**NCERT Solutions for Class 7 Maths Chapter 6 – Triangle and its properties**

### NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.1

**1. In ∆ PQR, D is the mid-point of QR .**

PM is __________

Ans: Altitude

PD is ___________

Ans: Median

Is QM = MR

Ans: No, because PM is not the median.

**2. Draw rough sketches for the following:**

**a) In ∆ABC, BE is a median.**

**b) In ∆PQR, PQ and PR are altitudes of the triangle.**

**c) In ∆XYZ, YL is an altitude in the exterior of the triangle.**

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

### NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.2

**1. Find the value of the unknown exterior angle x in the following diagrams:**

We know that,

Exterior angle = Sum of interior opposite angles

**(i)**

x = ∠CAB + ∠CBA

x = 50^{o} + 70^{o}x = 120^{o}

**(ii)**

x = ∠CBA + ∠CAB

x = 65^{o} + 45^{o}x = 110^{o}

**(iii)**

x = ∠CAB + ∠CBA

x = 30^{o} + 40^{o}x = 70^{o}

**(iv)**

x = ∠CAB + ∠CBA

x = 60^{o} + 60^{o}x = 120^{o}

**(v) **

x = ∠CAB + ∠CBA

x = 50^{o} + 50^{o}x = 100^{o}

**(vi) **

x = ∠CAB + ∠CBA

x = 30^{o} + 60^{o}x = 90^{o}

**2. Find the value of the unknown interior angle x in the following figures:**

**(i)**

We know that,

Exterior angle = Sum of interior opposite angles

i.e ∠ACD = ∠CAB + ∠ CBA

115^{o} = 50^{o} + x

x = 115^{o} – 50^{o}Therefore, x = 65^{o}

**(ii) **

We know that,

Exterior angle = Sum of interior opposite angles

i.e ∠DCB = ∠CAB + ∠ CBA

100^{o} = 70^{o} + x

x = 100^{o} – 70^{o}Therefore, x = 30^{o}

**(iii)**

We know that,

Exterior angle = Sum of interior opposite angles

i.e ∠DCA = ∠CAB + ∠ CBA

125^{o} = x + 90^{o}x = 125^{o} – 90^{o}Therefore, x = 35^{o}

**(iv)**

We know that,

Exterior angle = Sum of interior opposite angles

i.e ∠DCA = ∠CAB + ∠ CBA

120^{o} = 60^{o} + x

x = 120^{o} – 60^{o}Therefore, x = 60^{o}

**(v)**

We know that,

Exterior angle = Sum of interior opposite angles

i.e ∠DCB = ∠CAB + ∠ CBA

80^{o} = 30^{o} + x

x = 80^{o} – 30^{o}Therefore, x = 50^{o}

**(vi)**

We know that,

Exterior angle = Sum of interior opposite angles

i.e ∠DCA = ∠CAB + ∠ CBA

75^{o} = 35^{o} + x

x = 75^{o} – 35^{o}Therefore, x = 40^{o}

### NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.3

**1. Find the value of the unknown x in the following diagrams:**

**(i)**

By Angle Sum Property of a Triangle

∠A + ∠B + ∠C = 180^{o}x + 50^{o} + 60^{o} = 180^{o}x + 110^{o} = 180^{o}x = 180^{o} – 110^{o}x = 70^{o}

**(ii)**

By Angle Sum Property of a Triangle

∠P + ∠Q + ∠R = 180^{o}90^{o} + 30^{o} + x = 180^{o}120^{o} + x = 180^{o}x = 180^{o} – 120^{o}x = 60^{o}

**(iii)**

By Angle Sum Property of a Triangle

∠X + ∠Y + ∠Z = 180^{o}30^{o} + 110^{o} + x = 180^{o}140^{o} + x = 180^{o}x = 180^{o} – 140^{o}x = 40^{o}

**(iv)**

By Angle Sum Property of a Triangle

∠A + ∠B + ∠C = 180^{o}50^{o} + x + x = 180^{o}50^{o} + 2x = 180^{o}2x = 180^{o} – 50^{o}2x = 130^{o}x =

x = 65^{o}

**(v)**

By Angle Sum Property of a Triangle

∠A + ∠B + ∠C = 180^{o}x + x + x = 180^{o}3x = 180^{o}x =

x = 60^{o}

**(vi)**

By Angle Sum Property of a Triangle

∠A + ∠B + ∠C = 180^{o}2x + 90^{o} + x = 180^{o}3x + 90^{o} = 180^{o}3x = 180^{o} – 90^{o}3x = 90^{o}x =

x = 30^{o}

**2. Find the values of the unknowns x and y in the following diagrams:**

**(i)**

We know that,

Exterior angle = Sum of Interior opposite angles

i.e ∠BCD = ∠ABC + ∠BAC

120^{o} = x + 50^{o}x = 120^{o} – 50^{o}x = 70^{o}

∠BCA and ∠BCD are linear pairs, hence, they are supplementary

∠BCA + ∠BCD = 180^{o}∠BCA = 180^{o} – ∠BCD

y = 180^{o} – 120^{o}y = 60^{o}

**(ii)**

∠BCA = ∠ECD (Vertically opposite angles)

y = 80^{o}

By using Angle Sum Property, In ∆ABC

∠A + ∠B + ∠C = 180

x + 50^{o} + y = 180^{o}x + 50^{o} + 80^{o} = 180^{o}x + 130^{o} = 180^{o}x = 180^{o} – 130^{o}x = 50^{o}

