## The Triangles and its Properties Class 7 Extra Questions Maths Chapter 6

**Extra Questions for Class 7 Maths Chapter 6 The Triangles and its Properties**

### The Triangles and its Properties Class 7 Extra Questions Very Short Answer Type

Question 1.

In ∆ABC, write the following:

(a) Angle opposite to side BC.

(b) The side opposite to ∠ABC.

(c) Vertex opposite to side AC.

Solution:

(a) In ∆ABC, Angle opposite to BC is ∠BAC

(b) Side opposite to ∠ABC is AC

(c) Vertex opposite to side AC is B

Question 2.

Classify the following triangle on the bases of sides

Solution:

(i) PQ = 5 cm, PR = 6 cm and QR = 7 cm

PQ ≠ PR ≠ QR

Thus, ∆PQR is a scalene triangle.

(ii) AB = 4 cm, AC = 4 cm

AB = AC

Thus, ∆ABC is an isosceles triangle.

(iii) MN = 3 cm, ML = 3 cm and NL = 3 cm

MN = ML = NL

Thus, ∆MNL is an equilateral triangle.

Question 3.

In the given figure, name the median and the altitude. Here E is the midpoint of BC.

Solution

In ∆ABC, we have

AD is the altitude.

AE is the median.

Question 4.

In the given diagrams, find the value of x in each case.

Solution:

(i) x + 45° + 30° = 180° (Angle sum property of a triangle)

⇒ x + 75° – 180°

⇒ x = 180° – 75°

x = 105°

(ii) Here, the given triangle is right angled triangle.

x + 30° = 90°

⇒ x = 90° – 30° = 60°

(iii) x = 60° + 65° (Exterior angle of a triangle is equal to the sum of interior opposite angles)

⇒ x = 125°

Question 5.

Which of the following cannot be the sides of a triangle?

(i) 4.5 cm, 3.5 cm, 6.4 cm

(ii) 2.5 cm, 3.5 cm, 6.0 cm

(iii) 2.5 cm, 4.2 cm, 8 cm

Solution:

(i) Given sides are, 4.5 cm, 3.5 cm, 6.4 cm

Sum of any two sides = 4.5 cm + 3.5 cm = 8 cm

Since 8 cm > 6.4 cm (Triangle inequality)

The given sides form a triangle.

(ii) Given sides are 2.5 cm, 3.5 cm, 6.0 cm

Sum of any two sides = 2.5 cm + 3.5 cm = 6.0 cm

Since 6.0 cm = 6.0 cm

The given sides do not form a triangle.

(iii) 2.5 cm, 4.2 cm, 8 cm

Sum of any two sides = 2.5 cm + 4.2 cm = 6.7 cm

Since 6.7 cm < 8 cm

The given sides do not form a triangle.

Question 6.

In the given figure, find x.

Solution:

In ∆ABC, we have

5x – 60° + 2x + 40° + 3x – 80° = 180° (Angle sum property of a triangle)

⇒ 5x + 2x + 3x – 60° + 40° – 80° = 180°

⇒ 10x – 100° = 180°

⇒ 10x = 180° + 100°

⇒ 10x = 280°

⇒ x = 28°

Thus, x = 28°

Question 7.

One of the equal angles of an isosceles triangle is 50°. Find all the angles of this triangle.

Solution:

Let the third angle be x°.

x + 50° + 50° = 180°

⇒ x° + 100° = 180°

⇒ x° = 180° – 100° = 80°

Thus ∠x = 80°

Question 8.

In ΔABC, AC = BC and ∠C = 110°. Find ∠A and ∠B.

Solution:

In given ΔABC, ∠C = 110°

Let ∠A = ∠B = x° (Angle opposite to equal sides of a triangle are equal)

x + x + 110° = 180°

⇒ 2x + 110° = 180°

⇒ 2x = 180° – 110°

⇒ 2x = 70°

⇒ x = 35°

Thus, ∠A = ∠B = 35°

### The Triangles and its Properties Class 7 Extra Questions Short Answer Type

Question 9.

Two sides of a triangle are 4 cm and 7 cm. What can be the length of its third side to make the triangle possible?

Solution:

Let the length of the third side be x cm.

Condition I: Sum of two sides > the third side

i.e. 4 + 7 > x ⇒ 11 > x ⇒ x < 11

Condition II: The difference of two sides less than the third side.

i.e. 7 – 4 < x ⇒ 3 < x ⇒ x > 3

Hence the possible value of x are 3 < x < 11

i.e. x < 3 < 11

Question 10.

Find whether the following triplets are Pythagorean or not?

(a) (5, 8, 17)

(b) (8, 15, 17)

Solution:

(a) Given triplet: (5, 8, 17)

17^{2} = 289

8^{2} = 64

5^{2} = 25

8^{2} + 5^{2} = 64 + 25 = 89

Since 89 ≠ 289

5^{2} + 8^{2} ≠ 17^{2}

Hence (5, 8, 17) is not Pythagorean triplet.

(b) Given triplet: (8, 15, 17)

17^{2} = 289

15^{2} = 225

8^{2} = 64

15^{2} + 8^{2} = 225 + 64 = 289

17^{2} = 15^{2} + 8^{2}

Hence (8, 15, 17) is a Pythagorean triplet.

Question 11.

