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# NCERT Solution for Class 10th Maths Chapter 6 Triangles

This chapter introduces you to the fundamental concepts of triangles, including different types of triangles, congruency, and similarity.
The chapter covers various concepts related to triangles, such as the Pythagorean theorem, the properties of triangles, and the sum of angles in a triangle. The solutions provided by Swastik Classes offer step-by-step explanations of all the exercises and are designed to help you understand the concepts and solve problems effectively.
Triangles are an essential part of the mathematics curriculum and are widely used in various fields such as physics, engineering, and computer science. The solutions provided by Swastik Classes will help you to build a strong foundation in triangles and prepare you for more advanced mathematical concepts.
Overall, the NCERT Solutions for Class 10th Maths Chapter 6 – Triangles, provided by Swastik Classes, are an excellent resource for students who want to excel in the subject. The solutions will not only help them score well in their exams but also prepare them for more advanced topics in mathematics. It is crucial to have a clear understanding of the concepts covered in this chapter as it forms the foundation for more advanced concepts in triangles.

## Answers of Math NCERT solutions for class 10 Chapter 6 Triangles

Unit 6

Triangles

Exercise 6.1:

Q1:

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

1.     All circles are ……… (congruent, similar)
1.     All squares are …….. (similar, congruent)
1.     All ……… triangles are similar. (isosceles, equilateral)
1.     Two polygons of the same number of sides are similar, if their corresponding             angles are (a)……………… and their corresponding sides are (b)…………….(equal,         proportional)

(i)     Similar

(ii)    Similar

(iii)   Equilateral

(iv)    (a) Equal

(b) Proportional

Q2:

Give two different examples of pair of

(i)     Similar figures                 (ii)    Non-similar figures

1. Two equilateral triangles with sides 1 cm and 2 cm

Two squares with sides 1 cm and 2 cm

1. Trapezium and square

Triangle and parallelogram

Q3:

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.

Exercise 6.2

Q1:

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

(i)

(ii)

(i)

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

(ii)

It is given that DE || BC.

By using basic proportionality theorem, we obtain

Q2:

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.

1.         PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
1.     PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
1. PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

(i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

(ii)

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

(iii)

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

Q3:

In the following figure, if LM || CB and LN || CD, prove that

In the given figure, LM || CB

By using basic proportionality theorem, we obtain

Q4:

In the following figure, DE || AC and DF || AE. Prove that

In ΔABC, DE || AC

Q5:

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

In Δ POQ, DE || OQ

Q6:

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.

In Δ POQ, AB || PQ

*Q7:

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).

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that

Or, Q is the mid-point of AC.

*Q8:

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. (Recall that you have done it in Class IX).

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

Hence, by using basic proportionality theorem, we obtain

*Q9:

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show

that

Draw a line EF through point O, such that

By using basic proportionality theorem, we obtain

In ΔABD,

So, by using basic proportionality theorem, we obtain

From equations (1) and (2), we obtain

*Q10:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that Show that ABCD is a trapezium.

Let us consider the following figure for the given question.

Draw a line OE || AB

In ΔABD, OE || AB

By using basic proportionality theorem, we obtain

However, it is given that

[By the converse of basic proportionality theorem]

∴ ABCD is a trapezium.

Exercise 6.3

Q1:

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:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(i)     ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, [By AAA similarity criterion]

(ii)

(iii)        The given triangles are not similar as the corresponding sides are not proportional.

(iv)        In ≅ ΔMNL ≅  ΔQPR we observe that,

MNQP = MLQR = 12

Q2:

In the following figure, ΔODC ~ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC,

∠ DCO and ∠ OAB

DOB is a straight line.

∴ ∠ DOC + ∠ COB = 180°

⇒ ∠ DOC = 180° – 125°

= 55°

In ΔDOC,

∠ DCO + ∠ CDO + ∠ DOC = 180°

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

⇒ ∠ DCO + 70º + 55º = 180°

⇒ ∠ DCO = 55°

It is given that ΔODC ≅ΔOBA.

∴ ∠ OAB = ∠ OCD [Corresponding angles are equal in similar triangles.]

