∆ABD and ∆ECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Proved above)

∴ ∆ABD ∼ ∆ECF (By using AA similarity criterion) 

12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see the given figure). Show that ∆ABC – ∆PQR.

Ans: 

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that

Ans: 

In ∆ADC and ∆BAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ∆ADC ∼ ∆BAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in Proportion.

14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that 

Ans: Do yourself.

15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. 

Ans: 

Let AB and CD be a tower and a pole respectively.

Let the shadow of BE and DF be the shadow of AB and CD respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, ∠DCF = ∠BAE

And, ∠DFC = ∠BEA

∠CDF = ∠ABE (Tower and pole are vertical to the ground)

∴ ∆ABE ∼ ∆CDF (AAA similarity criterion)

Therefore, the height of the tower will be 42 metres.

16. If AD and PM are medians of triangles ABC and PQR, respectively where 

Ans: 

1.

Ans: 

2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Ans: 

Since AB || CD,

∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ∆AOB and ∆COD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∴ ∆AOB ∼ ∆COD (By AAA similarity criterion)

3. In the following figure, ABC and DBC are two triangles on the same base BC. If AD 

Ans: Let us draw two perpendiculars AP and DM on line BC. 

In ∆APO and ∆DMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ∆APO ∼ ∆DMO (By AA similarity criterion)

4. If the areas of two similar triangles are equal, prove that they are congruent

Ans: Let us assume two similar triangles as ∆ABC ∼ ∆PQR.

5. D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the area of ∆DEF and ∆ABC. 

Ans: 

D and E are the mid-points of ∆ABC. 

6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Ans: 

7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. 

Ans: 

Let ABCD be a square of side a.  

Therefore, its diagonal = √2a 

Two desired equilateral triangles are formed as ∆ABE an ∆DBF. 

Side of an equilateral triangle, ∆ABE, described on one of its sides = a

Side of an equilateral triangle, ∆DBF, described on one of its Diagonals

We know that equilateral triangles have all its angles as 60º and all its sides of the same length.

Therefore, all equilateral triangles are similar to each other.

Hence, the ratio between the areas of these triangles will be equal to the square 

of the ratio between the sides of these triangles.

8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is

(a) 2 : 1 

(b) 1 : 2

(c) 4 : 1

(d) 1 : 4

Ans: 

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. 

Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. 

Let side of ∆ABC = x

Hence, the correct answer is (C).

Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

(a) 2 : 3 

(b) 4 : 9 

(c) 81 : 16 

(d) 16 : 81

Ans: If two triangles are similar to each other, then the ratio of 

the areas of these triangles will be equal to the square of the 

ratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9.

Therefore, ratio between areas of these triangles =

Hence, the correct answer is (D). 

1. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse

(i) 7cm, 24 cm, 25 cm.

Ans: Therefore, given sides 7cm, 24cm, 25cm make a right triangle.

(ii) 3cm, 8 cm, 6 cm.

Ans: (3)² + (6)² ≠ (8)² ( not a right triangle).

(iii) 50cm, 80 cm, 100 cm.

Ans: (50)² + (80)² + (100)² Not a right triangle.

 (iv) 13cm, 12 cm, 5 cm.

Ans: (iv) (12)² + (5)² = 144 + 25 = 169 = (13)2

Therefore, given sides 13cm, 12cm, 5cm make a right triangle.

3. In fig a triangle right angled at A and ACBD. Show that:

4. ABC is an isosceles triangle right angled at C. Prove that 2AC²

Ans: 

5. ABC is an isosceles triangle with AC = BC If AB² = 2AC² prove that ABC is a right triangle.

Ans: 

6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Ans: In equilateral Delta*ABC each side is 2a.

7. Prove that the sum of square of the sides of a rhombus is equal to th sum of the square of its diagonals.

Ans: ABCD is a rhombus in which AB = BC = CD = DA = a(say) Its diagonals AC and BD are right bisectors of each other at O.

8. In fig, O is point the interior of a triangle ABC, ODBC and of OF 丄AB. Show that: 

(i) OA² + OB² + OC² – OD2 -OE² – OF² AF² + BD² + CE²

9. A ladder 10m long reaches a window 8m above the ground. Find the distance of the foot of the ladder from base of the wall.

10. A guy attached to a vertical pole of height 18m is 24m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the write will be taunted.

Ans: Let x metre be the distance of the stake from the base of the pole.

