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

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

Edited By Ramraj Saini | Updated on Nov 25, 2023 08:20 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 Triangles

NCERT Solutions for Exercise 6.5 Class 10 Maths Chapter 6 Triangles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 6.5 introduces the concept of similar triangles having an angle of 90 degree. In NCERT solutions for Class 10 Maths chapter 6 exercise 6.5 three theorems are given which explains some important concepts of the right-angled triangle. In NCERT book Class 10 Maths, theorem 6.7 states that “if a perpendicular is drawn from the vertex of a right-angle triangle to the hypotenuse, and then triangles on both sides of the perpendicular are comparable to each other and to the total triangle”. The above-stated theorem of NCERT syllabus Class 10 Maths chapter 6 can be used to prove the Pythagoras theorem.

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  1. NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 Triangles
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 6 exercise 6.5
  3. Assess NCERT Solutions for Class 10 Maths chapter 6 exercise 6.5
  4. More About NCERT Solutions for Class 10 Maths Exercise 6.5
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 6.5
  6. Also, See
  7. NCERT Solutions of Class 10 Subject Wise
  8. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 6.5 Class 10 Maths Chapter 6 - Triangles
NCERT Solutions for Exercise 6.5 Class 10 Maths Chapter 6 - Triangles

10th class Maths exercise 6.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 6 exercise 6.5

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Triangles Class 10 Chapter 6 Exercise: 6.5

Q1 (1) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 7 cm, 24 cm, 25 cm

Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 7 cm, 24 cm

By Pythagoras theorem,

h^2=7^2+24^2

h^2=49+576

h^2=625

h=25 = given third side.

Hence, it is the right triangle with h=25 cm.

Q1 (2) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 3 cm, 8 cm, 6 cm

Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 3 cm, 6 cm

By Pythagoras theorem,

h^2=3^2+6^2

h^2=9+36

h^2=45

h=\sqrt{45}\neq 8

Hence, it is not the right triangle.

Q1 (3) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 50 cm, 80 cm, 100 cm

Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 50 cm, 80 cm

By Pythagoras theorem,

h^2=50^2+80^2

h^2=2500+6400

h^2=8900

h=\sqrt{8900}\neq 100

Hence, it is not a right triangle.

Q1 (4) 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. 13 cm, 12 cm, 5 cm

Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 5cm, 12 cm

By Pythagoras theorem,

h^2=5^2+12^2

h^2=25+144

h^2=169

h=13 = given third side.

Hence, it is a right triangle with h=13 cm.

Q2 PQR is a triangle right angled at P and M is a point on QR such that PM \perp QR . Show that PM ^2 = QM . MR .

Answer:

1635933005578

Let \angle MPR be x

In \triangle MPR ,

\angle MRP=180 \degree-90 \degree-x

\angle MRP=90 \degree-x

Similarly,

In \triangle MPQ ,

\angle MPQ=90 \degree-\angle MPR

\angle MPQ=90 \degree-x

\angle MQP=180 \degree-90 \degree-(90 \degree-x)=x

In \triangle QMP\, and\, \triangle PMR,

\angle MPQ\, =\angle MRP

\angle PMQ\, =\angle RMP

\angle MQP\, =\angle MPR

\triangle QMP\, \sim \triangle PMR, (By AAA)

\frac{QM}{PM}=\frac{MP}{MR}

\Rightarrow PM^2=MQ\times MR

Hence proved.

Q3 (1) In Fig. 6.53, ABD is a triangle right angled at A and AC \perp BD. Show that AB^2 = BC . BD .

1635933025836

Answer:

In \triangle ADB\, and\, \triangle ABC,

\angle DAB\, =\angle ACB \, \, \, \, \, \, \, \, (Each 90 \degree)

\angle ABD\, =\angle CBA (common )

\triangle ADB\, \sim \triangle ABC (By AA)

\Rightarrow \frac{AB}{BC}=\frac{BD}{AB}

\Rightarrow AB^2=BC.BD , hence prooved .

Q3 (2) In Fig. 6.53, ABD is a triangle right angled at A and AC \perp BD. Show that AC^2 = BC . DC .

