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Chapter 6: The Triangles and its Properties

Exercise 6.1

Question 1:

In Δ ABC, X is the midpoint of BC.

AP is _____________

AX is _____________

Is BP = PC?

1

Answer:

Given:

BX = XC

Therefore, AP is altitude.

AX is a median.

No, BP ≠ PC as X is the midpoint of BC

 

Question 2:

Draw a sketch for the following:

(i) In Δ PQR, QX is a median

(ii) In Δ ABC, AB and AC are altitudes of a triangle.

(iii) In ABC, BP is an altitude in the exterior of a triangle.

Answer:

(i) Here, QX is a median in ΔPQR and PX = XR

2

(ii) Here, AB and AC are the altitudes of the ΔABC and CA ^ BA

3

(iii) BP is an altitude in the exterior of ΔABC

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Question 3:

Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same.

Answer:

Isosceles triangle means any two sides are same.

Take ΔPQR and draw the median when PQ= PR

PX is the median and altitude of the given triangle.

5

 

Exercise 6.2

Question 1:

Find the value of the unknown exterior angle a in the following diagrams:

6

Answers:

Since the exterior angle = sum of the interior opposite angles, then

(a) a = 500 + 70 = 1200

(b) a = 650 + 450 = 1100

( c) a = 300 + 400 = 700

(d ) a = 60 + 60= 1200

(E) a = 50 + 50= 1000

(F) a = 60+ 30= 900

 

Question 2:

Find the value of the unknown interior angle a in the following figures:

7

Answers:

Since the exterior angle = sum of the interior opposite angles, then

(a) a + 50o = 115o    =>  a = 115o – 50 o = 65 o

(b) 70 o  + a = 100o  => a = 100o – 70o = 30o

( c) a + 90o = 125o  => a = 120 o – 90 o = 35 o

(d) 60 o  + a = 120 o  => a = 120 o – 60 o  = 60 o

(e) 30 o  + a = 80 o  => a = 80 o +30 o = 50 o

(f) a + 35 o = 75 o   => a = 75 o – 35 o = 40 o

 

Exercise 6.3

Question 1:

Find the value of the unknown a in the following given diagrams.

8

Answers:

(a) In \(\Delta\) PQR

\(\angle\) QPR + \(\angle\) PRQ + \(\angle\) PQR = 180o [ By angle sum property of a triangle]

=> a + 50o + 60o = 180o

=> a + 110o = 180o

=> a = 180o – 110 = 70o

(b) In \(\Delta\) ABC,

\(\angle\) CAB + \(\angle\) ABC + \(\angle\) CAB = 90o [ By angle sum property of a triangle]

=> 90o + 30o + a = 180o

=> a + 120= 180o

=> a = 180o – 120o = 60o

(c ) In \(\Delta\) ABC ,

\(\angle\) CAB + \(\angle\) ABC + \(\angle\) BCA = 180o [ By angle of sum property of a triangle]

=> 30 + 110+ a  = 180o

=> a + 140o = 180o

=> a = 180– 140O = 40O

(d) In the given isosceles triangle,

a + a + 50o = 180o

=> 2a + 50 = 180o

=> 2a = 180o – 50o

=> 2a = 130o

=> \(a = \frac{130^{\circ}}{2}\) = 65o

(e) In the given equilateral triangle,

a + a + a = 180o                                              [ By angle sum property of a triangle]

=> 3 a = 180o

=> \(a = \frac{180^{\circ}}{3}\) = 60o

(f) In the given right angled triangle,

a + 2a + 90o = 180o      [ By the angle sum property of a triangle]

=> 3a + 90o = 180o

=> 3a = 180o – 90o

=> 3a = 90o

=> \(a = \frac{90^{\circ}}{3}\) = 30o

 

Question 2

Find the values a and b in the following diagrams that are given below:

9

Answers:

(a) 50+ a = 120o                                 [Exterior angle property of a \(\Delta\) ]

=> a = 120 o – 50 o = 70 o

Now, 50 o + a + b = 180 o                               [Angle sum property of a \(\Delta\) ]

=> 50 o + 70 o + b = 180 o

=> 120 o + b = 180 o

=> b = 180 o – 120 o = 60 o

(b) b = 80 o – – – – – – – – – – – (1)                         [ Vertically opposite angle]

Now, 50 o + a + b = 180 o                     [ Angle sum property of a \(\Delta\) ]

=> 50 o + 80 o + b = 180 o                      [From equation (1) ]

=> 130 o + b = 180 o

=> b = 180 o – 130 o = 50 o

( c) 50 o + 60 o = a                               [Exterior angle property of a \(\Delta\) ]

