• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • RBSE Model Papers
    • RBSE Class 12th Board Model Papers 2022
    • RBSE Class 10th Board Model Papers 2022
    • RBSE Class 8th Board Model Papers 2022
    • RBSE Class 5th Board Model Papers 2022
  • RBSE Books
  • RBSE Solutions for Class 10
    • RBSE Solutions for Class 10 Maths
    • RBSE Solutions for Class 10 Science
    • RBSE Solutions for Class 10 Social Science
    • RBSE Solutions for Class 10 English First Flight & Footprints without Feet
    • RBSE Solutions for Class 10 Hindi
    • RBSE Solutions for Class 10 Sanskrit
    • RBSE Solutions for Class 10 Rajasthan Adhyayan
    • RBSE Solutions for Class 10 Physical Education
  • RBSE Solutions for Class 9
    • RBSE Solutions for Class 9 Maths
    • RBSE Solutions for Class 9 Science
    • RBSE Solutions for Class 9 Social Science
    • RBSE Solutions for Class 9 English
    • RBSE Solutions for Class 9 Hindi
    • RBSE Solutions for Class 9 Sanskrit
    • RBSE Solutions for Class 9 Rajasthan Adhyayan
    • RBSE Solutions for Class 9 Physical Education
    • RBSE Solutions for Class 9 Information Technology
  • RBSE Solutions for Class 8
    • RBSE Solutions for Class 8 Maths
    • RBSE Solutions for Class 8 Science
    • RBSE Solutions for Class 8 Social Science
    • RBSE Solutions for Class 8 English
    • RBSE Solutions for Class 8 Hindi
    • RBSE Solutions for Class 8 Sanskrit
    • RBSE Solutions

RBSE Solutions

Rajasthan Board Textbook Solutions for Class 5, 6, 7, 8, 9, 10, 11 and 12

  • RBSE Solutions for Class 7
    • RBSE Solutions for Class 7 Maths
    • RBSE Solutions for Class 7 Science
    • RBSE Solutions for Class 7 Social Science
    • RBSE Solutions for Class 7 English
    • RBSE Solutions for Class 7 Hindi
    • RBSE Solutions for Class 7 Sanskrit
  • RBSE Solutions for Class 6
    • RBSE Solutions for Class 6 Maths
    • RBSE Solutions for Class 6 Science
    • RBSE Solutions for Class 6 Social Science
    • RBSE Solutions for Class 6 English
    • RBSE Solutions for Class 6 Hindi
    • RBSE Solutions for Class 6 Sanskrit
  • RBSE Solutions for Class 5
    • RBSE Solutions for Class 5 Maths
    • RBSE Solutions for Class 5 Environmental Studies
    • RBSE Solutions for Class 5 English
    • RBSE Solutions for Class 5 Hindi
  • RBSE Solutions Class 12
    • RBSE Solutions for Class 12 Maths
    • RBSE Solutions for Class 12 Physics
    • RBSE Solutions for Class 12 Chemistry
    • RBSE Solutions for Class 12 Biology
    • RBSE Solutions for Class 12 English
    • RBSE Solutions for Class 12 Hindi
    • RBSE Solutions for Class 12 Sanskrit
  • RBSE Class 11

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

April 18, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.6

Question 1.
In Fig., PS is the Wsector of ∠QPR of △PQR. Prove that \(\frac{\mathrm{QS}}{\mathrm{SR}}\) = \(\frac{\mathrm{PQ}}{\mathrm{PR}}\).
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 1
Solution:
Given : A △PQR in which PS is the bisector of QPR, i.e., ∠1 = ∠2
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 2
Construction : Through R draw a line parallel to PS which meets QP produced at T.
Proof: In △QRT,
PS || TR
∴ ∠2 = ∠3 (Alternate angles)
and ∠1 =∠4 (Corresponding angles)
But, ∠1 = ∠2 (Given)
∴ ∠3 = ∠4
In △PRT, ∠3 = ∠4 (Proved above)
∴ PT = PR (v Angles opposite to equal side are equal.)
In △QRT,
PS || TR
∴ \(\frac{\mathrm{QP}}{\mathrm{PT}}\) = \(\frac{\mathrm{QS}}{\mathrm{SR}}\)
(By Basic Proportionality Theorem)
\(\frac{\mathrm{QP}}{\mathrm{PR}}\) = \(\frac{\mathrm{QS}}{\mathrm{SR}}\) (∵ PT = PR)
or \(\frac{\mathrm{PQ}}{\mathrm{PR}}\) = \(\frac{\mathrm{QS}}{\mathrm{SR}}\) (Hence Proved)

