Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 Textbook Questions and Answers.

BSEB Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Bihar Board Class 10th Maths Solutions 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) Let a = 7 cm, b = 24 cm and c = 25 cm.
Here the largest side is c = 25 cm.
We have : a² + b² = 7² + 24² = 49 + 576 = 625 = c²
So, the triangle with the given sides is a right triangle. Its hypotenuse is 25 cm.

(ii) Let a – 3 cm, 6 = 8 cm and c = 6 cm.
Here, the largest side is 6= 8 cm.
We have : a² + c² = 3² + 6² = 9 + 36 = 45 ≠ b²
So, the triangle with the given sides is not a right triangle.

(iii) Let a = 50 cm, b – 80 cm and c = 100 cm.
Here, the largest side is c = 100 cm.
We have : a² + b² = 50² + 80² = 2500 + 6400
= 8900 ≠ c²
So, the triangle with the given sides is not a right triangle.

(iv) Let a = 13 cm, b – 12 cm and c = 5 cm.
Here, the largest side is a = 13 cm.
We have : b² + c² = 12² + 5² = 144 + 25 = 169 = a²
So, the triangle with the given sides is a right triangle. Its hypotenuse is 13 cm.

Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

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:
Given : PQR is a triangle right angled at P and PM ⊥ QR.
To Prove : PM² = QM.MR
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 1
Proof: Since PM ⊥ QR, therefore
∆ PQM ~ ∆ RPM
or \(\frac { PM }{ QM }\) = \(\frac { MR }{ PM }\)
So, PM² = QM.MR.

Question 3.
In the figure, ABD is triangle right angled at and AC ⊥ BD. Show that
(i) Ab² = BC.BD
(ii) Ac² = BC.DC
(iii) AD² = BD.CD
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 2
Solution:
Given : ABD is a triangle right angled at A and AC ⊥ BD.
To Prove :
(i) Ab² = BC.BD
(ii) Ac² = BC.DC
(iii) AD² = BD.CD
Proof:
(i) Since AC ⊥ BD, therefore
∆ ABC – ∆ DAC and each triangle is similar to the whole ∆ ABD.
∵ ∆ ABC – ∆ DBA
∴ \(\frac { AB }{ DB }\) = \(\frac { BC }{ AB }\) or AB² = BC.BD AB

(ii) Since ∆ ABC ~ ∆ DAC, therefore
\(\frac { AC }{ BC }\) = \(\frac { DC }{ AC }\) or Ac² = BC.DC AC

(iii) Since ∆ ACD ~ ∆ BAD, therefore
∴ \(\frac { AD }{ CD }\) = \(\frac { BD }{ AD }\) or AD² = BD. CD.

Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 4.
ABC is an isosceles triangle right angled at C. Prove that Ab² = 2Ac².
Solution:
Since ABC is an isosceles A right triangle, right angled at C, therefore
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 3
Ab² = Ac² + Bc²
or Ab² = Ac² + Ac² [∵ BC = AC, given]
So, Ab² = 2 Ac²

Question 5.
ABC is an isosceles triangle with AC = BC. If Ab² = 2Ac², prove that ABC is a right triangle.
Solution:
Since ABC is an isosceles triangle with AC = BC and Ab² = 2Ac², therefore
Ab² = Ac² + Ac²
or Ab² = Ac² + Bc² [∵ AC = BC, given]
∴ ∆ ABC is right angled at C.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Let ABC be an equilateral triangle of side 2a units. Draw AD ⊥ BC. Then, D is the mid-point of BC
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 4
So, BD = \(\frac { 1 }{ 2 }\) BC
= \(\frac { 1 }{ 2 }\) x 2a = a
Since ABD is a right triangle, right angled at D, therefore
Ab² = AD² + BD²
or (2a)² = AD² + (a)²
or AD² = 4a² – a² = 3a² or AD = \(\sqrt{3}\)a
∴ Each of its altitudes is V3a.

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum o diagonals.
Solution:
Let the diagonals AC and BD of a rhombus ABCD intersect each other at O. Since the diagonals of a rhombus bisect each other at right angles.
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 5
∠AOB = ∠BOC – ∠COD
= ∠DOA = 90°
and AO = CO, BO = OD
Since AOB is a right triangle right
AB² = Oa² + Ob²
or AB² = (\(\frac { 1 }{ 2 }\)AC)² +(\(\frac { 1 }{ 2 }\)BD)² [∵ OA = OC and OB = OD]
or 4AB² = Ac² + BD² … (1)
Similarly, we have :
4BC² = Ac² + BD² …(2)
4CD² = Ac² + BD² …(3)
and 4AD² = Ac² + BD²
Adding all these results, we get
4(AB² + BC² + CD² + AD)² = 4(AC² + BD²)
or AB² + BC² + CD² + DA² = AC² + BD².

Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 8.
In the 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 B = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF².
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 6
Solution:
Join AO, BO and CO.
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 7
(i) In right As OFA, ODB and OEC, we have :
OA² = AF² + OF²
Ob² = BD² + OD²
and OC² = CE² + OE²
Adding all these, we get
OA² + OB² + OC² = AF² + BD² + CE² + OF² + OD² + OE² or OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE².

(ii) In-right As ODB and ODC, we have :
OB² = OD² + BD²
and OC² = OD² + CD²
or OB² – OC² – BD² – CD² … (1)
Similarly, we have :
OC² – OA² = CE² – AE² … (2)
and OA² – OB² = AF² – BF² … (3)
Adding equations (1), (2) and (3), we get
(OB² – OC²) + (OC² – OA²) + (OA² – OB²)
= (BD² – CD²) + (CE² – AE²) + (AF² – BF²) or (BD² + CE² + AF²) – (AE² + CD² + BF²) = 0
or AF² + BD² + CE²
= AE² + BF2 + CD².