⇒ ∠ OAB = 55°

Q3:

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that

In ΔDOC and ΔBOA,

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

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

∠DOC = ∠BOA [Vertically opposite angles]

∴ ΔDOC ≅¼ ΔBOA [AAA similarity criterion]

Q4:

In the following figure, and Show that

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR (i)

Given,

Q5:

S and T are point on sides PR and QR of ΔPQR such that ∠ P = ∠ RTS. Show that ΔRPQ ~ ΔRTS.

In ΔRPQ and ΔRST,

∠ RTS = ∠ QPS (Given)

∠ R = ∠ R (Common angle)

(By AA similarity criterion)

Q6:

In the following figure, if ΔABE≅ΔACD, show that ΔADE ≅ ΔABC.

It is given that ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] (1) And, AD = AE [By CPCT] (2) In and

[Dividing equation (2) by (1)]

∠A = ∠A [Common angle]

∴ ΔADE ~ ΔABC [By SAS similarity criterion]

Q7:

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

(i)

(ii)

(iii)

(iv)

(i)

In ΔAEP and ΔCDP,

∠ AEP = ∠ CDP (Each 90°)

∠ APE = ∠ CPD (Vertically opposite angles) Hence, by using AA similarity criterion,

(ii)

In ΔABD and ΔCBE,

∠ ADB = ∠ CEB (Each 90°)

∠ ABD = ∠ CBE (Common)

Hence, by using AA similarity criterion,

(iii)

∠ AEP = ∠ ADB (Each 90°)

∠ PAE = ∠ DAB (Common)

Hence, by using AA similarity criterion,

(iv)

In ΔPDC and ΔBEC,

∠ PDC = ∠ BEC (Each 90°)

∠ PCD = ∠ BCE (Common angle)

Hence, by using AA similarity criterion,

Q8:

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE

In ΔABE and ΔCFB,

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

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

(By AA similarity criterion)

Q9:

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

1.         ΔABC ~ ΔAMP

In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ΔABC ΔAMP (By AA similarity criterion)

Q10:

CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of

ΔABC and ΔEFG respectively. If ΔABC ΔFEG show that:

(i)

(ii)        ΔDCB ~ ΔHGE

(iii)        ΔDCA ~ ΔHGF

It is given that ΔABC ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector) And, ∠DCB = ∠HGE (Angle bisector) In ΔACD and ΔFGH,

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴ ΔACDΔFGH (By AA similarity criterion)

In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ ΔDCB ΔHGE (By AA similarity criterion) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴ ΔDCA ΔHGF (By AA similarity criterion)

*Q11:

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

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ ABD = ∠ ECF

In ΔABD and ΔECF,

∠ADB = ∠ EFC (Each 90°)

∠ BAD = ∠ CEF (Proved above)

∴ ΔABD ΔECF  (By using AA similarity criterion)

*Q12:

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and

median PM of ΔPQR (see the given figure). Show that ΔABC ~ ΔPQR.

Median divides the opposite side.

Given that,

In ΔABD and ΔPQM,

(Proved above)

∴ ΔABD ΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

∴ ΔABC ΔPQR  (By SAS similarity criterion)

*Q13:

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

∠ACD = ∠BCA (Common angle)

∴ ΔADC ΔBAC (By AA similarity criterion)

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

*Q14:

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

Given that,

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides. Therefore, BD = DC and QM = MR

Also, AD = DE (By construction) And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR It was given that

∴ ΔABE ΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠BAE = ∠QPL … (1)

Similarly, it can be proved that ΔAEC ΔPLR and

∠CAE = ∠RPL … (2)

Adding equation (1) and (2), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒ ∠CAB = ∠RPQ … (3) In ΔABC and ΔPQR,

(Given)

∠CAB = ∠RPQ [Using equation (3)]

∴ ΔABC ΔPQR (By SAS similarity criterion)

*Q15:

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.

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)

∴ ΔABE ΔCDF (AAA similarity criterion)

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

*Q16:

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

It is given that ΔABC ΔPQR

We know that the corresponding sides of similar triangles are in proportion.

∴  … (1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PM are medians, they will divide their opposite sides.

∴  … (3)

From equations (1) and (3), we obtain

… (4)

In ΔABD and ΔPQM,

∠B = ∠Q [Using equation (2)]

[Using equation (4)]

∴ ΔABD ΔPQM (By SAS similarity criterion)

Exercise 6.4

Q1:

Let  and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Q2:

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.