Question 2.
In Fig., D is a point on hypotenuse AC of △ABC, DM ⊥ BC and DN ⊥ AB. Prove that:
(i) DM2 = DN. MC
(ii) DN2 = DM.AN
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 3
Solution:
Given : In right △ABC, DM ⊥ BC, DN ⊥ AB
To prove : (i) DM2 = DN . MC
(ii) DN2 = DM . AN
Proof: BD ⊥ AC (Given)
⇒ ∠BDC = 90°
⇒ ∠BDM + ∠MDC = 90° ..(i)
In △DMC, ∠DMC = 90°
[∵ DM ⊥ BC (Given)]
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 4
⇒ ∠C + ∠MDC = 90° ….(ii)
From (i) and (ii),
∠BDM + ∠MDC = ∠C + ∠MDC
⇒ ∠BDM = ∠C
Now in △BMD and △MDC,
∠BDM = ∠C (Proved above)
∠BDM = ∠MDC (Each 90°)

(i) ∴ △BMD ~ △MDC
(By AA similarity criterion)
⇒ \(\frac{\mathrm{DM}}{\mathrm{BM}}\) = \(\frac{\mathrm{MC}}{\mathrm{DM}}\) or DM2 = BM × MC
or DM2 = DN.MC (∵ BM = DM) (Hence proved)

(ii) Similarly,
△NDA ~ △NBD
⇒ \(\frac{\mathrm{DN}}{\mathrm{BN}}\) = \(\frac{\mathrm{AN}}{\mathrm{DN}}\) or DN2 = BN × AN
or DN2 = DM.AN (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

Question 3.
In Fig., ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC . BD.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 5
Solution:
Given : In △ABC, AD ⊥ BC when BC is produced.
∠ABC > 90°
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 6
To prove : AC2 = AB2 + BC2 + 2BC . BD
Let BC = a, CA = b, AB = c, AD = h
and BD = x
In right △ADC,
By Pythagoras Theorem
AC2 = CD2 + AD2
or b2 = (x + a)2 + h2 ….(i)
In right △ADB,
AB2 = DB2 + AD2
⇒ c2 = x2 + h2 ….(ii)
Subtracting (ii) from equation (i)
b2 – c2 = (x + a)2 + h2 – x2 – h2
⇒ b2 – c2 = x2 + 2ax + a2 + h2 – x2 – h2
⇒ b2 – c2 = a2 + 2ax
⇒ b2 = c2 + a2 + 2ax
So, AC2 = AB2 + BC2 + 2BC . BD
(Hence Proved)

4. In Fig., ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC . BD.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 7
Solution:
Given : In △ABC in which ∠ABC <90° and AD ⊥ BC
To prove : AC2 = AB2 + BC2 – 2BC . BD
Proof : AADC is a right triangle in which there is right angle at D.
∴ AC2 = CD2 + DA2
(By Pythagoras Theorem) ….(i)
Also △ADB is a right △ in which there is a right angle at D.
AB2 = AD2 + DB2
(By Pythagoras Theorem) ….(ii)
From (i) AC2 = AD2 + (CB – BD)2
= AD2 + CB2 + BD2 – 2CB × BD
or AC2 = (BD2 + AD2) + CB22 – 2CB × BD
AC2 – AB2 + BC2 – 2BC × BD
[Using (ii)] (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

Question 5.
In Fig., AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) AC2 = AD2 + BC . DM + \(\left(\frac{B C}{2}\right)^{2}\)
(ii) AB2 = AD2 – BC . DM + \(\left(\frac{B C}{2}\right)^{2}\)
(iii) AC2 + AB2 = 2AD2 + \(\frac{1}{2}\)BC2
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 8
Solution:
Given : In △ABC, AM ⊥ BC, AD is a median of △ABC.
To prove:
(i) AC2 = AD2 + BC DM + \(\left(\frac{B C}{2}\right)^{2}\)
(ii) AB2 = AD2 – BC . DM + \(\left(\frac{B C}{2}\right)^{2}\)
(iii) AC2 + AB2 = 2AD2 + \(\frac{1}{2}\)BC2
Proof: (i) In △AMC,
AC2 = AM2 + MC2 = AM2 + (MD + DC)2
AC2 = AM2 + MD2 + DC2 + 2MD × DC
AC2 = (AM2 + MD2) + \(\left(\frac{B C}{2}\right)^{2}\) + 2.MD \(\left(\frac{\mathrm{BC}}{2}\right)\) (∵ AD is a median)
AC2 = AD2 + BC × MD + \(\left(\frac{B C}{2}\right)^{2}\)
∴ AC2 = AD2 + BC . MD + \(\left(\frac{B C}{2}\right)^{2}\) …..(i)
(Hence Proved)