**(iii)**

We know that, exterior angle = sum of interior opposite angles

∠ACD = ∠CAB + ∠ CBA

x = 50^{o} + 60^{o}x = 110^{o}

∠ACD and ∠ACB are linear pairs, hence, they are supplementary

∠ACD + ∠ACB = 180^{o}∠ACB = 180^{o} – ∠ACD

y = 180^{o} – x

y = 180^{o} – 110^{o}y = 70^{o}

**(iv)**

∠ACB = ∠DCE (Vertically opposite angles)

x = 60^{o}

By using angle sum property, In ∆ABC

∠A + ∠B + ∠C = 180^{o}y + 30^{o} + x = 180^{o}y + 30^{o} + 60^{o} = 180^{o}y + 90^{o} = 180^{o}y = 180^{o} – 90^{o}y = 90^{o}

**(v)**

∠ACB = ∠DCE (Vertically opposite angles)

y = 90^{o}

By using angle sum property, In ∆ABC

∠A + ∠B + ∠C = 180^{o}x + x + y = 180^{o}2x + y = 180^{o}2x + 90^{o} = 180^{o}2x = 180^{o} – 90^{o}2x = 90^{o}x =

x = 45^{o}

**(vi)**

∠ICH = BCA (Vertically opposite angles)

i.e x = y

Similarly, ∠EBD =∠ABC = x

∠FAG = ∠BAC = x

By using angle sum property, In ∆ABC

∠A + ∠B + ∠C = 180^{o}x + x + y = 180^{o}x + x + x = 180^{o}3x = 180^{o}x =

x = 60^{o}

### NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.4

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

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

i) 2cm, 3cm, 5cm

2cm + 3cm = 5cm 5cm

Hence, this triangle is not possible.

ii) 3cm, 6cm, 7cm

6cm + 3cm = 9cm > 7cm

6cm + 7cm = 13cm > 3cm

7cm + 3cm = 10cm > 6cm

Hence, this triangle is possible.

iii) 6cm, 3cm, 2cm

3cm + 2cm = 5cm 6cm

Hence, this triangle is not possible.

**2. Take any point O in the interior of a triangle PQR. Is**

If O is a point in the interior of the triangle ∆PRQ, then, we can construct three triangles ∆OPR, ∆OPQ and ∆ORQ

In a triangle, the sum of the lengths of any two sides is greater than the third side.

i) OP + OQ > PQ?

Yes

ii) OQ + OR > QR?

Yes

iii) OR + OP > RP?

Yes

**3. AM is a median of a triangle ABC.**

**Is AB + BC + CA > 2 AM?**

(Consider the sides of triangles ∆ABM and ∆AMC.)

In a triangle, the sum of the lengths of any two sides is greater than the third side.

Consider ∆ABM

AB + BM > AM…… (i)

Consider ∆AMC

CA + CM > AM…. (ii)

Add equation (i) and (ii)

AB + BM + CA + CM > AM + AM

From the figure BM + CM = BC

Therefore AB + BC + CA > 2AM

**4. ABCD is a quadrilateral.**

**Is AB + BC + CD + DA > AC + BD?**

In a triangle, the sum of the lengths of any two sides is greater than the third side.

Consider ∆ABC

AB + BC > CA ….(i)

Consider ∆BCD

BC + CD > DB ….(ii)

Consider ∆CDA

CD + DA > AC ….(iii)

Consider ∆DAB

DA + AB > DB ….(iv)

On adding equation (i), (ii), (iii) and (iv)

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

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

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

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

**5. ABCD is quadrilateral.**

**Is AB + BC + CD + DA < 2 (AC + BD)?**

In a triangle, the sum of the lengths of any two sides is greater than the third side.

Consider ∆OAB

OA + OB > AB ….(i)

Consider ∆OBC

OB + OC > BC ….(ii)

Consider ∆OCD

OC + OD > CD ….(iii)

Consider ∆ODA

OD + OA > DA ….(iv)

On adding equation (i), (ii), (iii) and (iv)

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

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

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

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

Therefore, AB + BC + CD + DA < 2(AC + BD)

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

Given: Length of the two sides of the triangle are 12cm and 15cm

Let X cm be the length of the third side of the triangle.

