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

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.

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

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

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.

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

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

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.

∠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.

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.

Solution:

Join AO, BO and CO.

(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.

∴ 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.

∴ 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.

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.

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:

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.

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 :

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.

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

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.

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.

So, 9AD² = 7 AB², which is the required result.

Aliter:

DE = BE – BD

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 :

Hence, the result.

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 :

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.