The Triangle And Its Properties

In the above given question, we have

Also, given that:

D is the mid-point of QR

Therefore,

PM is altitude

Also,

PD is Median

And,

QM MR

(a) We have,

ΔABC,

And in this triangle, BE is a median

Therefore,

The rough sketch for the given figure is as follows:

Median is the line that divides the opposite side into two equal parts.

(b) We have,

ΔPQR,

And in this triangle, PQ and QR are the altitudes of the triangle

Therefore,

The rough sketch for the given figure is as follows:

(c) We have,

ΔXYZ,

And in this triangle YL is an altitude in the exterior of the triangle

Therefore,

The rough sketch for the given figure is as follows:

In this question, We have to prove that:

The median and altitude of an isosceles triangle are same

The diagram for the given question is as follows:

This can be proved as follows:

First of all we have to draw a line segment AD which is perpendicular to BC. This is an altitude for this triangle

Now,

From the figure it can also be observed that the length of BD and DC is also same

Therefore,

AD is also median of this isosceles triangle

Hence, proved

(i) It is given in the question that,

1^st interior angle = 50^o and 2^nd interior angle = 70^o

Note: According to exterior angle theorem,

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = x

Sum of interior angles = 50^o + 70^o

Thus, using exterior angle theorem,

x = 50^o + 70^o

x = 120^o

∴ The value of x is 120^o

(ii) It is given in the question that,

1^st interior angle = 65^o and 2^nd interior angle = 45^o

Note: According to exterior angle theorem,

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = x

Sum of interior angles = 65^o + 45^o

Using, exterior angle theorem, we have

x = 65^o + 45^o

x = 110^o

Hence, the value of x is 110^o

(iii) It is given in the question that,

1^st interior angle = 30^o and 2^nd interior angle = 40^o

Note: According to exterior angle theorem,

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = x

Sum of interior angles = 30^o + 40^o

Using exterior angle theorem, we have

x = 30^o + 40^o

x = 70^o

Hence, the value of x is 70^o

(iv) It is given in the question that,

1^st interior angle = 60^o and 2^nd interior angle = 60^o

Note: According to exterior angle theorem,

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = x

Sum of interior angles = 60^o + 60^o

Using exterior angle theorem, we have

x = 60^o + 60^o

x = 120^o

Hence, the value of x is 120^o

(v) It is given in the question that,

1^st interior angle = 50^o and 2^nd interior angle = 50^o

Note: According to exterior angle theorem:

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = x

Sum of interior angles = 50^o + 50^o

Using exterior angle theorem, we have

x = 50^o + 50^o

x = 100^o

Hence, the value of x is 100^o

(vi) It is given in the question that,

1^st interior angle = 30^o and 2^nd interior angle = 60^o

Note: According to exterior angle theorem:

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = x

Sum of interior angles = 30^o + 60^o

Using, exterior angle theorem, we have

x = 30^o + 60^o

x = 90^o

Hence, the value of x is 90^o.

(i) It is given in the question that,

1^st interior angle = x and 2^nd interior angle = 50^o

Note: According to exterior angle theorem:

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = 115^o

Sum of interior angles = x + 50^o

Using exterior angle theorem, we have

115^o = x + 50^o

x = 115^o – 50^o

x = 65^o

Hence, the value of x is 65^o

(ii) It is given in the question that,

1^st interior angle = 70^o and 2^nd interior angle = x

Note: According to exterior angle theorem:

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = 100^o

Sum of interior angles = 70^o + x

Using exterior angle theorem, we have

100^o = 70^o + x

x = 100^o – 70^o

x = 30^o

Hence, the value of x is 30^o

(iii) It is given in the question that,

1^st interior angle = x and 2^nd interior angle = 90^o

Now, according to exterior angle theorem:

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = 125^o

Sum of interior angles = x + 90^o

Using exterior angle theorem, we have

125^o = x + 90^o

x = 125^o – 90^o

x = 35^o

Hence, the value of x is 35^o

(iv) It is given in the question that,

1^st interior angle = x and, 2^nd interior angle = 60^o

Note: According to exterior angle theorem:

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = 120^o

Sum of interior angles = x + 60^o

Using exterior angle theorem, we have

120^o = x + 60^o

x = 120^o – 60^o

x = 60^o

Hence, the value of x is 60^o

(v) It is given in the question that,

1^st interior angle = x and 2^nd interior angle = 30^o

Note: According to exterior angle theorem:

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = 80^o

Sum of interior angles = x + 30^o

Using exterior angle theorem, we have

80^o = x + 30^o

x = 80^o – 30^o

x = 50^o

Hence, the value of x is 50^o

(vi) It is given in the question that,

1^st interior angle = x and 2^nd interior angle = 35^o

Note: According to exterior angle theorem:

The measure of an exterior angle of a triangle is equal to the sum of the measure of the two non-adjacent interior angles of the triangle.

