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.

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.

## Assess NCERT Solutions for Class 10 Maths chapter 6 exercise 6.5

Triangles Class 10 Chapter 6 Exercise: 6.5

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.

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.

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.

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.

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.

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 .

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

In $\triangle ACD\, and\, \triangle ABD,$

$\angle DCA\, =\angle DAB\, \, \, \, \, \, \, \, (Each 90 \degree)$

$\angle CDA\, =\angle ADB$ (common )

$\triangle ACD\, \sim \triangle ABD$ (By AA)

$\Rightarrow \frac{CD}{AD}=\frac{AD}{BD}$

$\Rightarrow AD^2=BD\times CD$

Hence proved.

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.

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.

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

AB=BC=AC=2a

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 $\sqrt{3}a$ .

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,

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

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

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

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

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,

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

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

Hence proved

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,

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

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

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

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.

OB is a pole.

In $\triangle$ AOB, by Pythagoras theorem

$AB^2=AO^2+BO^2$

$\Rightarrow 24^2=18^2+AO^2$

$\Rightarrow 576=324+AO^2$

$\Rightarrow 576-324=AO^2$

$\Rightarrow 252=AO^2$

$\Rightarrow AO=6\sqrt{7} m$

Hence, the distance of the stack from the base of the pole is $6\sqrt{7}$ m.

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 $300\sqrt{61}km$ .

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.

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

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

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.

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,

$AD^2=AE^2+DE^2$

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

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

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

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

$\Rightarrow 9AD^2=7AB^2$

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

$AB^2=AE^2+BE^2$

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

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

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

$\Rightarrow 3a^2=4AE^2$

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

(B) 60°

(C) 90°

(D) 45°

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.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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