Aliter for (ii) :
Taking triangles OFB, ODC and OAE, we have :
OA² + OB² + OC²- OD²- OE² – OF² = AE² + BF² + CD²
So, using result (i) above, we have :
AF² + BD² + CE² = AE² + BF² + CD²

Question 9.
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.
Solution:
Let AB be the ladder, B be the B window and CB be the wall. Then, ABC is a right triangle, right angled at C.
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 8
∴ AB² = AC² + Bc²
So, 10² = AC² + 8²
or AC² = 100 – 64
or AC²= 36
or AC = 6 .A C
Hence, the foot of the ladder is at a distance of 6 m from the base of the wall.

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:
Let AB (= 24 m) be a guy wire attached to a vertical pole BC of height 18 m. To keep the wire taut, let it be fixed to a stake at A. Then, ABC is a right triangle, right angled at C.
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 9
∴ AB² = AC²+ Bc²
So, 24² = AC²+ 182
or AC² = 576 – 324
or AC² = 552
or AC = \(\sqrt{252}\) = 6\(\sqrt{7}\)
Hence, the stake may be placed at a distance of 6\(\sqrt{7}\) m from the base of the pole.

Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

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:
Let the first aeroplane starts from O and goes upto A towards north, where OA = (1000 x \(\frac { 3 }{ 2 }\)) km = 1500 km.
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 10
Let the second aeroplane starts from O at the same time and goes upto B towards west, where OB = (1200 x \(\frac { 3 }{ 2 }\)) km = 1800 km.
According to the problem, the required distance = BA.
∆In right angled ∆ ABC, by Pythagoras theorem, we have :
AB² = OA² + OB²
= (1500)² + (1800)² = 2250000 + 3240000
= 5490000 = 9 x 61 x 100 x 100
or AB = 3 x 100 \(\sqrt{61}\) km = 300 \(\sqrt{61}\) km.

Question 12.
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 tops.
Solution:
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 11
Let AB = 11 m and CD = 6 m be the two poles such that BD = 12 m.
Draw CE ⊥ AB and join AC.
∴CE = DB = 12 m,
AE = AB – BE
= AB – CD
= (11 – 6) m = 5 m.
In right angled ∆ ACE, by Pythagoras theorem, we have :
AC² = CE² + AE²
= (12)² + (5)² = 144 + 25 = 169
So, AC = \(\sqrt{169}\) = 13
Hence, the distance between the tops of the two poles is 13 m.

Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

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:
In right angled ∆s ACE and DCB, we have :
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 12
AE² = AC²+ CE²
and BD² = DC²+ Bc²
So, AE² + BD² = (AC²+ CE²) + (DC²+ Bc²)
AE² + BD² = (AC²+ Bc²) + (DC²+ CE²)
AE² + BD² = AB² + DE² [By Pythagoras theorem, AC²+ BC²= AB² and DC²+ CE² = DE²]

Question 14.
The perpendicular from A on side BC of a ∆ ABC intersects BC at D such that DB = 3 CD (see figure). Prove that 2AB² = 2AC²+ BC².
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 13
Solution:
We have : DB = 3CD
Now, BC = DB + CD
i.e., BC = 3CD + CD [∵ BD = 3CD]
or BC = 4CD
∴ CD = \(\frac { 1 }{ 4 }\) BC and DB = 3CD = \(\frac { 3 }{ 4 }\) BC … (1)
Since ∆ ABD is a right triangle, right angled at D, therefore by Pythagoras theorem, we have :
AB² = AB² + DB²
Similarly, from ∆ACD, we have :
AC² = AD² + CD²
(1) – (2) gives
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 14

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:
Let ABC be an equilateral triangle and let D be a point on BC such that
BD = \(\frac { 1 }{ 3 }\)BC.
Draw AE ⊥ BC. Join AD.
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 15
In ∆s AEB and AEC, we have :
AB = AC [∴∆ABC is equilateral]
∠AEB = ∠AEC [∵ Each = 90°]
and AE = AE
∴ By SAS criterion of congruence, we have :
∆ AEB = ∆ AEC
So, BE = EC
Now, we have :
BD = \(\frac { 1 }{ 3 }\)BC, DC = \(\frac { 2 }{ 3 }\)BC
BE = EC = \(\frac { 1 }{ 2 }\)BC … (1)
Since ∠C = 60°, therefore triangle.
∆ ADC is an acute triangle.
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 16
So, 9AD² = 7 AB², which is the required result.
Aliter:
DE = BE – BD
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 17

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:
Let ABC be an equilateral triangle and let AD ⊥ BC. In ∆ ADB and ∆ ADC, we have :
AB = AC [Given]
AD = AD [Common]
and ∠ADB = ∠ADC [Each = 90°]
By RHS criterion of congruence, we have :
∆ ADB ≅ ∆ ADC
So, BD = DC or BD = DC = \(\frac { 1 }{ 2 }\)BC … (1)
Since ∆ ADB is a right triangle, right angled at D, by Pythagoras theorem, we have :
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 18
Hence, the result.

Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 17.
Tick the correct answer and justify :
In ∆ 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°
Solution:
In ∆ ABC, we have :
Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 19
AB = 6\(\sqrt{3}\) cm, AC = 12 cm
and BC = 6 cm … (1)
Now, AB² + BC²= (6\(\sqrt{3}\))² + (6)² [From (1)]
= 36 x 3 + 36
= 108 + 36 = 144
= (AC)²
∴ Thus, ∆ ABC is a right triangle, right angled at B.
∠B = 90°
∴ (C) is the correct choice.