=> a = 110 o

Now, 50 o + 60 o + b = 180 o                  [Angle sum property of a \(\Delta\) ]

=> 110 o + b = 180 o

=> b = 180 o – 110 o

=> b = 70 o

(d) a = 60 o   – – – – – – – – – – – – – – – (1)             [ Vertically opposite angle]

Now, 30 o + a + b = 180 o                                 [ Angle sum property of a \(\Delta\) ]

=> 50 o + 60 o + b = 180 o                                  [From equation (1) ]

=> 90 o + b = 180 o

=> b = 180 o – 90 o = 90 o

(e) b = 90 o – – – – – – – – – – – – – (1)                    [ Vertically opposite angle]

Now, b + a + a = 180 o                                                 [ Angle sum property of a \(\Delta\) ]

=> 90 o + 2a = 180                                          [From equation (1) ]

=> 2a = 180 o – 90 o

=> 2a = 90 o

=> \(a=\frac{90^{\circ}}{2}\) = 45o

(f) a = b – – – – – – – – – – – (1)                             [ Vertically opposite angle]

Now, a + a + b = 180 o                                                 [ Angle sum property of a \(\Delta\) ]

=> 2a + a = 180 o                                             [From equation (1) ]

=> 3a = 180 o

=> \(a=\frac{180^{\circ}}{3}\) = 60o

 

Exercise 6.4

Question 1:

Is it possible to have a triangle with the following sides?

(a) 2cm, 3cm 5cm

(b) 3cm, 6cm, 7cm

(c) 6cm, 3cm, 2cm

Answer:

Since the triangle is possible having its sum of the lengths of any two sides would be greater than that length of the third side.

(a) 2cm , 3cm 5cm

2 + 3 > 5          No

2 + 5 > 3          Yes

3 + 5 > 2          Yes

This triangle is not possible.

(b) 3cm, 6cm and 7cm

3 +  6> 7         Yes

6 + 7 > 3          Yes

3 + 7 > 6          Yes

This triangle is possible.

( c) 6cm, 3cm and 2cm

6 + 3 > 2          Yes

6 + 2 > 3          Yes

2 + 3 > 6          No

This triangle is not possible.

 

Question 2:

Take any point P in the interior of a triangle ABC is:

(a) PA + PB > AB?

(b) PB + PC > BC?

(C ) PC + PA > CA?

10

Answers:

Join PC, PB and PA

(a) Is PA + PB > AB?

Yes, APB forms a triangle.

(b) Is PB + PC > BC?

Yes, CBP forms a triangle.

( c) Is PC + PA > CA?

Yes, CPA forms a triangle.

 

Question 3:

PO is a median of a triangle PQR. Is PQ + QR + RP > 2PO? (Consider the sides of a triangle as \(\Delta\) PQO and \(\Delta\) POR )

11

Answer:

Since, the sum of the lengths of any two sides in a triangle should be greater than the length of the third side.

Therefore,

In \(\Delta\) PQO, PQ + QO > PO – – – – – – – – – – (1)

In \(\Delta\) POR, PR + OQ > PO – – – – – – – – – – – (2)

Adding equation (1) and (2) we get,

PQ + PO + PR + OR > PO + PO

=> PQ + PR + ( QO + OR ) > 2PO

=> PQ + PR + QR > 2 PO

Hence, it is true.

 

Question 4:

PQRS is a quadrilateral. Is PQ + QR + RS + RP > PR + QS ?

12

Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of the third side.

13

Therefore,

In \(\Delta\) PQR, PQ + QR > PR – – – – – – – – – – (1)

In \(\Delta\) PSR, PS + SR > PR – – – – – – – – – (2)

In \(\Delta\) SRQ, SR + RQ > SQ – – – – – – – – – (3)

In \(\Delta\) PSQ, PS + PQ > SQ – – – – – – – – – – (4)

Adding the equations (1) , (2) , (3) and (4) we get,

PQ + QR + PS + SR + RQ + PS + PQ > PS + PS + SQ + SQ

=>( PQ + PQ ) + ( QR + QR ) + ( PS + PS ) + ( SR + SR ) > 2 PR + 2 SQ

=> 2 PQ + 2 QR + 2 PS + 2 SR > 2 PR + 2 SQ

=> 2 ( PQ + QR + PS + SR) > 2 ( PR + SQ )

=> PQ + QR + PS + SR > PR + SQ

=> PQ + QR + RS + SP > PR + SQ

Hence, it is true.

 

Question 5:

PQRS is a quadrilateral. Is PQ + QR + RS + SA < 2 (PR + QS)?

Answer:

Since the sum of the lengths of any two sides of a triangle should be greater than the length of the third side.

