## 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

**Triangle And Its Properties Class 7 Worksheets With Answers Question 1.**

In âˆ†ABC, write the following:

(a) Angle opposite to side BC.

(b) The side opposite to âˆ ABC.

(c) Vertex opposite to side AC.

Solution:

(a) In âˆ†ABC, Angle opposite to BC is âˆ BAC

(b) Side opposite to âˆ ABC is AC

(c) Vertex opposite to side AC is B

**Triangle And Its Properties Class 7 Extra Questions Question 2.**

Classify the following triangle on the bases of sides

Solution:

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

PQ â‰ PR â‰ QR

Thus, âˆ†PQR is a scalene triangle.

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

AB = AC

Thus, âˆ†ABC is an isosceles triangle.

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

MN = ML = NL

Thus, âˆ†MNL is an equilateral triangle.

**Class 7 Maths Chapter 6 Extra Questions Question 3.**

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

Solution

In âˆ†ABC, we have

AD is the altitude.

AE is the median.

**Triangle And Its Properties Class 7 Worksheets Pdf With Answers Question 4.**

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

Solution:

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

â‡’ x + 75Â° – 180Â°

â‡’ x = 180Â° – 75Â°

x = 105Â°

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

x + 30Â° = 90Â°

â‡’ x = 90Â° – 30Â° = 60Â°

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

â‡’ x = 125Â°

**Triangles And Its Properties Class 7 Worksheet 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.

**Class 7 Maths Chapter 6 Worksheet With Answers Question 6.**

In the given figure, find x.

Solution:

In âˆ†ABC, we have

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

â‡’ 5x + 2x + 3x – 60Â° + 40Â° – 80Â° = 180Â°

â‡’ 10x – 100Â° = 180Â°

â‡’ 10x = 180Â° + 100Â°

â‡’ 10x = 280Â°

â‡’ x = 28Â°

Thus, x = 28Â°

**Class 7 Maths Triangle And Its Properties Extra Questions Question 7.**

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

Solution:

Let the third angle be xÂ°.

x + 50Â° + 50Â° = 180Â°

â‡’ xÂ° + 100Â° = 180Â°

â‡’ xÂ° = 180Â° – 100Â° = 80Â°

Thus âˆ x = 80Â°

**Class 7 Maths Ch 6 Extra Questions Question 8.**

In Î”ABC, AC = BC and âˆ C = 110Â°. Find âˆ A and âˆ B.

Solution:

In given Î”ABC, âˆ C = 110Â°

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

x + x + 110Â° = 180Â°

â‡’ 2x + 110Â° = 180Â°

â‡’ 2x = 180Â° – 110Â°

â‡’ 2x = 70Â°

â‡’ x = 35Â°

Thus, âˆ A = âˆ B = 35Â°

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

**Extra Questions For Class 7 Maths Triangle And Its Properties 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 â‡’ 11 > x â‡’ x < 11

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

i.e. 7 – 4 < x â‡’ 3 < x â‡’ x > 3

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

i.e. x < 3 < 11

**Class 7 Triangle And Its Properties Worksheet 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 â‰ 289

5^{2} + 8^{2} â‰ 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.

**Class 7 Triangle And Its Properties Extra Questions Question 11.**

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

Solution:

In Î”ABC, âˆ B = 90Â°

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

(5)^{2} + (x – 3)^{2} = (x + 2)^{2}

â‡’ 25 + x^{2} + 9 – 6x = x^{2} + 4 + 4x

â‡’ -6x – 4x = 4 – 9 – 25

â‡’ -10x = -30

â‡’ x = 3

Hence, the required value of x = 3

**Class 7 Maths Chapter 6 Test Paper Question 12.**

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

Solution:

In Î”ABD,

AB + BD > AD …(i)

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

Similarly, In Î”ADC, we have

AC + DC > AD …(ii)

Adding (i) and (ii), we have

AB + BD + AC + DC > 2AD

â‡’ AB + (BD + DC) + AC > 2AD

â‡’ AB + BC + AC > 2AD

Hence, proved.

**Triangles And Its Properties Class 7 Extra Questions 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}

â‡’ 20^{2} + 21^{2} = AB^{2}

â‡’ 400 + 441 = AB^{2}

â‡’ 841 = AB^{2}

â‡’ AB = âˆš841 = 29 cm.

Perimeter of the rhombus = 4 Ã— side = 4 Ã— 29 = 116 cm

Hence, the required perimeter = 116 cm

**Triangle And Its Properties Class 7 Extra Questions And Answers 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}

â‡’ 25x^{2} = 9x^{2} + 16x^{2}

â‡’ 25x^{2} = 25x^{2}

Hence, the given triangle is a right-angled.

**Triangle And Its Properties Class 7 Worksheet 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, âˆ†OBA is right angled triangle

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

â‡’ OB^{2} = 320^{2} + 240^{2}

â‡’ OB^{2} = 102400 + 57600

â‡’ OB^{2} = 160000

â‡’ OB = âˆš160000 = 400 km.

Hence the required shortest distance = 400 km.

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

**Triangle And Its Properties Class 7 Worksheets Pdf Question 16.**

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

Solution:

Here, l || m

âˆ c = 110Â° (Corresponding angles)

âˆ c + âˆ a = 180Â° (Linear pair)

â‡’ 110Â° + âˆ a = 180Â°

â‡’ âˆ a = 180Â° – 110Â° = 70Â°

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

â‡’ âˆ b = 40Â° + 70Â° = 110Â°

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

**The Triangle And Its Properties Class 7 Extra Questions Question 17.**

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

Solution:

(i) In âˆ†ADC, we have

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

â‡’ âˆ c + 130Â° = 180Â°

â‡’ âˆ c = 180Â° – 130Â° = 50Â°

âˆ c + âˆ b = 180Â° (Linear pair)

â‡’ 50Â° + âˆ b = 180Â°

â‡’ âˆ b = 180Â° – 50Â° = 130Â°

In âˆ†ABD, we have

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

â‡’ âˆ a + âˆ 130Â° + 30Â° = 180Â°

â‡’ âˆ a + 160Â° = 180Â°

â‡’ âˆ a = 180Â° – 160Â° = 20Â°

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

(ii) In âˆ†PQS, we have

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

â‡’ âˆ a + 115Â° = 180Â°

â‡’ âˆ a = 180Â° – 115Â°

â‡’ âˆ a = 65Â°

âˆ a + âˆ b = 180Â° (Linear pair)

â‡’ 65Â° + âˆ b = 180Â°

â‡’ âˆ b = 180Â° – 65Â° = 115Â°

In âˆ†PSR, we have

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

â‡’ 115Â° + âˆ c + 40Â° = 180Â°

â‡’ âˆ c + 155Â° = 180Â°

â‡’ âˆ c = 180Â° – 155Â° = 25Â°

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

Question 18.

I have three sides. One of my angle measure 15Â°. Another has a measure of 60Â°. 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 15Â° and 60Â°.

Third angle = 180Â° – (15Â° + 60Â°)

= 180Â° – 75Â° (Angle sum property)

= 105Â°

which is greater than 90Â°.

Hence, it is an obtuse triangle.