NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles. Furthermore, here we’ve provided you with the latest solution for Class 10 CBSE NCERT Maths Chapter 6 – Triangles. As a result here you’ll find solutions to all the exercises. This NCERT Class 10 solution will help you to score good marks in your exam.

Students can refer to our solution for NCERT Class 10 Maths Chapter 6 – Triangles. The Chapter 6 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise wise latest solution.

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Pdf Download

NCERT Class 10 Maths Chapter 6 – Triangles Exercise wise solution

Exercise 6.1 – Page 122 of NCERT
Exercise 6.2 – Page 128 of NCERT
Exercise 6.3 – Page 138 of NCERT
Exercise 6.4 – Page 143 of NCERT
Exercise 6.5 – Page 150 of NCERT
Exercise 6.6 – Page 152 of NCERT

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.1

Here you’ll find NCERT Chapter 6 – Triangles Exercise 6.1 Solution.
Exercise 6.1: Solutions of Questions on Page Number: 122.

Q1: Fill in the blanks using correct word given in the brackets:

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

Q2: Give two different examples of pair of
(i) Similar figures
(ii)Non-similar figures

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

Two squares with sides 1 cm and 2 cm

Trapezium and square

Triangle and parallelogram

Q3: State whether the following quadrilaterals are similar or not:

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

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.2

Here you’ll find NCERT Chapter 6 – Triangles Exercise 6.1 Solution.
Exercise 6.2: Solutions of Questions on Page Number: 128.

Q1: In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)
(i)

(ii)

Answer:
(i)

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain

(ii)

Let AD = x cm
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.

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

Answer:

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

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

Answer:

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

Q4: In the following figure, DE || AC and DF || AE. Prove that

Answer:

In ΔABC, DE || AC

Q5: In the following figure, DE || OQ and DF || OR, show that EF || QR

Answer:

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

Answer:

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).
Answer:

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ || BC
By using proportionality theorem, we obtain

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).

Answer:

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 PQ || BC.

Q9: ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that

Answer:

Draw a line EF through point O, such that
In ΔADC
EF || CD
EO || CD
By using basic proportionality theorem, we obtain

In ΔABD
OE || AB
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.

Answer:
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

EO || DC [By the converse of basic proportionality theorem]
AB || OE || DC
AB || CD ∴ ABCD is a trapezium.

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.3

Here you’ll find NCERT Chapter 6 – Triangles Exercise 6.3 Solution.
Exercise 6.3: Solutions of Questions on Page Number: 138.

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)

Answer:

(i) ∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
Therefore, ΔABC ~ ΔPQR [By AAA similarity criterion]

(ii)

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

(iv) The given triangles are similar By SAS similarity criteria
∴ ∆MNL ∼ ∆QPR

(v) The given triangles are not similar as the corresponding angle is not
contained by the two corresponding sides.

Q2: In the following figure, ∆ODC ∼ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

DOB is a straight line.
∴ ∠DOC + ∠COB = 180°
∠DOC = 180° − 125°
=55°
In ∆DOC,
∠DCO + ∠CDO + ∠DOC = 180° (Sum of the measures of the angles of a triangle is 180^o.)
∠DCO + 70^o + 55^o = 180°
∠DCO = 55°
It is given that ∆ODC ∼ ∆OBA.
∴∠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 AO/OC = OB/OD

Answer: In ∆DOC and ∆BOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ∆DOC ∼ ∆BOA [AAA similarity criterion]

Q4: In the following figure Show that

Answer:

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 ∼ ∆RTS.

Answer:
In ∆RPQ and ∆RST,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ∆RPQ ∼ ∆RTS (By AA similarity criterion)

Q6: In the following figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ∼ ∆ABC

Answer:

It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By CPCT] … (1)
And, AD = AE [By CPCT]…. (2)
In ΔADE and ΔABC

[Dividing equation (2) by (1)]

∠A = ∠A [Common angle
∴ ∆ADE ∼ ∆ABC [By SAS similarity criterion]

Q7: In the following figure, altitudes AD and CE of ∆ABC intersect each other at
the point P. Show that:

(i) ∆AEP ∼ ∆CDP
(ii) ∆ABD ∼ ∆CBE
(iii) ∆AEP ∼ ∆ADB
(v) ∆PDC ∼ ∆BEC

Answer:

(i)