Exterior angle = 75^o

Sum of interior angles = x + 35^o

Using exterior angle theorem, we have

75^o = x + 35^o

x = 75^o – 35^o

x = 40^o

Hence, the value of x is 40^o

(i) In the above question, we have

1^st interior angle = x, 2^nd interior angle = 50^o and 3^rd interior angle = 60^o

We have to find out the value of x

Thus,

x + 50^o + 60^o = 180^o (angle sum property of triangle)

x + 110^o = 180^o

x = 180^o – 110^o

x = 70^o

Hence, the value of x is 70^o

(ii) In the above question, we have

1^st interior angle = x, 2^nd interior angle = 90^o and 3^rd interior angle = 30^o

We have to find out the value of x

Thus,

x + 90^o + 30^o = 180^o (angle sum property of triangle)

x + 120^o = 180^o

x = 180^o – 120^o

x = 60^o

Hence, the value of x is 60^o

(iii) In the above question, we have

1^st interior angle = x, 2^nd interior angle = 30^o and 3^rd interior angle = 110^o

We have to find out the value of x

Thus,

x + 30^o + 110^o = 180^o (angle sum property of triangle)

x + 140^o = 180^o

x = 180^o – 140^o

x = 40^o

Hence, the value of x is 40^o

(iv) In the above question, we have

1^st interior angle = 50^o, 2^nd interior angle = xand 3^rd interior angle = x

We have to find out the value of x

Thus,

50^o + x + x = 180^o (angle sum property of triangle)

2x + 50^o = 180^o

2x = 180^o – 50^o

2x = 130^o

x =

x = 65^o

Hence, the value of x is 65^o

(v) In the above question, we have

1^st interior angle = x, 2^nd interior angle = x and 3^rd interior angle = x

We have to find out the value of x

And,

x + x + x = 180^o (angle sum property of triangle)

3x = 180^o

x =

x = 60^o

Hence, the value of x is 60^o

(vi) In the above question, we have

1^st interior angle = x, 2^nd interior angle = 2x and 3^rd interior angle = 90^o

We have to find out the value of x

Thus,

x + 2x + 90^o = 180^o (angle sum property of triangle)

3x + 90^o = 180^o

3x = 180^o – 90^o

3x = 90^o

x =

x = 30^o

Hence, the value of x is 30^o

(i)

From the given figure, we have

y + 120^o = 180^o (Linear pair)

Therefore,

y = 180^o – 120^o

y = 60^o

x + y + 50^o = 180^o (angle sum property of triangle)

x + 110^o = 180^o

x = 180^o – 110^o

x = 70^o

Hence,

The value of x = 70^o

And,

Value of y = 60^o

(ii)

From the given figure, we have

y = 80^o (Vertically opposite angle)

y + x + 50^o = 180^o (angle sum property of triangle)

80^o + x + 50^o = 180^o

130^o + x = 180^o

x = 180^o – 130^o

x = 50^o

Hence,

The value of x = 50^o

And,

The value of y = 80^o

(iii)


From the given figure,

y + 50^o + 60^o = 180^o (angle sum property of triangle)

y + 110^o = 180^o

y = 180^o – 110^o

y = 70^o

x and y are on a straight line and forming a linear pair

Therefore,

x + y = 180^o

x + 70^o = 180^o

x = 180^o – 70^o

x = 110^o

Hence,

The value of x = 110^o

And,

Value of y = 70^o

(iv)


From the given figure, we have

x = 60^o (Vertically opposite angle)

30^o + x + y = 180^o (angle sum property of triangle)

30^o + 60^o + y = 180^o

90^o + y = 180^o

y = 180^o – 90^o

y = 90^o

Hence,

The value of x = 60^o

And,

Value of y = 90^o

(v)


From the given figure, we have

y = 90^o (Vertically opposite angle)

x + x + y = 180^o (angle sum property of triangle)

2x + y = 180^o

2x + 90^o = 180^o

2x = 180^o – 90^o

2x = 90^o

x =

x = 45^o

Hence, The value of x = 45^o

And,

The valuue of y = 90^o

(vi)


From the given figure, we have

y = x (Vertically opposite angles)

a = x (Vertically opposite angles)

b = x (Vertically opposite angles)

a + b + x = 180^o (angle sum property of triangle)

x + x + x = 180^o

3x = 180^o

x =

x = 60^o

y = x (Vertically opposite angle)

Hence,

The value of y = x = 60^o

(i) We know that,

In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side

The given sides of triangle in this question are:

2 cm, 3 cm and 5 cm

Now, 2 + 3 + = 5 cm

5 cm = 5 cm

Hence,

The triangle is not possible as the sum of the length of either two sides of the triangle is not greater than the third side

(ii) We know that,

In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side

The given sides of triangle in this question are:

3 cm, 6 cm and 7 cm

Now, 3 + 6 = 9 cm and 9 cm > 7 cm

6 + 7 = 13 cm and, 13 cm > 3 cm

3 + 7 = 10 cm and, 10 cm > 6 cm

Hence,

The triangle is possible as the sum of the length of either two sides of the triangle is greater than the third side

(iii) We know that,

In a triangle,

The sum of the length of either two sides of the triangle is always greater than the third side

Therefore,

The given sides of triangle in this question are:

6 cm, 3 cm and 2 cm

Now,

6 + 3 = 9 cm and 9 cm > 2 cm

3 + 2 = 5 cm But, 5 cm < 6 cm

Hence,

The triangle is not possible as the sum of the length of either two sides of the triangle is not greater than the third side

(i) According to the given condition in the question,

If O is a point in the interior of the given triangle

Then,

Three triangles can be constructed, these are:

and

We know that,

In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side

Therefore,

is a triangle having sides OP, OQ and PQ

As,

OP + OQ > PQ

(ii) According to the given condition in the question,

We have:

If O is a point in the interior of the given triangle

Then,

Three triangles can be constructed, these are:

and

We know that,

In a triangle,

The sum of the length of either two sides of the triangle is always greater than the third side

Therefore,

is a triangle having sides OR, OQ and QR

As,

OQ + OR > QR

(iii) According to the given condition in the question,

We have:

If O is a point in the interior of the given triangle

Then,

Three triangles can be constructed, these are:

and

We know that,

In a triangle,

The sum of the length of either two sides of the triangle is always greater than the third side

Therefore,

is a triangle having sides OR, OP and PR

As,

OR + OP > PR

Concept Used: In a triangle,

The sum of the length of either two sides of the triangle is always greater than the third side

To Prove: AB + BC + CA > 2 AM

Proof:

In Δ ABM, we have:

AB + BM > AM (i)

Same is the case with Δ ACM, in which:

AC + CM > AM (ii)

Adding (i) and (ii), we get

AB + BM + MC + AC > AM + AM

AB + BC + AC > 2 AM

Hence, Proved.

It is given in the question that,

A quadrilateral ABCD, in which we have to show that:

Whether AB + BC + CD + DA > AC + BD or not

We know that,

In a triangle,

The sum of the length of either two sides of the triangle is always greater than the third side

Therefore,

In ∆ABC, we have

AB + BC > CA (i)

In ∆BCD, we have

BC + CD > DB (ii)

In ∆CDA, we have

CD + DA > AC (iii)

In ∆DAB, we have

DA + AB > DB (iv)

Adding equations (i), (ii), (iii) and (iv), we get:

AB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD

2 AB + 2 BC + 2 CD + 2 DA > 2 AC + 2 BD

2 (AB + BC + CD + DA) > 2 (AC + BD)

(AB + BC + CD + DA) > (AC + BD)

Hence, the expression given in the question is true.

It is given in the question that,

A quadrilateral ABCD, in which we have to show that:

Whether AB + BC + CD + DA < 2 (AC + BD) or not

We know that,

In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side

In ∆OAB, we have

OA + OB > AB ... (i)

In ∆OBC, we have

OB + OC > BC ..... (ii)

In ∆OCD, we have

OC + OD > CD ..... (iii)

In ∆ODA, we have

OD + OA > DA ..... (iv)

Adding equations (i), (ii), (iii) and (iv), we get:

OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA

2 OA + 2 OB + 2 OC + 2 OD > AB + BC + CD + DA

2 OA + 2 OC + 2 OB + 2 OD > AB + BC + CD + DA

2 (OA + OC) + 2 (OB + OD) > AB + BC + CD + DA

2 (AC) + 2 (BD) > AB + BC + CD + DA

2 (AC + BD) >AB + BC + CD + DA

Hence, the expression given in the question is true.

It is given in the question that,

Two sides of a triangle are 12 cm and 15 cm respectively

We know that,

In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side

In a triangle, the difference of the length of either two sides of the triangle is always lesser than the third side

The third side will be lesser than the sum of these two sides:

12 + 15 = 27 cm

The third side will be greater than the difference between these two sides:

15 – 12 = 3 cm

Hence, the third side should be greater than 3 cm and less than 27 cm.