1635933077069

Answer:

Let \angle CAB be x

In \triangle ABC ,

\angle CBA=180 \degree-90 \degree-x

\angle CBA=90 \degree-x

Similarly,

In \triangle CAD ,

\angle CAD=90 \degree-\angle CAB

\angle CAD=90 \degree-x

\angle CDA=180 \degree-90 \degree-(90 \degree-x)=x

In \triangle ABC\, and\, \triangle ACD,

\angle CBA\, =\angle CAD

\angle CAB\, =\angle CDA

\angle ACB\, =\angle DCA ( Each right angle)

\triangle ABC\, \sim \triangle ,ACD (By AAA)

\frac{AC}{DC}=\frac{BC}{AC}

\Rightarrow AC^2=BC\times DC

Hence proved

Q4 ABC is an isosceles triangle right angled at C. Prove that AB^2 = 2AC ^2

Answer:

1635933142785

Given: ABC is an isosceles triangle right angled at C.

Let AC=BC

In \triangle ABC,

By Pythagoras theorem

AB^2=AC^2+BC^2

AB^2=AC^2+AC^2 (AC=BC)

AB^2=2.AC^2

Hence proved.

Q5 ABC is an isosceles triangle with AC = BC. If AB ^ 2 = 2 AC ^ 2 , prove that ABC is a right triangle.

Answer:

1635933162430

Given: ABC is an isosceles triangle with AC=BC.

In \triangle ABC,

AB^2=2.AC^2 (Given )

AB^2=AC^2+AC^2 (AC=BC)

AB^2=AC^2+BC^2

These sides satisfy Pythagoras theorem so ABC is a right-angled triangle.

Hence proved.

Q6 ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer:

Given: ABC is an equilateral triangle of side 2a.

1635933173577

AB=BC=AC=2a

AD is perpendicular to BC.

We know that the altitude of an equilateral triangle bisects the opposite side.

So, BD=CD=a

In \triangle ADB,

By Pythagoras theorem,

AB^2=AD^2+BD^2

\Rightarrow (2a)^2=AD^2+a^2

\Rightarrow 4a^2=AD^2+a^2

\Rightarrow 4a^2-a^2=AD^2

\Rightarrow 3a^2=AD^2

\Rightarrow AD=\sqrt{3}a

The length of each altitude is \sqrt{3}a .

Q7 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer:

1635933187293

In \triangle AOB, by Pythagoras theorem,

AB^2=AO^2+BO^2..................1

In \triangle BOC, by Pythagoras theorem,

BC^2=BO^2+CO^2..................2

In \triangle COD, by Pythagoras theorem,

CD^2=CO^2+DO^2..................3

In \triangle AOD, by Pythagoras theorem,

AD^2=AO^2+DO^2..................4

Adding equation 1,2,3,4,we get

AB^2+BC^2+CD^2+AD^2=AO^2+BO^2+BO^2+CO^2+CO^2+DO^2+AO^2+DO^2

AB^2+BC^2+CD^2+AD^2=2(AO^2+BO^2+CO^2+DO^2)

\Rightarrow AB^2+BC^2+CD^2+AD^2=2(2.AO^2+2.BO^2) (AO=CO and BO=DO)

\Rightarrow AB^2+BC^2+CD^2+AD^2=4(AO^2+BO^2)

\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC}{2})^2+(\frac{BD}{2})^2)

\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC^2}{4})+(\frac{BD^2}{4}))

\Rightarrow AB^2+BC^2+CD^2+AD^2=AC^2+BD^2

Hence proved .

Q8 (1) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD \perp BC, OE \perp AC and OF \perp AB. Show that OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2,

1635933201520

Answer:


1635933215218

Join AO, BO, CO

In \triangle AOF, by Pythagoras theorem,

OA^2=OF^2+AF^2..................1

In \triangle BOD, by Pythagoras theorem,

OB^2=OD^2+BD^2..................2

In \triangle COE, by Pythagoras theorem,

OC^2=OE^2+EC^2..................3

Adding equation 1,2,3,we get

OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2 \Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4

Hence proved

Q8 (2) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD \perp BC, OE \perp AC and OF \perp AB. AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2.