In ∆AEP and ∆CDP,
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
∆AEP ∼ ∆CDP

(ii)

In ∆ABD and ∆CBE,
∠ADB = ∠CEB (Each 90°)
∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
∆ABD ∼ ∆CBE

(iii)

In ∆AEP and ∆ADB,
∠AEP = ∠ADB (Each 90°)
∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
∆AEP ∼ ∆ADB

(iv)

In ∆PDC and ∆BEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
∆PDC ∼ ∆BEC

Q8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ∼ ∆CFB

Answer:

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)

Q9: In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

(i) ∆ABC ∼ ∆AMP
(ii) CA/PA = BC/MP
Answer:

(i) In ∆ABC and ∆AMP,
∠ABC = ∠AMP (Each 90°)
∠A = ∠A (Common)
∴ ∆ABC ∼ ∆AMP (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 ∆ABC and ∆EFG respectively. If ∆ABC ∼ ∆FEG, Show that:

(i) CD/GH = AC/FG
(ii) ∆DCB ∼ ∆HGE
(iii) ∆DCA ∼ ∆HGF

Answer:
It is given that ∆ABC ∼ ∆FEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
Since, ∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ∆ACD and ∆FGH,
∠A = ∠F (Proved above)
∠ACD = ∠FGH (Proved above)
∴ ∆ACD ∼ ∆FGH (By AA similarity criterion)

(ii) In ∆DCB and ∆HGE,
∠DCB = ∠HGE (Proved above)
∠B = ∠E (Proved above)
∴ ∆DCB ∼ ∆HGE (By AA similarity criterion)

(iii)

In ∆DCA and ∆HGF,
∠ACD = ∠FGH (Proved above)
∠A = ∠F (Proved above)
∴ ∆DCA ∼ ∆HGF (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 ∆ABD ∼ ∆ECF

Answer:
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)

Q12: 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.

Median divides the opposite side

∴ ∆ABC ∼ ∆PQR (By SAS similarity criterion)

Q13: D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD

Answer:

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.

= CA² = CB.CD

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 ∆ABC ∼ ∆PQR.

Answer:

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

∆ABE ∼ ∆PQL (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
∴ ∠BAE = ∠QPL … (1)
Similarly, it can be proved that ∆AEC ∼ ∆PLR and
∠CAE = ∠RPL … (2)
Adding equation (1) and (2), we obtain
∠BAE + ∠CAE = ∠QPL + ∠RPL
⇒ ∠CAB = ∠RPQ … (3)
In ∆ABC and ∆PQR,
AB AC
PQ PR

(Given)
∠CAB = ∠RPQ [Using equation (3)]
∴ ∆ABC ∼ ∆PQR (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.
Answer

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.

Q16: If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ∼ ∆PQR Prove that AB/PQ = AD/PM
Answer:

It is given that ∆ABC ∼ ∆PQR
We know that the corresponding sides of similar triangles are in proportion.

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.4

Here you’ll find NCERT Chapter 6 – Triangles Exercise 6.4 Solution.
Exercise 6.4: Solutions of Questions on Page Number: 143.

Q1: Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Answer:

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.
Answer:

∠AOB = ∠COD (Vertically opposite angles)
∠OAB = ∠OCD (Alternate interior angles)
∠OBA = ∠ODC (Alternate interior angles)
∴ ∆AOB ∼ ∆COD (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 .

Answer: We know that area of a triangle = 1/2 × Base × Height.

In ∆APO and ∆DMO,
∠APO = ∠DMO (Each = 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ∆APO ∼ ∆DMO (By AA similarity criterion)

Q4: If the areas of two similar triangles are equal, prove that they are congruent.
Answer:
Let us assume two similar triangles as ∆ABC ∼ ∆PQR.

Q5: 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.
Answer:

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.

Answer:

Let us assume two similar triangles as ∆ABC ∼ ∆PQR.
Let AD and PS be the medians of these triangles.
∆ABC ∼ ∆PQR

∴ ∆ABD ∼ ∆PQS (SAS similarity criterion)
Therefore, it can be said that

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.
Answer:

Let ABCD be a square of side a
Therefore, its diagonal = √2a
Two desired equilateral triangles are formed as ∆ABE and ∆DBF.
Side of an equilateral triangle, ∆ABE, described on one of its sides = √2a
Side of an equilateral triangle, ∆DBF, described on one of its diagonals.
We know that equilateral triangles have all its angles as 60° 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

Answer:

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 ∆ABC = x
Therefore, side of ∆BDE = x/2.