It is given in the question that,

PQ = 10 cm

Also,

PR = 24 cm

We have to find QR

We know that,

According to Pythagoras theorem,

In a right angled triangle:

(Hypotenuse)² = (Perpendicular)² + (Base)²

By applying Pythagoras theorem in the given question, we get:

(QR)² = (PQ)² + (PR)²

(QR)² = (10)² + (24)²

(QR)² = 100 + 576

(QR)² = 676

Hence,

QR =

QR = 26 cm

Given: AB = 25 cm and AC = 7 cm

According to Pythagoras theorem, In a right angled triangle:

(Hypotenuse)² = (Perpendicular)² + (Base)²

By applying Pythagoras theorem in ABC, we get:

(AB)² = (AC)² + (BC)²

(BC)² = (AB)² - (AC)²

(BC)² = (25)² – (7)²

(BC)² = 625 – 49

(BC)² = 576

Hence,

BC = √576 cm

BC = 24 cm

It is given in the question that,

Length of ladder = 15 m

Height of ladder = 12 m

We have to find the distance of the foot of the ladder from the wall

Let the distance be a

According to Pythagoras theorem,

In a right angled triangle:

(Hypotenuse)² = (Perpendicular)² + (Base)²

By applying Pythagoras theorem in the given question, we get:

(15)² = (12)² + (a)²

225 = 144 + (a)²

(a)² = 225 - 144

(a)² = 81

Hence,

a =

a = 9 m

(i) In the given question, we have to identify the right angles

It is given in the question that,

Sides of triangle = 2.5 cm, 6.5 cm and 6 cm

(2.5)² = 6.25

(6.5)² = 42.25

(6)² = 36

Now, it can be observed that,

36 + 6.25 = 42.25

(6)² + (2.5)² = (6.5)²

∴ The square of the length of one side of the triangle is the sum of the squares of other two sides

Hence,

These are the sides of the right angled triangle

The right angle of the triangle will be in the front of the side 6.5 cm

(ii) In the given question, we have to identify the right angles

It is given in the question that,

Sides of triangle = 2 cm, 2 cm and 5 cm

(2)² = 4

(2)² = 4

(5)² = 25

Now, it can be observed that,

(2)² + (2)² (5)²

∴ The square of the length of one side of the triangle is not equal to the sum of the squares of other two sides

Hence,

These are not the sides of the right angled triangle

(iii) In the given question, we have to identify the right angles

It is given in the question that,

Sides of triangle = 1.5 cm, 2 cm and 2.5 cm

(1.5)² = 2.25

(2)² = 4

2.5)² = 6.25

Now, it can be observed that,

2.25 + 4 = 6.25

Hence,

(1.5)² + (2)² = (2.5)²

∴ The square of the length of one side of the triangle is the sum of the squares of other two sides

Hence,

These are the sides of the right angled triangle

The right angle of the triangle will be in the front of the side 2.5 cm

From the figure, we have

BC is the unbroken part of the tree

Point C represents that point where the tree broke

CA represents the broken part of the tree

is right angled at B

By applying pythagoras theorem, we get:

AC² = BC² + AB²

AC² = (5)² + (12)²

AC² = 25 + 144

AC² = 169 m²

AC =

AC = 13 m

Hence,

The original height of the tree = AC + CB

= 13 m + 5 m

= 18 m

In the above question, it is given that:

In PQR, we have

∠Q = 25^o and ∠R = 65^o

We know that,

The sum of interior angles of a triangle = 180^o

∠PQR + ∠PRQ + ∠QPR = 180^o

25^o + 65^o + ∠QPR = 180^o

90^o + ∠QPR = 180^o

∠QPR = 180^o – 90^o

= 90^o

Hence, is right angled at point P

Thus,

By using Pythagoras theorem, we get:

(PR)² + (PQ)² = (QR)²

Hence,

Option (ii) is true

Here,

It is given in the question that,

Length of a rectangle = 40 cm

Diagonal of a rectangle = 41 cm

We know that,

Diagonal of a rectangle divides it into two right-angled triangles

By using Pythagoras theorem, we get

(Diagonal)² = (Length)² + (Breadth)²

(41)² = (40)² + b²

b² = (41)² – (40)²

b²= 1681 – 1600

b² = 81

b =

b = 9 cm

Thus,

Perimeter of rectangle = 2 (l + b)

= 2 (40 + 9)

= 2 (49)

= 98 cm

Let us assume a rhombus ABCD having diagonals AC and BD

Diagonals of this rhombus ABCD intersects each other at a point O.

We know that,

Diagonals of rhombus bisects each other at 90^o

Now,

AO = = = 8 cm

BO = = = 15 cm

By applying Pythagoras theorem, we get:

OA² + OB² = AB²

8² + 15² = AB²

64 + 225 = AB²

AB² = 289

AB =

AB = 17 cm

∴ The length of the side of Rhombus = 17 cm

Perimeter of rhombus = 4 × Side of rhombus

= 4 × 17

= 68 cm