Since AB || CD,

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

In ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∴ ΔAOB ΔCOD (By AAA similarity criterion)

Q3:

In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

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

We know that area of a triangle =

.

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ΔDMO (By AA similarity criterion)

Q4:

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

Let us assume two similar triangles as ΔABC ΔPQR.

Q5:

D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

D and E are the mid-points of ΔABC.

*Q6:

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

Let us assume two similar triangles as ΔABC ΔPQR. Let AD and PS be the medians of these triangles. ΔABC ΔPQR

…(1)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R     …(2)

Since AD and PS are medians,

and,

Equation (1) becomes

… (3)

In ΔABD and ΔPQS,

∠B = ∠Q [Using equation (2)]

And,  [Using equation (3)]

∴ ΔABD ΔPQS (SAS similarity criterion)

Therefore, it can be said that

… (4)

From equations (1) and (4), we may find that

And hence,

*Q7:

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.

Let ABCD be a square of side a. Therefore, its diagonal

Two desired equilateral triangles are formed as ΔABE and ΔDBF.

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

Side of an equilateral triangle, ΔDBF, described on one of its diagonals

We know that equilateral triangles have all its angles as 60 º 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.

*Q8:

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

We know that equilateral triangles have all its angles as 60 º 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 ΔABC = x

Therefore, side of

Hence, the correct answer is (C).

*Q9:

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

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).

Exercise 6.5

Q1:

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.

1.         7 cm, 24 cm, 25 cm
2.         3 cm, 8 cm, 6 cm
3.     50 cm, 80 cm, 100 cm
4.     13 cm, 12 cm, 5 cm

1.     It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the     lengths of these sides, we will obtain 49, 576, and

Or,

The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a     right triangle.

We know that the longest side of a right triangle is the hypotenuse. Therefore, the     length of the hypotenuse of this triangle is 25 cm.

1.     It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm. Squaring the             lengths of these sides, we will obtain 9, 64, and 36. However,

Or,

Clearly, the sum of the squares of the lengths of two sides is not equal to the     square of the length of the third side. Therefore, the given triangle is not satisfying     Pythagoras theorem.

Hence, it is not a right triangle.

1.     Given that sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.

However,

Or,

Clearly, the sum of the squares of the lengths of two sides is not equal to the     square of the length of the third side. Therefore, the given triangle is not satisfying     Pythagoras theorem.

Hence, it is not a right triangle.

1.     Given that sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will obtain 169, 144, and 25. Clearly,

Or,

The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a     right triangle.

We know that the longest side of a right triangle is the hypotenuse. Therefore, the     length of the hypotenuse of this triangle is 13 cm.

Q2:

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM x MR.

Q-3:

In the given figure, ABD is a triangle right angled at A and AC i. BD. Show that

(i)

(ii)

(iii)

Q4:

ABC is an isosceles triangle right angled at C. prove that AB2 = 2 AC2.

Given that ΔABC is an isosceles triangle.

∴ AC = CB

Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain

Hence proved.

Q5:

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

Given that,

Q6:

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

Let AD be the altitude in the given equilateral triangle, ΔABC.

We know that altitude bisects the opposite side.

∴ BD = DC = a

In an equilateral triangle, all the altitudes are equal in length. Therefore, the length of each altitude will be

Q7:

Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain

Hence proved.

Q8:

In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

1.         OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
1.     AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Join OA, OB, and OC.

1.     Applying Pythagoras theorem in ΔAOF, we obtain

Similarly, in ΔBOD,

Similarly, in ΔCOE,

(ii)    From the above result,

Q9:

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

Let OA be the wall and AB be the ladder.

Therefore, by Pythagoras theorem,

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

Q10:

A guy wire attached to a vertical pole of height 18 m is 24 m 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 wire will be taut?

Let OB be the pole and AB be the wire. By Pythagoras theorem,

OA2+OB2=AB2

OA2+182=242

OA2=252

Therefore, the distance from the base is

Q11:

An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be

the two planes after hours?

Distance travelled by the plane flying towards north inSimilarly, distance travelled by the plane flying towards west in Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distance between these planes after

Therefore, the distance between these planes will be    km after

Q12:

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

Let CD and AB be the poles of height 11 m and 6 m. Therefore, CP = 11 – 6 = 5 m

From the figure, it can be observed that AP = 12m

Applying Pythagoras theorem for ΔAPC, we obtain

Therefore, the distance between their tops is 13 m.

