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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-6-ex-6-5/

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

Question 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) 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
Solution:
(i) 7 cm, 24 cm,-25 cm
(7)² + (24)² = 49 + 576 = 625 = (25)² = 25
∴ The given sides make a right angled triangle with hypotenuse 25 cm

(ii) 3 cm, 8 cm, 6 cm
(8)² = 64
(3)² + (6)² = 9 + 36 = 45
64 ≠ 45
The square of larger side is not equal to the sum of squares of other two sides.
∴ The given triangle is not a right angled.

(iii) 50 cm, 80 cm, 100 cm
(100)²= 10000
(80)² + (50)² = 6400 + 2500
= 8900
The square of larger side is not equal to the sum of squares of other two sides.
∴The given triangle is not a right angled.

(iv) 13 cm, 12 cm, 5 cm
(13)² = 169
(12)² + (5)²= 144 + 25 = 169
= (13)² = 13
Sides make a right angled triangle with hypotenuse 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM • MR.

Solution:
In right angled ∆QPR,
∠P = 90°, PM ⊥ QR
∴ ∆PMQ ~ ∆RMP
[If ⊥ is drawn from the vertex of right angle to the hypotenuse then triangles on both sides of perpendicular are similar to each other, and to whole triangle]

⇒ [Corresponding sides of similar
⇒ PM x MP = RM x MQ ⇒ PM2 = QM.MR

Question 3.
In the given figure, ABD is a triangle right angled at A and AC i. BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD

Solution:

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
Given: In ∆ABC, ∠C = 90° and AC = BC
To Prove: AB² = 2AC²
Proof: In ∆ABC,
AB²= BC² + AC²
AB² = AC² + AC² [Pythagoras theorem]
= 2AC²

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC² , Prove that ABC is a right triangle.
Solution:

Question 6.
ABC is an equilateral triangle of side la. Find each of its altitudes.
Solution:

Given: In ∆ABC, AB = BC = AC = 2a
We have to find length of AD
In ∆ABC,
AB = BC = AC = 2a
and AD ⊥ BC
BD = \(\frac { 1 }{ 2 }\) x 2 a = a
In right angled triangle ADB,
AD² + BD² = AB²
⇒ AD² = AB² – BD²= (2a)² – (a)² = 4a²– a²= 3a²
AD = √3a

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Given: ABCD is a rhombus. Diagonals AC and BD intersect at O.
To Prove: AB²+ BC²+ CD²+ DA² = AC²+ BD²

Question 8.
In the given 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.

Solution:

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

Let AC be the ladder of length 10 m and AB = 8 m
In ∆ABC, BC² + AB² = AC²
⇒ BC²= AC² – AB²= (10)² – (8)²
BC² = 100-64 – 36 BC = √36 = 6 m
Hence distance of foot of the ladder from base of the wall is 6 m.

Question 10.
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?
Solution:

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1\(\frac { 1 }{ 2 }\) hours?
Solution:

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

Length of poles is 6 m and 11m.
DE = DC – EC = 11m-6m = 5m
In ∆DAE,
AD² = AE² + DE² [ ∵AE = BC]
= (12)² + (5)² =144 + 25 = 169
AD = √l69 = 13

Question 13.
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.
Solution:

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

Solution:

Question 15.
In an equilateral triangle ABC, D is a point on side BC, such that BD = \(\frac { 1 }{ 3 }\)BC. Prove that 9AD² = 7AB².
Solution:

Question 16.
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.
Solution:

Question 17.
Tick the correct answer and justify : In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:
(a) 120°
(b) 60°
(c) 90°
(d) 45
Solution:

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-6-ex-6-4/

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4

Question 1.
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
Since, ∆ABC ~ ∆DEF
The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

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

Question 3.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that


Solution:

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

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

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

Question 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.
Solution:

Question 8.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(a) 2 :1
(b) 1:2
(c) 4 :1
(d) 1:4
Solution:

Question 9.
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
Solution:
Justification: Areas of two similar triangles are in the ratio of the squares of their corresponding sides.

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-6-ex-6-2/

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

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

Solution:

Question 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
(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
Solution:

Question 3.
In the given figure, if LM || CB and LN || CD.


Solution:

Question 4.
In the given figure, DE || AC and DF || AE.


Solution:

Question 5.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Solution:

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

Solution:

Question 7.
Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:
Given: A ∆ABC in which D is the mid-point of AB and DE || BC

Question 8.
Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:
Given: A ΔABC in which D and E are mid-points of sides AB and AC respectively.
To Prove: DE || BC

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the
Solution:

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) Show that ABCD is a trapezium.
Solution:

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2, drop a comment below and we will get back to you at the earliest.