Q.1 In \(\triangle \)PQR, D is the mid-point of
\(\overline{QR} \) ,

(i) \(\overline{PM} \) is..........

(ii) \(\overline{PD} \) is...........

(iii) Is QM = MR?

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

(i) \(\overline{PM} \) is altitude.

(ii) \(\overline{PD} \) is median.

(iii) No, QM \(\neq \) MR.

Q.2 Draw rough sketches for the following:

(a) In \(\triangle \)ABC, BE is a median.

(ib) In \(\triangle \)PQR, PQ and PR are altitudes of the triangle.

(c) In \(\triangle \)XYZ, YL is an altitude in the exterior of the triangle.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

Draw a Line segment AD \(\perp BC \). It is an altitude for this triangle. Here we observe that length of QS and SR is also same. So PS is also a median of this triangle.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

(i) \(\angle \)x = 50° + 70° = 120° (Exterior angle is equal to sum of its interior opposite angles)

(ii) \(\angle \)x = 65°+ 45° = 110° (Exterior angle is equal to sum of its interior opposite angles)

(iii) \(\angle \)x = 30° + 40° = 70° (Exterior angle is equal to sum of its interior opposite angles)

(iv) \(\angle \)x = 60° + 60° = 120° (Exterior angle is equal to sum of its interior opposite angles)

(v) \(\angle \)x = 50° + 50° =100° (Exterior angle is equal to sum of its interior opposite angles)

(vi) \(\angle \)x = 30° + 60° = 90° (Exterior angle is equal to sum of its interior opposite angle)

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

(i) \(\angle \)x + 50° = 115° (Exterior angle of a triangle)

\(\therefore \) \(\angle \)x = 115°- 50° = 65°

(ii) \(\angle \)x + 70° = 110° (Exterior angle of a triangle)

\(\therefore \) \(\angle \)x = 110° – 70° = 40°

(iii) \(\angle \) x + 90° = 125° (Exterior angle of a right triangle)

\(\therefore \) \(\angle \)x = 125° – 90° = 35°

(iv) \(\angle \)x + 60° = 120° (Exterior angle of a triangle)

\(\therefore \) \(\angle \)x = 120° – 60° = 60°

(v)\(\angle \) X + 30° = 80° (Exterior angle of a triangle)

\(\therefore \) \(\angle \)x = 80° – 30° = 50°

(vi) \(\angle \) x + 35° = 75° (Exterior angle of a triangle)

\(\therefore \) \(\angle \) x = 75° – 35° = 40°

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

(i) we know that sum of all angle of a triangle = 180

\(\angle \) x + 50° + 60° = 180°

Or, \(\angle \) x + 110° = 180°

\(\therefore \) \(\angle \)x = 180° – 110° = 70°

(ii) we know that sum of all angle of a triangle = 180

\(\angle \)x + 90° + 30 = 180° [\(\triangle \) is right angled triangle]

Or, \(\angle \)x + 120° = 180°

Or, \(\angle \)x – 180° – 120° = 60°

(iii) we know that sum of all angle of a triangle = 180

\(\angle \)x + 30° + 110° =180°

Or, \(\angle \)x + 140° = 180°

\(\therefore \) \(\angle \)x = 180° – 140° = 40°

(iv ) we know that sum of all angle of a triangle = 180

\(\angle \)x + \(\angle \)x + 50° = 180°

Or, 2x + 50° = 180°

Or, 2x = 180° – 50°

Or, 2x = 130°

\(\therefore \) x=130/2=65°

(v) we know that sum of all angle of a triangle = 180

\(\angle \)x + \(\angle \)x +\(\angle \)x =180°

Or, 3 \(\angle \)x = 180°

\(\therefore \) \(\angle \)x=180°/3=60°

(vi) we know that sum of all angle of a triangle = 180

x + 2 x + 90° = 180° (\(\triangle \) is right angled triangle)

Or, 3x + 90° = 180°

Or, 3x = 180° – 90°

Or, 3x = 90°

\(\therefore \) x=90/3=30°

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

(i) \(\angle \)x + 50° = 120° (Exterior angle of a triangle)

\(\therefore \) \(\angle \)x = 120°- 50° = 70°

\(\angle \)x + \(\angle \)y + 50° = 180° (Angle sum property of a triangle)

70° + \(\angle \)y + 50° = 180°

\(\angle \)y + 120° = 180°

\(\angle \)y = 180° – 120°

\(\therefore \) \(\angle \)y = 60°

Thus \(\angle \)x = 70 and \(\angle \)y – 60°

(ii) \(\angle \)y = 80° (Vertically opposite angles are same)

\(\angle \)x + \(\angle \)y + 50° = 180° (Angle sum property of a triangle)

\(\therefore \) \(\angle \)x + 80° + 50° = 180°

\(\therefore \) \(\angle \)x + 130° = 180°

\(\therefore \)\(\angle \)x = 180° – 130° = 50°

Thus, \(\angle \)x = 50° and \(\angle \)y = 80°

(iii) \(\angle \)y + 50° + 60° = 180° (Angle sum property of a triangle)