*Q13:

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

Applying Pythagoras theorem in ΔACE, we obtain

*Q14:

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

Applying Pythagoras theorem for ΔACD, we obtain

Applying Pythagoras theorem in ΔABD, we obtain

Hence  proved.

*Q15:

In an equilateral triangle ABC, D is a point on side BC such thatBC. Prove that 9

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

and

Given that,

∴ BD =

Applying Pythagoras theorem in ΔADE, we obtain

Hence proved.

*Q16:

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

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

Applying Pythagoras theorem in ΔABE, we obtain

AB2 = AE2 + BE2

4AE2 = 3a2

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

Hence proved.

*Q17:

Tick the correct answer and justify: In The angle B is:

(A) 120°         (B) 60°        (C) 90°         (D) 45°

Given thatcm, AC = 12 cm, and BC = 6 cm It can be observed that

AB2 = 108

AC2 = 144 And, BC2 = 36

AB2 +BC2 = AC2

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct answer is (C).

Exercise 6.6

*Q1:

In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T. Given that, PS is the angle bisector of ∠QPR.

∠QPS = ∠SPR … (1)

By construction,

∠SPR = ∠PRT (As PS || TR) … (2)

∠QPS = ∠QTR (As PS || TR) … (3)

Using these equations, we obtain

∠PRT = ∠QTR

∴ PT = PR

By construction, PS || TR

By using basic proportionality theorem for ΔQTR,

QSSR=QPPT

⇒ QSSR

*Q2:

In the given figure, D is a point on hypotenuse AC of ΔABC, DM ⊥ BC and DN ⊥ AB, Prove that:

1.         DM2 = DN.MC
1.     DN2 = DM.AN

1. Let us join DB.

We have, DN || CB, DM || AB, and ∠B = 90°

∴ DMBN is a rectangle.

∴ DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90°

⇒ ∠2 + ∠3 = 90° … (1) In ΔCDM,

∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° … (2)

In ΔDMB,

∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° … (3)

From equation (1) and (2), we obtain

∠1 = ∠3

From equation (1) and (3), we obtain

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

∴ ΔDCM ΔBDM AA similarity criterion)

⇒ DM2 = DN × MC

Hence proved.

(ii) In right triangle DBN,

∠5 + ∠7 = 90° … (4)

In right triangle DAN,

∠6 + ∠8 = 90° … (5)

D is the foot of the perpendicular drawn from B to AC.

⇒ ∠5 + ∠6 = 90° … (6)

From equation (4) and (6), we obtain

∠6 = ∠7

From equation (5) and (6), we obtain

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

∴ ΔDNA ΔBND (AA similarity criterion)

⇒ DN2 = AN × NB

⇒ DN2 = AN × DM (As NB = DM)

Hence proved.

*Q3:

In the given figure, ABC is a triangle in which ∠ ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.

Applying Pythagoras theorem in ΔADB, we obtain

AB2 = AD2 + DB2 … (1)

Applying Pythagoras theorem in ΔACD, we obtain

AC2 = AD2 + (DB + BC)2

AC2 = AD2 + DB2 + BC2 + 2DB x BC

AC2 = AB2 + BC2 + 2DB x BC [Using equation (1)]

*Q4:

In the given figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that

AC2 = AB2 + BC2 – 2BC.BD.

Applying Pythagoras theorem in ΔADB, we obtain

⇒ AD2 = AB2 – DB2 … (1)

Applying Pythagoras theorem in ΔADC, we obtain

AB2 – BD2 + DC2 = AC2 [Using equation (1)] AB2 – BD2 + (BC – BD)2 = AC2

AC2 = AB2 – BD2 + BC2 + BD2 -2BC x BD

= AB2 + BC2 – 2BC x BD

Hence proved.