(ii) In right triangle AMB,
AB2 = AM2 + BM2 = AM2 + (BD – MD)2
= AM2 + BD2 + MD2 – 2BD × MD
= (AM2 + MD2) + BD2 – 2 (\(\frac{1}{2}\)BC) MD
=AD2+ (\(\frac{1}{2}\)BC)2 – BC.MD
AB2 = AD2 – BC.MD + \(\left(\frac{B C}{2}\right)^{2}\) ….(ii)
(Hence Proved)

(iii) Adding equations (i) and (ii),
AB2 + AC2 = AD2 + BC . MD + \(\left(\frac{B C}{2}\right)^{2}\) + AD2 – BC.MD + \(\left(\frac{B C}{2}\right)^{2}\)
= 2D2 + \(\frac{\mathrm{BC}^{2}}{4}\) + \(\frac{\mathrm{BC}^{2}}{4}\)
= 2AD2 = 2\(\frac{\mathrm{BC}^{2}}{4}\)
AB2 + AC2 = 2AD2 + \(\frac{1}{2}\)BC2 (HenceProved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

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.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 9
Solution:
Given : Let ABCD be a parallelogram in which the diagonals AC and BD intersect each other at the point 0.
To prove : AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Proof : The diagonals of a parallelogram bisect each other.
i.e., OB and OD are respectively the medians of △ABC and △ADC.
∴ AB2 + BC2 = 2BO2 + \(\frac{1}{2}\)AC2 ….(i)
and AD2 + CD2 = 2DO2 + \(\frac{1}{2}\)AC2 ….(ii)
Adding (i) and (ii),
AB2 + BC2 + AD2 + CD2
= 2 (BO2 + DO2) + \(\frac{1}{2}\)(AC2 + AC2)
AB2 + BC2 + AD2 + CD2
= 2(\(\frac{1}{4}\)BD2 + \(\frac{1}{4}\)BD2] + AC2
[∵ BO = \(\frac{1}{2}\)BD and OD = \(\frac{1}{2}\)BD]
AB2 + BC2 = 2 × \(\frac{1}{2}\)BD2 + AC2
or AB2 + BC2 + AD2 + CD2 = BD2 + AC2
i.e., the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its side. (Hence Proved)

Question 7.
In Fig., two chords AB and CD intersect each other at the point P. Prove that:
(i) △APC ~ △DPB
(ii) AP. PB = CP. DP
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 10
Solution:
Given : Two chords AB and CD of a circle intersect each other at the point P.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 11
To prove : (i) △APC ~ △DPB
(ii) AP . PB = CP . DP
Proof: (i) In △APC and △DPB,
∠1 = ∠2 (Vertically opposite angles)
∠3 = ∠4 (Angles of the same segment)
∴ △APC ~ △DPB [By AA similarity criterion] (Hence Proved)

(ii) △APC ~ △DPB (Proved above)
So \(\frac{\mathrm{AP}}{\mathrm{DP}}\) = \(\frac{\mathrm{PC}}{\mathrm{PB}}\)
(If two triangles are similar then their corresponding sides are proportional.)
⇒ AP . PB = PC . DP (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

Question 8.
In Fig., two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) △PAC ~ △PDB
(ii) PA. PB = PC . PD.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 12
Solution:
Given : Two chords AB and CD of a circle intersect each other at P when produced.
To prove : (i) △PAC ~ △PDB
(ii) PA. PB = PC . PD
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 13
Proof: (i) In △PAC and △PDB,
∠P = ∠P (Common angle)
∠PAC = ∠PDB
(∵ Exterior angle of a cyclic quadrilateral is equal to its interior opposite angle)
∴ △PAC ~ △PDB
(By AA similarity criterion) (Hence Proved)