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

Answer:

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 = (4/9)² = 16/81
Hence, the correct answer is (D)

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.5

Here you’ll find NCERT Chapter 6 – Triangles Exercise 6.5 Solution.
Exercise 6.5: Solutions of Questions on Page Number: 150.

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.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Answer:
(i)

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 625.
49 + 576 = 625 Or
7² + 24² = 25²
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.

(ii) 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, 9 + 36 ≠ 64
Or, 3² + 6² ≠ 8²
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.

(iii) 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, 2500 + 6400 ≠ 1000
Or, 50² + 80² ≠ 100²
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.

(iv) 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, 144 +25 = 169
Or, 12² + 5² = 13²
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 × MR.

Let ∠MPR = x
In ∆MPR,
∠MRP = 180° – 90° – x
∠MRP = 90° – x
Similarly, In ∆MPQ,
∠MPQ = 90° −ÐMPR
= 90° – x
∠MQP = 180° – 90° – (90° – x)
∠MQP = x
In ∆QMP and ∆PMR
∠MPQ = ∠MRP
∠PMQ = ∠RMP
∠MQP = ∠MPR
∴ ∆QMP ~ ∆PMR

Q3: ABC is an isosceles triangle right angled at C. prove that AB²= 2 AC².
Answer:

Given that ΔABC is an isosceles triangle.

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

Q4: ABC is an isosceles triangle with AC = BC. If AB²= 2 AC², prove that ABC is a right triangle.
Answer:

Q5: ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Answer:

Let AD be the altitude in the given equilateral triangle, ∆ABC. We know that altitude bisects the opposite side.
∴BD = DC = a
In DADB,
∠ADB = 90
Applying Pythagoras theorem, we obtain
∠AD² = DB² + AB²
∠AD² + a² = (2a)2
∠AD² + a² = 4a
∠AD² = 3a²
∠AD = a√3
In an equilateral triangle, all the altitudes are equal in length. Therefore, the length of each altitude will be √3a.

Q6: Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
Answer:

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,
Applying Pythagoras theorem, we obtain

Q7: In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA² + OB² + OC² − OD² − OE² − OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF²

Answer: Join OA, OB, and OC

(i)

Applying Pythagoras theorem in ∆AOF, we obtain
OA² = OF² + AF² …..(i)
Similarly, in ∆BOD,
OB² = OD² + BD² …(ii)
Similarly, in ∆COE,
OC² = OE² + EC² … (iii)
Adding these equations
OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²
OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + EC²

(ii) From the above result,
AF² + BD² + EC² = (OA² – OE² ) + (OC² – OD² ) + (OB² – OF² )
AF² + BD² + EC² = AE² + CD² + BF²

Q8: 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.

Answer:

Let OA be the wall and AB be the ladder. Therefore, by Pythagoras theorem,
AB² = OA² + BO²
(10m)² = (8m)² + OB² 100 m² = 64 m² + OB² OB² = 36m²
OB = 6 m
Therefore, the distance of the foot of the ladder from the base of the wall is 6m.

Q9: 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?

Answer:

Let OB be the pole and AB be the wire. By Pythagoras theorem,
AB² = OB² + OA²
(24m)² = (18m)² + OA²
OA² = (576 – 324)m² = 252m²
OA = 6√7m
Therefore, the distance from the base is 6√7m.

Q10:

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?

Answer:

Q11: 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.

Answer:

Let CD and AB be the poles of height 11 m and 6 m. Therefore, CP = 11 − 6 = 5m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for ∆APC, we obtain
AP² + PC² = AC²
(12m)² + (5m)² = AC²
AC² = (144 + 25)m² = 169 m²
AC = 13m.