14

Therefore,

In \(\Delta\) PXQ, PQ < XP + XQ – – – – – – – – – – (1)

In \(\Delta\) QXR, QR < XQ + XR – – – – – – – – – – – (2)

In \(\Delta\) RXS, RS < XR + XS – – – – – – – – – – – (3)

In \(\Delta\) PXS, SP < XS + XP – – – – – – – – – – (4)

Adding the equations (1) , (2) , (3) and (4) We get,

PQ + QR + RS + SP < XP + XQ + XQ + XR + XR + XS + XS + XP

=> PQ + QR + RS + SP < 2XP + 2 XQ + 2 XR + XS

=> PQ + QR + RS + SP < 2[(PX + XR) + (XS + XQ)]

=> PQ + QR + RS + SP < 2[PR + QS]

Hence it is proved.

 

Question 6:

The lengths of the two sides of a given triangle are 12cm and 15cm respectively. Between what two measures should the length of the third side fall?

Answer:

Since, the sum of the lengths of any two sides of a triangle should be greater than the length of the third side.

It is given that the two sides of a triangle are 15cm and 12cm.

Therefore, the third side should be less than 15+ 12 = 27cm

Also that, the third side cannot be less than the difference of the other two sides.

Therefore, the third side has to be greater than 15-12 = 3cm.

Hence, the third side could be the length more than 3cm and less than 27cm.

 

 

Exercise. 6.5

Question 1:

 ABC is a right angled triangle with \(\angle B = 90^{\circ}\). If AB = 12mm and BC = 5mm, find AC.

Answer:

15

Here AB = 12mm and BC = 5mm (given)

Assuming AC =  y mm.

\(AC^{2} = AB^{2} + BC^{2}\)In \(\triangle ABC\) by applying Pythagoras theorem, we get

\( y^{2} = 12^{2} + 5^{2}\) \(\ therefore y^{2} = 144 + 25\) \(∴y^{2} = 169\) \(∴ y = \sqrt{169} = 13mm\)

Thus, AC = 13mm

 

Question 2:

XYZ is a right angled triangle with \(\angle Y = 90^{\circ}\). If XZ = 5cm and YZ = 3cm, find XY.

Answer:

Assuming XY =  acm.

16

Here XZ = 5cm and YZ = 3cm (given)

In \(\triangle XYZ\) by applying

Pythagoras theorem, we get

\(XZ^{2} = XY^{2} + YZ^{2}\) \(\ therefore a^{2} = 5^{2} – 3^{2}\) \(\ therefore a^{2} = 25 – 9 \) \(∴ a^{2} = 16\) \(∴ a = \sqrt{16} = 4cm\)

Thus, XY = 4cm

 

Question 3:

A 25cm long stick is rested on a wall at a point 24cm high on the wall. If the stick is at a distance x cm from wall on the ground, find the distance x.

17

Answer:

 18

 

Let PR be the stick and P be the point where the stick touches the wall.

So, it becomes right angled triangle with \(\angle Q = 90^{\circ}\)

In \(\triangle PQR\) by applying Pythagoras theorem, we get

\( x^{2} =25^{2} – 24^{2}\)\(PR^{2} = PQ^{2} + QR^{2}\)

\(\ therefore x^{2} = 625 – 576 \) \(∴x^{2} = 49\) \(∴ x = \sqrt{49} = 7cm\)

Thus, QR = 7cm

 

Question 4:

Which of the given below could be the lengths of the side of a right angled triangle?

  1. 5mm, 1.2mm, .9mm
  2. 3mm, 2mm, 1mm
  3. 1mm, 4mm, 9mm

 

Answer:

19

Assuming that the hypotenuse is the side with largest length so as per Pythagoras Theorem,

\((Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2}\)

5mm, 1.2mm, 0.9mm

Let \(\triangle PQR\) be the right angles triangle with \(\angle Q = 90^{\circ}\).

So, by Pythagoras theorem
\( 1.2^{2} + 0.9^{2} = 2.25\)\(PR^{2} = PQ^{2} + QR^{2}\)

\(∴ 1.5^{2} = 2.25 \)

R.H.S = L.H.S

Therefore, 1.5mm, 1.2mm, 0.9mmare

The lengths of the side of a right angled triangle.

3mm, 2mm, 1mm

Let \(\triangle PQR\) be the right angles triangle with \(\angle Q = 90^{\circ}\).

So, by Pythagoras theorem

\(PR^{2} = PQ^{2} + QR^{2}\) \(\ therefore 2^{2} + 1^{2} = 5\) \(∴ 3^{2} = 9 \) \(R.H.S \neq L.H.S\)

Therefore, 3mm, 2mm, 1mm are not

the lengths of the side of a right angled triangle.