In a triangle, the sum of the lengths of any two sides is greater than the third side.

i.e the third side has to be less than the sum of the two sides

12cm + 15cm > X cm

27cm > Xcm

X cm < 27cm

Also, the side cannot be less than the difference of the two sides

Xcm > 15cm – 12cm

Xcm > 3cm

3 cm < X < 27 cm

The length of the third side could be any length greater than 3 cm and less than 27 cm.

### NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.5

**1. PQR is a triangle, right-angled at P. If PQ = 10cm and PR = 24 cm, find QR.**

By using Pythagoras Theorem, In ∆PQR

(QP)^{2} + (PR)^{2} = (QR)^{2}(10)^{2} + (24)^{2} = (QR)^{2}100 + 576 = (QR)^{2}QR =

QR =

QR = 26 cm

**2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.**

By using Pythagoras Theorem, In ∆ABC

(AC)^{2} + (BC)^{2} = (AB)^{2}(7)^{2} + (BC)^{2} = (25)^{2}49 + (BC)^{2} = 625

(BC)^{2} = 625 – 49

BC =

BC =

BC = 24 cm

**3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.**

By using Pythagoras Theorem

a^{2} + (12)^{2 }= (15)^{2}

a^{2} = (15)^{2} – (12)^{2}a^{2} = 225 – 144

a^{2} = 81

a =

a = 9 m

Hence, the distance of the foot of the ladder from the wall is 9 m.

**4. Which of the following can be the sides of a right triangle?**

**i) 2.5 cm,6.5 cm, 6 cm.**(2.5)

^{2}= 6.25

(6.5)

^{2}= 42.25

(6)

^{2}= 36

36 + 6.25 = 42.25

(6)^{2} + (2.5)^{2} = (6.5)^{2}

Pythagoras Theorem holds true; therefore, these sides are of a right triangle.

**(ii) 2 cm, 2 cm, 5 cm**(2)

^{2}= 4

(2)

^{2}= 4

(5)

^{2}= 25

(2)

^{2}+ (2)

^{2}≠ (5)

^{2}Therefore, these sides are not of a right triangle.

**(iii) 1.5 cm, 2cm, 2.5 cm.**(1.5)

^{2}= 2.25

(2)

^{2}= 4

(2.5)

^{2}= 6.25

2.25 + 4 = 6.25

(1.5)^{2} + (2)^{2} = (2.5)^{2}

Pythagoras Theorem holds true; therefore, these sides are of a right triangle.

**5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.**

Let AB be the unbroken part of the tree

Let AC be the broken part of the tree

This implies height of the tree is AB + AC

Using Pythagoras Theorem

(AC)^{2} = (AB)^{2} + (BC)^{2}= (5)^{2} + (12)^{2}= 25 + 144

= 169

AC =

AC = 13 m

Therefore, original height of the tree = 5 + 13 = 18 m

**6. Angles Q and R of a ∆PQR are 25º and 65º. Write which of the following is true: **

We have the total measure of the three angles of a triangle equal to 180°.

Thus,

∠P = 180^{o} – (25^{o }+65^{o} )

∠P = 90^{o}

According to Pythagoras Theorem, if P is right angled, then the hypotenuse is QR, therefore QR^{2} = PQ^{2} + PR^{2 }

(i) PQ^{2} + QR^{2} = RP^{2}False

(ii) PQ^{2} + RP^{2} = QR^{2}True

(iii) RP^{2} + QR^{2} = PQ^{2}False

**7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.**

Since the edges of the rectangle form right angles

Using Pythagoras Theorem, In ∆ADC

(AD)^{2} + (DC)^{2} = (AC)^{2}(AD)^{2} = (AC)^{2} – (DC)^{2}= 41^{2} – 40^{2}= 1681 – 1600

= 81

AD =

AD = 9 cm

Perimeter of Rectangle = 2 ( Length + Breadth)

= 2( DC + AD)

= 2(40 + 9)

=2(49)

Perimeter of Rectangle = 98 cm

**8. Th****e diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.**

Let ABCD be a rhombus with all sides having equal length

Given AD = 16 cm

Therefore, AO = = = 8 cm

Given BC = 30 cm

Therefore, OC = = = 15 cm

The diagonals of a rhombus bisect each other at 90^{o}Consider the right triangle ∆AOC

Using Pythagoras Theorem,

(AC)^{2} = (AO)^{2} + (OC)^{2}= 8^{2} + 15^{2}= 64 + 225

= 289

AC =

AC = 17 cm

The length of the side of rhombus is 17 cm

Thus, the perimeter of rhombus = 4 x Side of the rhombus

Perimeter of Rhombus = 4 x 17 = 68 cm

**NCERT Solutions for Class 7 Maths Chapter 6 – Triangle and its properties**

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