1635933229014

Answer:

1635933270968

Join AO, BO, CO

In \triangle AOF, by Pythagoras theorem,

OA^2=OF^2+AF^2..................1

In \triangle BOD, by Pythagoras theorem,

OB^2=OD^2+BD^2..................2

In \triangle COE, by Pythagoras theorem,

OC^2=OE^2+EC^2..................3

Adding equation 1,2,3,we get

OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2 \Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4

\Rightarrow (OA^2-OE^2)+(OC^2-OD^2)+(OB^2-OF^2)=AF^2+BD^2+EC^2 \Rightarrow AE^2+CD^2+BF^2=AF^2+BD^2+EC^2

Q9 A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Answer:

1635933282458

OA is a wall and AB is a ladder.

In \triangle AOB, by Pythagoras theorem

AB^2=AO^2+BO^2

\Rightarrow 10^2=8^2+BO^2

\Rightarrow 100=64+BO^2

\Rightarrow 100-64=BO^2

\Rightarrow 36=BO^2

\Rightarrow BO=6 m

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

Q11 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 1/2 hours?

Answer:

1635933311960

Distance travelled by the first aeroplane due north in 1\frac{1}{2} hours.

=1000\times \frac{3}{2}=1500 km

Distance travelled by second aeroplane due west in 1\frac{1}{2} hours.

=1200\times \frac{3}{2}=1800 km

OA and OB are the distance travelled.

By Pythagoras theorem,

AB^2=OA^2+OB^2

\Rightarrow AB^2=1500^2+1800^2

\Rightarrow AB^2=2250000+3240000

\Rightarrow AB^2=5490000

\Rightarrow AB^2=300\sqrt{61}km

Thus, the distance between the two planes is 300\sqrt{61}km .

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 top

Answer:

1635933349053

Let AB and CD be poles of heights 6 m and 11 m respectively.

CP=11-6=5 m and AP= 12 m

In \triangle APC,

By Pythagoras theorem,

AP^2+PC^2=AC^2

\Rightarrow 12^2+5^2=AC^2

\Rightarrow 144+25=AC^2

\Rightarrow 169=AC^2

\Rightarrow AC=13m

Hence, the distance between the tops of two poles 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 AE^2 + BD^2 = AB^2 + DE^2.

Answer:

1635933368584

In \triangle ACE, by Pythagoras theorem,

AE^2=AC^2+CE^2..................1

In \triangle BCD, by Pythagoras theorem,

DB^2=BC^2+CD^2..................2

From 1 and 2, we get

AC^2+CE^2+BC^2+CD^2=AE^2+DB^2..................3

In \triangle CDE, by Pythagoras theorem,

DE^2=CD^2+CE^2..................4

In \triangle ABC, by Pythagoras theorem,

AB^2=AC^2+CB^2..................5

From 3,4,5 we get

DE^2+AB^2=AE^2+DB^2

Q14 The perpendicular from A on side BC of a \Delta ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that 2 AB^2 = 2 AC^2 + BC^2.

1635933387695

Answer:

In \triangle ACD, by Pythagoras theorem,

AC^2=AD^2+DC^2

AC^2-DC^2=AD^2..................1

In \triangle ABD, by Pythagoras theorem,

AB^2=AD^2+BD^2

AB^2-BD^2=AD^2.................2

From 1 and 2, we get

AC^2-CD^2=AB^2-DB^2..................3

Given : 3DC=DB, so

CD=\frac{BC}{4}\, \, and\, \, BD=\frac{3BC}{4}........................4

From 3 and 4, we get

AC^2-(\frac{BC}{4})^2=AB^2-(\frac{3BC}{4})^2

AC^2-(\frac{BC^2}{16})=AB^2-(\frac{9BC^2}{16})

16AC^2-BC^2=16AB^2- 9BC^2

16AC^2=16AB^2- 8BC^2

\Rightarrow 2AC^2=2AB^2- BC^2

2 AB^2 = 2 AC^2 + BC^2.

Hence proved.

Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9 AD^2 = 7 AB^2

Answer:

1635933424958

Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.