 

1mm, 4mm, 0.9mm

Let \(\triangle PQR\) be the right angles triangle with \(\angle Q = 90^{\circ}\).

So, by Pythagoras theorem

\(PR^{2} = PQ^{2} + QR^{2}\) \(\ therefore 4^{2} + 0.9 ^{2} = 16.81\) \(∴ 4.1^{2} = 16.81 \)

R.H.S = L.H.S

20

Therefore, 4.1mm, 4mm, 0.9mm are

the lengths of the side of a right angled triangle.

 

Question 5:

A pole is broken at the height of 500cm from the footpath and its peak point is in contact with the footpath at a distance of 1200cm from the base of the pole. Find the actual length when it was in normal condition.

Answer: Let \(\triangle PQR\)be the right angles triangle with \(\angle Q = 90^{\circ}\) representing the pole broken at point P and touches the footpath at point R and Q is the base of the pole.

21

Here, PQ = 500cm, QR = 1200cm ,

Let PR = x cm

In \(\triangle PQR\) by applying Pythagoras theorem, we get

\(PR^{2} = PQ^{2} + QR^{2}\) \(\ therefore x^{2} =500^{2} + 1200^{2}\) \(∴x^{2} = 250000 + 1440000 \) \(∴x^{2} = 1690000\) \(∴ x = \sqrt{49} = 1300cm\)

Thus, PR = 1300cm

Now the actual length of the pole  = PQ + PR

= 500 + 1300

= 1800cm

Thus, the actual length of the pole is 1800cm.

 

Question 6:

In \(\triangle XYZ\)

Find out the correct one from equations given below, \(\angle Y\; = 35^{\circ} and\;\angle Z=55^{\circ}\)

  1. \(XZ^{2} = XY^{2} + YZ^{2}\)
  2. \(XY^{2} = YZ^{2} + XZ^{2}\)
  3. \(YZ^{2} = XY^{2} + XZ^{2}\)

 22

Answer: In \(\triangle XYZ\)

\(\angle XYZ + \angle YZX + \angle YXZ = 180^{\circ}\)
  • \(35^{\circ} + 55^{\circ} +\angle YXZ = 180^{\circ}\)
  • \(\angle YXZ = 180^{\circ} – (35^{\circ} + 55^{\circ})\)
  • \(\angle YXZ = 180^{\circ} – 90^{\circ}\)
  • \(\angle YXZ = 90^{\circ}\)

So, \(\triangle XYZ\) is a right angled triangle with  \(\angle X = 90^{\circ}\).

Thus, \((Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2}\)

  • \(YZ^{2} = XY^{2} + XZ^{2}\)

Thus, (c) is the correct option.

 

Question 7:

For a rectangle it is given that, width of a rectangle is 15mm and diagonal is 17 mm, then find the perimeter of the rectangle.

Answer:  Let WXYZ be a rectangle.

Let the length of a rectangle is y mm.

23

In \(\triangle XYZ\) by applying

Pythagoras theorem, we get

\(XZ^{2} = YZ^{2} + XY^{2}\) \(∴ 17^{2} =15^{2}  + y^{2}\) \(∴y^{2} = 289 – 225  \) \(∴y^{2} = 64\) \(∴ y = \sqrt{64} = 8mm\)

Thus, XY = 8mm

Thus, the length of a rectangle is 8mm.

Now, perimeter of rectangle = 2 (XY + YZ) = 2(8 + 15)

= 46mm

Thus, the perimeter of the rectangle is 46mm.

 

Question 8:

For a rhombus the length of diagonals are given 80mm and 18mm.Then determine the perimeter of the rhombus.

Answer: Let PQRS be the rhombus.

Let the length of side of rhombus is x mm.

Also PR = 18mm,  SQ = 80mm

24

As the diagonals bisect each other at right angle in case of rhombus,

So, OP = PR/2 = 18/2 = 9mm

And  SO = SQ/2 = 80/2 = 40mm

In \(\triangle POS\) by applying

Pythagoras theorem, we get

\(PS^{2} = OP^{2} + SO^{2}\) \(∴x^{2} =9^{2}  + 40^{2}\) \(∴x^{2} = 81 + 1600 \) \(∴x^{2} = 1681\) \(∴ x = \sqrt{1681} = 41mm\)

Thus, PS = 41mm

Now, the perimeter of the rhombus is = 4\(\times\)Length of side of rhombus = 4\(\times\)PS

= 4\(\times\)41 = 164mm

Thus, the perimeter of the rhombus is 164mm.

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