(ii) △PAC ~ △PDB (Proved above)
∴ \(\frac{\mathrm{PA}}{\mathrm{PD}}\) = \(\frac{\mathrm{PC}}{\mathrm{PB}}\)
(If two triangles are similar then their corresponding sides are proportional.)
∴ PA × PB = PC × PD (Hence Proved)

Question 9.
In Fig., D is a point on side BC of △ABC such that \(\frac{\mathrm{BD}}{\mathrm{CD}}\) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\). Prove that AD is the bisector of ∠BAC.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 14
Solution:
Given : In △ABC, D is a point on side BC of △ABC such that \(\frac{\mathrm{BD}}{\mathrm{CD}}\) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
To prove : AD is the bisector of ∠BAC.
i.e., ∠1 = ∠2
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 15
Construction : Through C draw CE || DA which meets BA produced at E.
Proof: In △BCE,
AD || CE (By Construction)
∴ By Basic Proportionality Theorem,
∴ \(\frac{\mathrm{AB}}{\mathrm{AE}}\) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ AE =AC
In △ACE, AE =AC
⇒ ∠3 = ∠4
(Angles opposite to equal sides)
∵ CE || DA and BAE intersect these, therefore
∠2 = ∠4 (Alternate angles)
and CE || DA and BAE intersects these, therefore
∠1 = ∠3 (Corresponding angles)
Thus, ∠3 = ∠4
⇒ ∠4 = ∠1
∠3 =∠1
But, ∠4 = ∠3
⇒ ∠1 =∠2
Hence, AD bisects △BAC. (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

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.)? If she pulls in the string at the rate of 5 cm per second, what will be the hori∠ontal distance of the fly from her after 12 seconds?
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 16
Solution:
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 17
In right triangle ABC,
AB = 1.8 m, BC = 2.4 m, ∠B = 90°
By Pythagoras Theorem
AC22 = AB2 + BC2
AC2 = (1.8)2 + (2.4)2
AC2 = 3.24 + 5.76 = 9
AC2 = (3)2
AC = 3 m
Now, if Nazima pulls the string at the rate of 5 cm per second, then the length of the string is reduced by
= 5 × 12 = 60 cm
= 0.6 m in 12 seconds
Let the position of the fly after 12 seconds be D.
∴ AD = AC – (distance covered in 12 seconds)
= (3 – 0.6) m = 2.4 m
Now, in right triangle △ABD,
By Pythagoras Theorem,
AD2 = AB2 + BD2
⇒ (2.4)2 = (1.8)2 + BD2
⇒ BD2 = 5.76 – 3.24
⇒ BD2 = 2.52 m ∴ BD = \(\sqrt {2.52}\)
⇒ BD = 1.587 m ≃ 1.59 m
∴ The horizontal distance covered by the Nazima
= BD + 1.2 m
= (1.59+ 1.2) m = 2.79 m
Hence, the length of the string and the distance covered by Nazima are 3 m and 2.79 m respectively.

Share this:

  • Click to share on WhatsApp (Opens in new window)
  • Click to share on Twitter (Opens in new window)
  • Click to share on Facebook (Opens in new window)

Related

Filed Under: Class 10

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Recent Posts

  • RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions
  • RBSE Solutions for Class 11 Psychology in Hindi Medium & English Medium
  • RBSE Solutions for Class 11 Geography in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Hindi
  • RBSE Solutions for Class 3 English Let’s Learn English
  • RBSE Solutions for Class 3 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Maths in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 in Hindi Medium & English Medium
  • RBSE Solutions for Class 4 Hindi
  • RBSE Solutions for Class 4 English Let’s Learn English
  • RBSE Solutions for Class 4 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium

Footer

RBSE Solutions for Class 12
RBSE Solutions for Class 11
RBSE Solutions for Class 10
RBSE Solutions for Class 9
RBSE Solutions for Class 8
RBSE Solutions for Class 7
RBSE Solutions for Class 6
RBSE Solutions for Class 5
RBSE Solutions for Class 12 Maths
RBSE Solutions for Class 11 Maths
RBSE Solutions for Class 10 Maths
RBSE Solutions for Class 9 Maths
RBSE Solutions for Class 8 Maths
RBSE Solutions for Class 7 Maths
RBSE Solutions for Class 6 Maths
RBSE Solutions for Class 5 Maths
RBSE Class 11 Political Science Notes
RBSE Class 11 Geography Notes
RBSE Class 11 History Notes

Copyright © 2023 RBSE Solutions

 

Loading Comments...