11. An aeroplane leaves an airport and flies due north a speed of 1000 km per 1000 other aeroplanes leave the speed of 1000 and flies due west speed of 1200 km per hour. How far apart will be the 

Ans: 

12. Two poles of height 6m and 11m stand on a plane ground. If the distance between the feet of the poles is 12m, find the distance between the tops.

Ans: x = (12)2 + 52 

= 144 + 25 = 169          

⇒ x = 13

∴  Distance between the tops = 13m

13. D and E are point on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².

14. The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD. Prove that 2AB² = 2AC² + BC²

Ans: BD = 3CD

15.MIn an equilateral triangle ABC, D is a point on side B such that BD = 1/3 * BC Prove that:

Ans:

16. In an equilateral triangle, prove that three times the square of on one side is equal to four times the square of one of its altitudes.

Ans: ∆ABC is an equilateral triangle of side a (say)

Ans: ADBC, AD = x (say)

17. Tick the correct answer and justify : In ∆ABC, AB = 6√3 cm, AC = 12 and BC = 6 cm. The triangle B is:

(a) 120°

(b) 60°

(c) 90°

(d) 450

Ans: Correct is (C) Justification: Here AC² = AB2 + BC2

⇒ ∆ABC is right angled at B ⇒ ∠B = 90° Also BC = 6 cm and AB = 6√3cm ⇒ BC <AB

⇒ ∠ACB can not be more than 450 ⇒ ∠A = 30°

AC2 = (12)² = 144 BC2 + AB² = 62 + (6√3) = 36 + 108 = 144

1. In fig, PS is the bisector of QS ∠PQ ߡPQR of APQR. Prove that QS/SR = PQ/PR

2. In fig, D is a point on hypotenuse AC of ∆ABC, such that BD⊥AC, DM⊥BC and DN⊥AB. Prove that:

Ans: 

3. In fig., ABC is a triangle in which ∠ABC > 90° and ADICB produced. Proved that AC² = AB² + BC2 + 2BC.BD

Ans: An obtuse triangle, with obtuse angle at B and ADICB produced.

To prove : AC² = AB² + BC² + 2BC. BD

Proof : Since AADB is a triangle, right angle at D. By Pythagorus Theorem, AB² = AD² + BD2 → (1)

Again AADC is a right triangle with right angle at D.

4. In fig., ABC is a triangle in which ∠ABC <90° and ADBC. Prove that AC² = AB² + BC-2BC. BD.

Ans: given ΔABC in which ABC< 90° AD丄BC

To prove: AC² = AB² + BC-2BC. BD.

Proof: Since ADB is a right triangle with right angle at D. By Pythagorus Theorem, AB² = AD² + BD² … (i)

Again AADC is a right triangle with ∠D = 90°

5. In fig., AD is a median of a triangle ABC and AM1BC. Prove that:

6. Prove that the sum of the square of the diagonals of a parallelo- gram is equal to the sum of the square of its sides.

Ans: Let ABCD be a parallelogram with ∠B as obtuse and ∠A as acute.

7. In fig., two chords AB and CD intersect each at the point P. Prove.

Ans: (i) AAPC – ADPB

(ii) ΔΑΡ.ΡΒ = CP. DP.

Ans: In AAPC and ADPB

CAP=/PDB

[Angles in the same segments]

8. In fig., two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that :

Ans:  

9. In fig., D is a point of side BC of △ABC such that BD CD AB AC Prove that AD is the bisector of ∠BAC.

Ans: In a right angled ΔABC, 

AB = 1.8 BC = 2. 4m

AC2 = (I.8)2 + (2.4)2

= 3.24 + 5.57 = 9 ⇒ AC = √9.= 3m.

Hence the original length of the string AC (when taut) is 3m. (ii) When Nazima pulls 

in the string at the rate of 5 cm/sec.

then the length of the string decrease = 5 x 12 = 60cm = 0.6m in 12 seconds 

Remaining length of the string (AD) after 12 seconds pull =(3-0.60)=2.40 m 

Rightarrow AD = 2.4m . 

AD² + DB² + AB² ⇒ DB² 

= AD2 – AB2 

= (2040)2 – (1.80)2

= 1.587 = 1 . 59m

∴ DB = 2.52m 

Horizontal distance (DE) of the from Nazima = (1.59 + 1.2)m = 2.79.

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