To prove : 9 AD^2 = 7 AB^2

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, BE=CE=\frac{a}{2}

In \triangle AEB, by Pythagoras theorem

AB^2=AE^2+BE^2

a^2=AE^2+(\frac{a}{2})^2

\Rightarrow a^2-(\frac{a^2}{4})=AE^2

\Rightarrow (\frac{3a^2}{4})=AE^2

\Rightarrow AE=(\frac{\sqrt{3}a}{2})

Given : BD = 1/3 BC.

BD=\frac{a}{3}

DE=BE=BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}

In \triangle ADE, by Pythagoras theorem,

AD^2=AE^2+DE^2

\Rightarrow AD^2=(\frac{\sqrt{3}a}{2})^2+(\frac{a}{6})^2

\Rightarrow AD^2=(\frac{3a^2}{4})+(\frac{a^2}{36})

\Rightarrow AD^2=(\frac{7a^2}{9})

\Rightarrow AD^2=(\frac{7AB^2}{9})

\Rightarrow 9AD^2=7AB^2

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.

Answer:

1635933437532

Given: An equilateral triangle ABC.

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, BE=CE=\frac{a}{2}

In \triangle AEB, by Pythagoras theorem

AB^2=AE^2+BE^2

a^2=AE^2+(\frac{a}{2})^2

\Rightarrow a^2-(\frac{a^2}{4})=AE^2

\Rightarrow (\frac{3a^2}{4})=AE^2

\Rightarrow 3a^2=4AE^2

\Rightarrow 4.(altitude)^2=3.(side)^2

Q17 Tick the correct answer and justify : In \Delta ABC AB = 6 \sqrt 3 cm, AC = 12 cm and BC = 6 cm.
The angle B is :
(A) 120°

(B) 60°

(C) 90°

(D) 45°

Answer:

In \Delta ABC AB = 6 \sqrt 3 cm, AC = 12 cm and BC = 6 cm.

AB^2+BC^2=108+36

=144

=12^2

=AC^2

It satisfies the Pythagoras theorem.

Hence, ABC is a right-angled triangle and right-angled at B.

Option C is correct.

More About NCERT Solutions for Class 10 Maths Exercise 6.5

Most of the questions are based on right-angled triangles and the concept of similarity of triangles, but we should be also aware of some basic geometry which is used to solve problems of NCERT solutions for Class 10 Maths exercise 6.5. Pythagoras theorem is most important to solve questions of Class 10 Maths chapter 6 exercise 6.5. We should understand the properties of the trapezium and the criterion of similarity of triangles. Questions given in the exercise 6.5 Class 10 Maths are very much important for the board exam. Students can also access Triangles Class 10 Notes here and use them for quickly revision of the concepts related to Triangles.

Benefits of NCERT Solutions for Class 10 Maths Exercise 6.5

  • Class 10 Maths chapter 6 exercise 6.5 is used to find the altitude of the triangle.
  • NCERT Class 10 Maths chapter 6 exercise 6.5 is used to find the unknown side of a right-angle triangle.
  • Exercise 6.5 Class 10 Maths, is founded on irrational numbers and the Fundamental Theorem of Arithmetic, both of which are key concepts in the chapter.

Also, See

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions


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CBSE Class 10 compartment result 2024 declared; 48.83% pass
August 05, 2024 - 03:14 PM IST
CBSE Class 10 compartment result 2024 soon; passing marks, websites to check
August 05, 2024 - 02:44 PM IST
CBSE Class 10 compartment result 2024 expected soon at cbseresults.nic.in; grading system, passing marks
August 04, 2024 - 04:20 PM IST
CBSE Class 10 compartment result 2024 soon at cbseresults.nic.in; websites to check
August 03, 2024 - 04:38 PM IST
CBSE 10th, 12th revaluation lot 2 result 2024 out at cbseresults.nic.in; passing marks
July 30, 2024 - 06:39 PM IST
CBSE: No open book exams for Classes 10, 12 this year; pilot only in a few schools
June 12, 2024 - 10:52 AM IST
CBSE Class 10 result 2024 revaluation application begins at cbse.gov.in; apply by May 24
May 21, 2024 - 02:20 PM IST

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