RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

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 \(\frac{\mathrm{QS}}{\mathrm{SR}}\) = \(\frac{\mathrm{PQ}}{\mathrm{PR}}\).

Solution:
Given : A △PQR in which PS is the bisector of QPR, i.e., ∠1 = ∠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

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)]

⇒ ∠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)

Question 3.
In Fig., ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC . BD.

Solution:
Given : In △ABC, AD ⊥ BC when BC is produced.
∠ABC > 90°

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.

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)

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

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)

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.

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

Solution:
Given : Two chords AB and CD of a circle intersect each other at the point P.

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)

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.

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

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.

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

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)

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?

Solution:

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.