*Q5:

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that: (i)

(ii)

(iii)

1.         Applying Pythagoras theorem in ΔAMD, we obtain

AM2 + MD2 = AD2 … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM2 + MC2 = AC2

AM2 + (MD + DC)2 = AC2

(AM2 + MD2) + DC2 + 2MD.DC = AC2

AD2 + DC2 + 2MD.DC = AC2 [Using equation (1)] Using the result, we     obtain

1.     Applying Pythagoras theorem in ΔABM, we obtain

AB2 = AM2 + MB2

= (AD2 – DM2) + MB2

= (AD2 – DM2) + (BD – MD)2

= AD2 – DM2 + BD2 + MD2 – 2BD × MD

= AD2 + BD2 – 2BD × MD

1. Applying Pythagoras theorem in ΔABM, we obtain

AM2 + MB2 = AB2 … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM2 + MC2 = AC2 … (2)

Adding equations (1) and (2), we obtain

2AM2 + (BD – DM)2 + (MD + DC)2 = AB2 + AC2

2AM2+BD2 + DM2 – 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2

2AM2 + 2MD2 + BD2 + DC2 + 2MD ( – BD + DC) = AB2 + AC2

*Question 6:

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Let ABCD be a parallelogram.

Let us draw perpendicular DE on extended side AB, and AF on side DC. Applying Pythagoras theorem in ΔDEA, we obtain

DE2 + EA2 = DA2                                 … (i)

Applying Pythagoras theorem in ΔDEB, we obtain DE2 + EB2 = DB2

DE2 + (EA + AB)2 = DB2

… (ii)

Applying Pythagoras theorem in ΔADF, we obtain AD2 = AF2 + FD2

Applying Pythagoras theorem in ΔAFC, we obtain

AC2 = AF2 + FC2

= AF2 + (DC − FD)2

= AF2 + DC2 + FD2 − 2DC × FD

….(iii)

Since ABCD is a parallelogram, AB = CD                 … (iv)

And, BC = AD                                 … (v)

∠DEA = ∠AFD (Both 90°)

∴ ΔEAD ΔFDA (AAS congruence criterion)

EA = DF                                     … (vi)

Adding equations (i) and (iii), we obtain

[Using equations (iv) and (vi)]

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

Hence proved.

*Question 7:

In the given ﬁgure, two chords AB and CD intersect each other at the point P. prove that:

(i)

(ii)

Let us join CB.

1.         In ΔAPC and ΔDPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB) ΔAPC ~ ΔDPB (By AA     similarity criterion)

Hence proved

1.     We have already proved that

ΔAPC ~ ΔDPB

We know that the corresponding sides of similar triangles are proportional.

∴ AP. PB = PC. DP

Hence proved

*Question 8:

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

(i)

(ii)

1.         In ΔPAC and ΔPDB,

∠P = ∠P (Common)

∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to     the opposite interior angle)

∴ ΔPAC ~ ΔPDB

Hence proved.

1. We know that the corresponding sides of similar triangles are proportional.

∴ PA.PB = PC.PD

Hence proved

*Question 9:

In the given ﬁgure, D is a point on side BC of ΔABC such that (i)

(ii)    . Prove that AD is the bisector of ∠BAC.

Let us extend BA to P such that AP = AC. Join PC.

It is given that,

By using the converse of basic proportionality theorem, we obtain AD || PC

∠BAD = ∠APC (Corresponding angles) … (1)

And, ∠DAC = ∠ACP (Alternate interior angles) … (2)

By construction, we have AP = AC

∠APC = ∠ACP … (3)

On comparing equations (1), (2), and (3), we obtain

AD is the bisector of the angle BAC.

*Question 10:

Nazima is ﬂy ﬁshing in a stream. The tip of her ﬁshing rod is 1.8 m above the surface of the water and the ﬂy at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the ﬂy) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the ﬂy from her after 12 seconds?

Let AB be the height of the tip of the ﬁshing rod from the water surface.

Let BC be the horizontal distance of the ﬂy from the tip of the ﬁshing rod.

Then, AC is the length of the string.

AC can be found by applying Pythagoras theorem in ΔABC.

Thus, the length of the string out is 3 m.

She pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds

Let the ﬂy be at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

(1.8 m)2 + BD2 = (2.4 m)2

Horizontal distance of ﬂy = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m

## NCERT Solutions for Class 10 Maths Chapter 6- Triangles

NCERT Solutions Class 10 Maths Chapter 6, Triangle, is part of the Unit Geometry which constitutes 15 marks of the total marks of 80. On the basis of the CBSE Class 10 syllabus, this chapter belongs to the unit that has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and the problem-solving methods in this chapter is mandatory to score well in the Board examination of class 10 maths.