\(\angle \)y + 110° = 180°

\(\therefore \)\(\angle \)y = 180°- 110° = 70°

\(\angle \)x + \(\angle \)y = 180° (Linear pairs)

Or, \(\angle \)x + 70° = 180°

\(\therefore \) \(\angle \)x = 180° – 70° = 110°

Thus, \(\angle \)x = 110° and y = 70°

(iv) \(\angle \)x = 60° (Vertically opposite angles)

\(\angle \)x + \(\angle \)y + 30° = 180° (Angle sum

Or, \(\angle \)y + 90° = 180°

Or, \(\angle \)y = 180° – 90° = 90°

Thus, \(\angle \)x = 60° and \(\angle \)y = 90°

(v) \(\angle \)y = 90° (Vertically opposite angles)

\(\angle \)x + \(\angle \)x + \(\angle \)y = 180° (Angle sum property of a triangle)

Or, 2 \(\angle \)x + 90° = 180°

Or, 2\(\angle \)x = 180° – 90°

Or, 2\(\angle \)x = 90°

\(\therefore \) \(\angle \)x=90°/2=45°

Thus, \(\angle \)x = 45° and \(\angle \)y = 90

(vi) From the given figure, we have

\(\angle \)y = \(\angle \)x

\(\angle \)1 = \(\angle \)x

\(\angle \)2 = \(\angle \)x

Adding both sides, we have
Adding both sides, we have
Adding both sides, we have
\(\angle \)y + \(\angle \)1 + \(\angle \)2 = 3\(\angle \)x

Or, 180° = 3\(\angle \)x [Angle sum property of a triangle]

\(\therefore \) \(\angle \)x=180°/3=60°

\(\angle \)x = 60°, \(\angle \)y = 60°

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

(i) 2 cm, 3 cm, 5 cm

(ii) 3 cm, 6 cm, 7 cm

(iii) 6 cm, 3 cm, 2 cm

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

We know that for a triangle, the sum of any two sides must be greater than the third side.

(i) Given sides are 2 cm, 3 cm, 5 cm

(2 + 3) = 5

5 = 5

Hence, the triangle is not possible.

(i) Given sides are 3 cm, 6 cm, 7 cm

(3 + 6) = 9 > 7

(6 + 7) = 13 > 3

(7 + 3) = 10 > 6

Hence, the triangle is possible.

(iii) Given sides are 6 cm, 3 cm, 2 cm

(3 + 2) = 5 < 6

Hence, the triangle is not possible.

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

(i) OP + OQ > PQ?

(ii) OQ + OR > QR?

(iii) OR + OP > RP?

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

(i) Yes, \(\triangle \)OPQ has sides OP, OQ and PQ.

So, OP + OQ > PQ

(ii) Yes, \(\triangle \)OQR has sides OR, OQ and QR.

So, OQ + OR > QR

(iii) Yes, \(\triangle \)ORP has sides OR, OP and PR.

So, OR + OP > RP

Q.3 AM is a median of a \(\triangle \) ABC.

Is AB + BC + CA > 2 AM?

(Consider the sides of triangles \(\triangle \)ABM and \(\triangle \)AMC.)

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

We know that,

The sum of the length of any two sides is always greater than the third side.

Now In \(\triangle \)ABM,

Here, AB + BM > AM … [equation i]

Then, In \(\triangle \)ACM

Here, AC + CM > AM … [equation ii]

By adding equation [i] and [ii] we get,

AB + BM + AC + CM > AM + AM

From the figure we have, BC = BM + CM

AB + BC + AC > 2 AM

Hence, the given expression is true.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

In \(\triangle \)ABC, we have

AB + BC > AC …(i)
[Sum of any two sides is greater than the third side]

In \(\triangle \)BDC, we have

BC + CD > BD …(ii)

In \(\triangle \)ADC, we have
CD + DA > AC .... (iii)

In \(\triangle \)DAB, we have
DA + AB > BD …(iv)

Adding eq. (i), (ii), (iii) and (iv), we get

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

Or, AB + BC + CD + DA > AC + BD [Dividing both sides by 2]

Hence, proved

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

\(\triangle \)AOB, we have AB < AO + BO …(i)
[Any side of a triangle is less than the sum of other two sides]

In \(\triangle \)BOC, we have

BC < BO + CO …(ii)

In \(\triangle \)COD, we have

CD < CO + DO …(iii)

In \(\triangle \)AOD, we have

DA < DO + AO …(iv)

Adding eq. (i), (ii), (iii) and (iv), we have

AB + BC + CD + DA < 2AO + 2BO + 2CO + 2DO

Or, AB + BC + CD + DA < 2(AO + BO + CO + DO)

Or, AB + BC + CD + DA < 2 [(AO + CO) + (BO + DO)]

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

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

Hence, proved.