Q12: 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

Answer:

Applying Pythagoras theorem in ∆ACE, we obtain
AC² + CE² = AE² …. (1)
Applying Pythagoras theorem in ∆BCD, we obtain
BC² + CD² = BD² … (2)
Using equation (1) and equation (2), we obtain
AC² + CE² + BC² + CD² = AE² + BD² … (iii)
Applying Pythagoras theorem in ∆CDE, we obtain
DE² = CD² + CE²
Applying Pythagoras theorem in ∆ABC, we obtain
AB² = AC² + CB²
Putting the values in equation (3), we obtain
DE² + AB² = AE² + BD²

Q13: 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

Answer:

Applying Pythagoras theorem for ∆ACD, we obtain
AC² = AD² + DC²
AD² = AC² – DC² ….(1)
Applying Pythagoras theorem in ∆ABD, we obtain
AB² = AD² + DB²
AD² = AB² – DB² ….. (2)
From equation (1) and equation (2), we obtain
AC² – DC² = AB² – DB²
It is given that 3DC = DB

Q14 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that
9AD² = 7AB²
Answer:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC
BE = EC = BC/2 = a/2
AE = (a√3)/2
Given that BD = 1/3 BC.

Q15: 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.
Answer:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC
BE = EC = BC/2 = a/2
Applying Pythagoras theorem in ΔABE,
we obtain AB²= AE²+ BE²

4AE²= 3a²
=4 × (Square of altitude) = 3 × (Square of one side).

Q16: Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:
(A) 120° (B) 60°
(C) 90° (D) 45°

AB² = 108
AC² = 144 And, BC² = 36
AB² +BC² = AC²
The given triangle, ∆ABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct answer is (C).

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.6

Here you’ll find NCERT Chapter 6 – Triangles Exercise 6.6 Solution.
Exercise 6.6: Solutions of Questions on Page Number: 152.

Q1: In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that

Answer:

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 ∆QTR,

Q2: In the given figure, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB,
Prove that:
(i) DM² = DN.MC
(ii) DN² = DM.AN

Answer:
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 ∆CDM,
∠1 + ∠2 + ∠DMC = 180°
⇒ ∠1 + ∠2 = 90° … (2)
In ∆DMB,
∠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 ∆DCM and ∆BDM,
∠1 = ∠3 (Proved above)
∠2 = ∠4 (Proved above)
∴ ∆DCM ∼ ∆BDM (AA similarity criterion)

= DM² = DN.MC

(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.
∴ ∠ADB = 90°
⇒ ∠5 + ∠6 = 90° … (6)
From equation (4) and (6), we obtain
∠6 = ∠7
From equation (5) and (6), we obtain
∠8 = ∠5
In ∆DNA and ∆BND,
∠6 = ∠7 (Proved above)
∠8 = ∠5 (Proved above)
∴ ∆DNA ∼ ∆BND (AA similarity criterion)
DN² = AN × NB

DN² = AN × DM (As NB = DM)

Q3: In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB
produced. Prove that AC² = AB² + BC² + 2BC.BD.

Answer:
Applying Pythagoras theorem in ∆ADB, we obtain
AB² = AD² + DB² … (1)
Applying Pythagoras theorem in ∆ACD, we obtain
AC² = AD² + DC²
AC² = AD² + (DB + BC)²
AC² = AD² + DB² + BC² + 2DB × BC
AC² = AB² + BC² + 2DB × 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

Answer:
Applying Pythagoras theorem in ∆ADB, we obtain
AD² + DB² = AB²
AD² = AB² − DB² … (1)
Applying Pythagoras theorem in ∆ADC, we obtain
AD² + DC² = AC²
AB² − BD² + DC² = AC² [Using equation (1)]
AB² − BD² + (BC − BD) ² = AC²
AC² = AB² − BD² + BC² + BD² −2BC × BD
= AB² + BC² − 2BC × BD

Q5: In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i)

(ii)

(iii)

Applying Pythagoras theorem in ∆AMD,
we obtain AM² + MD² = AD² … (1)
Applying Pythagoras theorem in ∆AMC, 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 results DC = BC/2, We obtain

(ii) Applying Pythagoras theorem in ∆ABM, we obtain
AB² = AM² + MB²
=(AD² − DM²) + MB²
= (AD² − DM²) + (BD − MD)²
=AD² − DM² + BD² + MD² − 2BD × MD
=AD² + BD² − 2BD × MD

(iii) Applying Pythagoras theorem in ∆ABM, we obtain AM² + MB² = AB² … (1)
Applying Pythagoras theorem in AMC, we obtain AM² + MC² = AC² … (2)
Adding equations (1) and (2), we obtain 2AM² + MB² + MC² = AB² + AC²
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

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles help you. If you have any queries regarding chapter 6. Drop a comment below and we will get back to you at the earliest.