# NCERT Solutions For Class 7 Maths Chapter 6

## NCERT Solutions Class 7 Maths The Triangles and its Properties

### Ncert Solutions For Class 7 Maths Chapter 6 PDF Free Download

NCERT Solutions For Class 7 Maths Chapter 6 Triangles and its Properties are given here in a simple and detailed way. These NCERT Solutions for class 7 maths Triangles and its Properties can be extremely helpful for the students to clear all their doubts easily and understand the basics of this chapter in a better and detailed way.

In the NCERT class 10 textbook, several exercise problems are included to help the students to have a better understanding of the concepts and let them develop their problem-solving abilities. For class 7 chapter 6 also, several practice questions are included for the students. Students often face doubts in the questions and eventually pile them up. These class 7 maths NCERT solutions for chapter 6 can help them to clear their doubts instantly.

NCERT solutions Triangles and its Properties class 7 maths given here are very easily understandable so that students do not face any difficulties regarding any of the solutions. The NCERT solutions for chapter 6 of class 7 maths PDF is also available here that the students can download and study.

### NCERT Solutions For Class 7 Maths Chapter 6 Exercises

Exercise 6.1

Question 1:

In Δ ABC, X is the midpoint of BC.

AP is _____________

AX is _____________

Is BP = PC?

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.

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

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

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

Question 3:

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

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.

Exercise 6.2

Question 1:

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

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:

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.

(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:

(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

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?

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 )

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 ?

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

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)?

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

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?

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.

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.

Assuming XY =  acm.

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.

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

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

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.

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. $YZ^{2} = XY^{2} + XZ^{2}$

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.

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

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.

Triangles and its properties chapter introduces some of the most crucial concepts from geometry that are extremely important as they are used in several other mathematics topics in the later grades. So, students need to be completely thorough with all the concepts from this chapter to be able to comprehend higher level concepts in the higher classes easily.

Some of the most important concepts covered in this chapter include median of a triangle, exterior angle of a triangle and its properties, angle sum property of a triangle, equilateral and isosceles triangles, right-angles triangles and Pythagoras property, etc.

NCERT has included several examples to help the students understand the concepts better and know how to solve various related questions. Students are also suggested to practice all the practice questions given in the book to develop their respective problem-solving abilities. Students can simply refer to these NCERT Solutions Class 7 Maths Chapter 6 to clear all their doubts instantly.

Students can also check the full NCERT Solutions For Class 7 Maths here and clear all the doubts from any chapter instantly. These NCERT solutions will not only help the students to clear their respective doubts but will also help them to learn the in-depth class 7 maths concepts in a more efficient way.