In the given right-angled triangle ABC, ∠B = 90°. Find the value of x.

Solution:

In ΔABC, ∠B = 90°

AB^{2} + BC^{2} = AC^{2} (By Pythagoras property)

(5)^{2} + (x – 3)^{2} = (x + 2)^{2}

⇒ 25 + x^{2} + 9 – 6x = x^{2} + 4 + 4x

⇒ -6x – 4x = 4 – 9 – 25

⇒ -10x = -30

⇒ x = 3

Hence, the required value of x = 3

Question 12.

AD is the median of a ΔABC, prove that AB + BC + CA > 2AD (HOTS)

Solution:

In ΔABD,

AB + BD > AD …(i)

(Sum of two sides of a triangle is greater than the third side)

Similarly, In ΔADC, we have

AC + DC > AD …(ii)

Adding (i) and (ii), we have

AB + BD + AC + DC > 2AD

⇒ AB + (BD + DC) + AC > 2AD

⇒ AB + BC + AC > 2AD

Hence, proved.

Question 13.

The length of the diagonals of a rhombus is 42 cm and 40 cm. Find the perimeter of the rhombus.

Solution:

AC and BD are the diagonals of a rhombus ABCD.

Since the diagonals of a rhombus bisect at the right angle.

AC = 40 cm

AO = \(\frac { 40 }{ 2 }\) = 20 cm

BD = 42 cm

OB = \(\frac { 42 }{ 2 }\) = 21 cm

In right angled triangle AOB, we have

AO^{2} + OB^{2} = AB^{2}

⇒ 20^{2} + 21^{2} = AB^{2}

⇒ 400 + 441 = AB^{2}

⇒ 841 = AB^{2}

⇒ AB = √841 = 29 cm.

Perimeter of the rhombus = 4 × side = 4 × 29 = 116 cm

Hence, the required perimeter = 116 cm

Question 14.

The sides of a triangle are in the ratio 3 : 4 : 5. State whether the triangle is right-angled or not.

Solution:

Let the sides of the given triangle are 3x, 4x and 5x units.

For right angled triangle, we have

Square of the longer side = Sum of the square of the other two sides

(5x)^{2} = (3x)^{2} + (4x)^{2}

⇒ 25x^{2} = 9x^{2} + 16x^{2}

⇒ 25x^{2} = 25x^{2}

Hence, the given triangle is a right-angled.

Question 15.

A plane flies 320 km due west and then 240 km due north. Find the shortest distance covered by the plane to reach its original position.

Solution:

Here, OA = 320 km

AB = 240 km

OB = ?

Clearly, ∆OBA is right angled triangle

OB^{2} = OA^{2} + AB^{2} (By Pythagoras property)

⇒ OB^{2} = 320^{2} + 240^{2}

⇒ OB^{2} = 102400 + 57600

⇒ OB^{2} = 160000

⇒ OB = √160000 = 400 km.

Hence the required shortest distance = 400 km.

### The Triangles and its Properties Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 16.

In the following figure, find the unknown angles a and b, if l || m.

Solution:

Here, l || m

∠c = 110° (Corresponding angles)

∠c + ∠a = 180° (Linear pair)

⇒ 110° + ∠a = 180°

⇒ ∠a = 180° – 110° = 70°

Now ∠b = 40° + ∠a (Exterior angle of a triangle)

⇒ ∠b = 40° + 70° = 110°

Hence, the values of unknown angles are a = 70° and b = 110°

Question 17.

In figure (i) and (ii), Find the values of a, b and c.

Solution:

(i) In ∆ADC, we have

∠c + 60° + 70° = 180° (Angle sum property)

⇒ ∠c + 130° = 180°

⇒ ∠c = 180° – 130° = 50°

∠c + ∠b = 180° (Linear pair)

⇒ 50° + ∠b = 180°

⇒ ∠ b = 180° – 50° = 130°

In ∆ABD, we have

∠a + ∠b + 30° = 180° (Angle sum property)

⇒ ∠a + ∠130° + 30° = 180°

⇒ ∠a + 160° = 180°

⇒ ∠a = 180° – 160° = 20°

Hence, the required values are a = 20°, b = 130° and c = 50°

(ii) In ∆PQS, we have

∠a + 60° + 55° = 180°(Angle sum property)

⇒ ∠a + 115° = 180°

⇒ ∠a = 180° – 115°

⇒ ∠a = 65°

∠a + ∠b = 180° (Linear pair)

⇒ 65° + ∠b = 180°

⇒ ∠b = 180° – 65° = 115°

In ∆PSR, we have

∠b + ∠c + 40° = 180° (Angle sum property)

⇒ 115° + ∠c + 40° = 180°

⇒ ∠c + 155° = 180°

⇒ ∠c = 180° – 155° = 25°

Hence, the required angles are a = 65°, b = 115° and c = 25°

Question 18.

I have three sides. One of my angle measure 15°. Another has a measure of 60°. What kind of a polygon am I? If I am a triangle, then what kind of triangle am I? [NCERT Exemplar]

Solution:

Since I have three sides.

It is a triangle i.e. three-sided polygon.

Two angles are 15° and 60°.

Third angle = 180° – (15° + 60°)

= 180° – 75° (Angle sum property)

= 105°

which is greater than 90°.

Hence, it is an obtuse triangle.