Main topics covered in this chapter include:

6.1 Introduction

From your earlier classes, you are familiar with triangles and many of their properties. In Class 9, you have studied congruence of triangles in detail. In this chapter, we shall study about those figures which have the same shape but not necessarily the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.

6.2 Similar Figures

In Class 9, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. The topic explains similarity of figures by performing relevant activity. Similar figures are two figures having the same shape but not necessarily the same size.

6.3 Similarity of Triangles

The topic recalls triangles and its similarities. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). It explains Basic Proportionality Theorem and different theorems are discussed performing various activities.

6.4 Criteria for Similarity of Triangles

In the previous section, we stated that two triangles are similar (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). The topic discusses the criteria for similarity of triangles referring to the topics we have studied in earlier classes. It also contains different theorems explained with proper examples.

6.5 Areas of Similar Triangles

You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. The topic Areas of Similar Triangles consists of theorem and relatable examples to prove the theorem.

6.6 Pythagoras Theorem

You are already familiar with the Pythagoras Theorem from your earlier classes. You have also seen proof of this theorem in Class 9. Now, we shall prove this theorem using the concept of similarity of triangles. Hence, the theorem is verified through some activities and make use of it while solving certain problems.

6.7 Summary

The summary contains the points you have studied in the chapter. Going through the points mentioned in the summary will help you to recollect all the important concepts and theorems of the chapter.

Triangle is one of the most interesting and exciting chapters of the unit Geometry as it takes us through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles including the detailed explanation of similar figure, different theorems related to the similarities of triangles with proof, and the areas of similar triangles. The chapter concludes by explaining the Pythagoras theorem and the ways to use it in solving problems. Read and learn the Chapter 6 of NCERT textbook to learn more about Triangles and the concepts covered in it. Ensure to learn the NCERT Solutions for Class 10 effectively to score high in the board examination.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

• Helps to ensure that the students use the concepts in solving the problems.
• Encourages the children to come up with diverse solutions to problems.
• Hints are given for those questions which are difficult to solve.
• Helps the students in checking if the solutions they gave for the questions are correct or not.

Conclusion

The NCERT Solutions for Class 10th Maths Chapter 6 – Triangles, provided by Swastik Classes, offer a comprehensive understanding of the fundamental concepts of triangles. The chapter covers various topics such as the properties of triangles, the Pythagorean theorem, and the sum of angles in a triangle.
The solutions provided by Swastik Classes are well-structured and easy to understand, with step-by-step explanations for all the exercises. The solutions are designed to help students develop their problem-solving skills and gain a deeper understanding of the concepts covered in the chapter.
Triangles are an essential part of the mathematics curriculum and are widely used in various fields such as physics, engineering, and computer science. The solutions provided by Swastik Classes will help students build a strong foundation in triangles and prepare them for more advanced mathematical concepts.
Overall, the NCERT Solutions for Class 10th Maths Chapter 6 – Triangles, provided by Swastik Classes, are an excellent resource for students who want to excel in the subject. The solutions will not only help them score well in their exams but also prepare them for more advanced topics in mathematics. It is crucial to have a clear understanding of the concepts covered in this chapter as it forms the foundation for more advanced concepts in triangles.

### Why should we learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6?

The concepts covered in Chapter 6 of NCERT Solution Maths provide questions based on the exam pattern and model question paper. So it is necessary to learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6.

### List out the important topics present in NCERT Solutions for Class 10 Maths Chapter 6?

The topics covered in the chapters are Introduction to the triangles, similar figures, similarity of triangles, criteria for similarity of triangles, areas of similar triangles and Pythagoras Theorem. These concepts are important from an exam perspective. It is strictly based on the latest syllabus of CBSE board, and also depends on CBSE question paper design and marking scheme.

### How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 6?

There are 6 exercises are there in NCERT Solutions for Class 10 Maths Chapter 6. First exercise contains 3 questions, second exercise contains 10 questions, third exercise has 16 questions, fourth exercise has 9 questions, fifth exercise has 17 questions and last or sixth exercise has ten questions based on the concepts of triangles.