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

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

Sum of two sides

= 12 cm + 15 cm = 27 cm

Difference of the two sides

= 15 cm – 12 cm = 3 cm

\(\therefore \) The measure of third side should fall between 3 cm and 27 cm.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

In right angled triangle PQR, we have

\(QR^2 = PQ^2 + PR^2 \) [From Pythagoras property]

\(QR^2 = 10^2 + 24^2 \)

\(QR^2 = 100 + 576 = 676 \)

\(\therefore QR = \sqrt{676} = 26 cm \)
The, the required length of QR = 26 cm.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

In right angled?ABC, we have

\( BC^2 + 7^2 = 25^2 \)(By Pythagoras property)

Or, \( BC^2 + 49 = 625 \)

Or, \( BC^2 = 625 – 49 \)

Or, \( BC^2 = 576 \)

\(\therefore BC = \sqrt{576} = 24 cm \)

Thus, the required length of BC = 24 cm.

Q.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.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

Here, the ladder forms a right angled triangle.

\(\therefore \) \(a^2 + 12^2 = 15^2 \)(By Pythagoras property)

Or, \(a^2+ 144 = 225\)

Or, \(a^2 = 225 – 144 \)

Or, \(a^2 = 81 \)

\(\therefore \) a = \(\sqrt{81}= 9 \) m

Thus, the distance of the foot from the ladder = 9m

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

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

(ii) 2 cm, 2 cm, 5 cm.

(iii) 1.5 cm, 2 cm, 2.5 cm

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

(i) Given sides are 2.5 cm, 6.5 cm, 6 cm.

Square of the longer side = \((6.5)^2 = 42.25 \) cm.

Sum of the square of other two sides
= \( (2.5)^2 + (6)^2 = 6.25 + 36 \)
= 42.25 cm.

Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.

\(\therefore \)The given sides form a right triangle.

(ii) Given sides are 2 cm, 2 cm, 5 cm .

Square of the longer side =\( (5)^2 = 25 \)cm

Sum of the square of other two sides
=\( (2)^2 + (2)^2 =4 + 4 = 8 cm \)
Since 25 cm \(\neq \)8 cm

\(\therefore \)The given sides do not form a right triangle.

(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm

Square of the longer side = \( (2.5)^2 = 6.25 \)cm

Sum of the square of other two sides
= \((1.5)^2 + (2)^2 = 2.25 + 4 = 6.25cm = 6.25 cm = 6.25 cm \)

Since the square of longer side in a triangle is equal to the sum of square of other two sides.

\(\therefore \)The given sides form a right triangle.

Q.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.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

Let AB be the original height of the tree and broken at C touching the ground at D such that

AC = 5 m

and AD = 12 m

In right triangle \(\triangle \)CAD,

\(AD^2 + AC^2 = CD^2 \) (By Pythagoras property)

Or, \( (12)^2 + (5)^2 = CD^2 \)

Or, 144 + 25 =\( CD^2 \)
Or, 169 = \( CD^2\)

\(\therefore \)CD =\(\sqrt{169} \) = 13 m

But CD = BC

AC + CB = AB

5 m + 13 m = AB

\(\therefore \) AB = 18 m .

Thus, the original height of the tree = 18 m.

Q.6 Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.

(i) \( PQ^2 + QR^2 = RP^2 \)

(ii)\( PQ^2 + RP^2 = QR^2 \)

(iii)\( RP^2 + QR^2 = PQ^2 \)

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

We know that

\(\angle P + \angle Q + \angle R = 180° \)(Angle sum property)

\(\angle \)P + 25° + 65° = 180°

\(\angle \)P + 90° = 180°

\(\angle \)P = 180° – 90° – 90°

\(\triangle \)PQR is a right triangle, right angled at P

(i) Not True

\(\therefore PQ^2 + QR^2 \neq RP^2 \) (By Pythagoras property)

(ii) True

\(\therefore \) \(PQ^2 + RP^2 = QP^2 \) (By Pythagoras property)

(iii) Not True

\(\therefore RP^2 + QR^2 \neq PQ^2 \)(By Pythagoras property)

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

Length of rectangle = 40cm

Diagonal = 41 cm

\(Breath^2 = 41^2 - 40^2 \)

Or, \( breath^2= 1681 – 1600 \)

Or, breath = \(\sqrt{81} \) cm

Breath = 9 cm

Perimeter of rectangle = 2×(40+9) = 2×49 = 98 cm

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Answer :

Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm

Since, the diagonals of a rhombus bisect each other at 90°.

\(\therefore \) OA = OC = 8 cm and OB = OD = 15 cm

In right \(\triangle \)OAB,

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

= \((8)^2+ (15)^2 = 64 + 225 \)
= 289

\(\therefore \)AB =\(\sqrt{289} \)= 17 cm

Since AB = BC = CD = DA (Property of rhombus)

\(\therefore \)Required perimeter of rhombus
= 4 × side = 4 × 17 = 68 cm.

There are total 21 questions present in ncert solutions for class 7 maths chapter 6 the triangle and its properties

There are total 3 long question/answers in ncert solutions for class 7 maths chapter 6 the triangle and its properties

There are total 5 exercise present in ncert solutions for class 7 maths chapter 6 the triangle and its properties