Mathematics NCERT Grade 7, Chapter 6: The Triangle and its Properties- This chapter talks about the following topics-
  • Medians of A Triangle
​The line segment joining a vertex of triangles to the midpoint of its opposite sides is called a median of the triangle.
  • Altitudes of A Triangle
    • An altitude has one end point at a vertex of the triangle and the other on the line containing the opposite side. 
    • ​A triangle has 3 altitudes. 
​Exercise 6.1 is a short exercise consisting of 3 exercises.
  • Exterior Angle of A Triangle And Its Properties
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
  • Angle Sum Property of A Triangle
The total measure of the three angles of a triangle is 180°.
Before moving onto next concept, students will have to solve exercise 6.3.
  • Two Special Triangles: Equilateral and Isosceles.
    • A triangle in which all the three sides are of equal lengths is called an equilateral triangle
    • A triangle in which two sides are of equal lengths is called an isosceles triangle
  • Sum of the Length of Two sides of A Triangle
​In this section it will be observed that the sum of the lengths of any two sides of a triangle is greater than the third side. 
  • Right- Angled Triangles And Pythagoras Property
In a right-angled triangle, the side opposite to the right angle is referred to as hypotenuse and the other two sides are called the legs of the triangle.

Pythagoras property of triangles is stated as:
  • In a right-angled triangle, the square on the hypotenuse is equal sum of the squares on the legs.
Assessment of the chapter is done through the 5 unsolved exercises given. Various solved examples associated with the chapter are given before every unsolved exercise. 
Important points of the chapter are discussed in the summary of the chapter- The Triangle and its Properties.

Page No 116:

Question 1:

In ΔPQR, D is the mid-point of .

is __________.

PD is __________.

Is QM = MR?

Answer:

(i) Altitude

(ii) Median

(iii) No

Page No 116:

Question 2:

Draw rough sketches for the following:

(a) In ΔABC, BE is a median.

(b) In ΔPQR, PQ and PR are altitudes of the triangle.

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

Answer:

(a)

(b)

(c)

Here, it can be observed that for ΔXYZ, YL is an altitude drawn exterior to side XZ which is extended up to point L.

Page No 116:

Question 3:

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

Answer:

Draw a line segment AD perpendicular to BC. It is an altitude for this triangle. It can be observed that the length of BD and DC is also same. Therefore, AD is also a median of this triangle.



Page No 118:

Question 1:

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

Answer:

(i) x = 50° + 70° (Exterior angle theorem)

x = 120°

(ii) x = 65° + 45° (Exterior angle theorem)

= 110°

(iii) x = 40° + 30° (Exterior angle theorem)

= 70°

(iv) x = 60° + 60° (Exterior angle theorem)

= 120°

(v) x = 50° + 50° (Exterior angle theorem)

= 100°

(vi) x = 30° + 60° (Exterior angle theorem)

= 90°



Page No 119:

Question 2:

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

Answer:

(i) x + 50° = 115° (Exterior angle theorem)

x = 115° − 50° = 65°

(ii) 70° + x = 100° (Exterior angle theorem)

x = 100° − 70° = 30°

(iii) x + 90° = 125° (Exterior angle theorem)

x = 125° − 90° = 35°

(iv) x + 60° = 120° (Exterior angle theorem)

x = 120° − 60° = 60°

(v) x + 30° = 80° (Exterior angle theorem)

x = 80° − 30° = 50°

(vi) x + 35° = 75° (Exterior angle theorem)

x = 75º − 35º = 40°



Page No 121:

Question 1:

Find the value of the unknown x in the following diagrams:

Answer:

The sum of all interior angles of a triangle is 180°. By using this property, these problems can be solved as follows.

(i) x + 50° + 60° = 180°

x + 110° = 180°

x = 180° − 110° = 70°

(ii) x + 90° + 30° = 180°

x + 120° = 180°

x = 180° − 120° = 60°

(iii) x + 30° + 110° = 180°

x + 140° = 180°

x = 180° − 140° = 40°

(iv) 50° + x + x = 180°

50° + 2x = 180°

2x = 180° − 50° = 130°

(v) x + x + x = 180°

3x = 180°

(vi) x + 2x + 90° = 180°

3x = 180° − 90° = 90º


Video Solution for The Triangle and its Properties (Page: 121 , Q.No.: 1)

NCERT Solution for Class 7 math - The Triangle and its Properties 121 , Question 1

Page No 121:

Question 2:

Find the value of the unknowns x and y in the following diagrams:

Answer:

(i) y + 120° = 180° (Linear pair)

y = 180° − 120º = 60º

x + y + 50° = 180° (Angle sum property)

x + 60° + 50° = 180°

x + 110° = 180°

x = 180° − 110° = 70°

(ii) y = 80° (Vertically opposite angles)

y + x + 50° = 180° (Angle sum property)

80° + x + 50° = 180°

x + 130º = 180°

x = 180° − 130º = 50°

(iii) y + 50° + 60° = 180° (Angle sum property)

y = 180° − 60° − 50° = 70°

x + y = 180° (Linear pair)

x = 180° − y = 180° − 70° = 110°

(iv) x = 60º (Vertically opposite angles)

30° + x + y = 180°

30° + 60° + y = 180°

y = 180° − 30° − 60° = 90°

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

x + x + y = 180° (Angle sum property)

2x + y = 180°

2x + 90° = 180°

2x = 180° − 90° = 90°


(vi)

y = x (Vertically opposite angles)

a = x (Vertically opposite angles)

b = x (Vertically opposite angles)

a + b + y = 180° (Angle sum property)

x + x + x = 180°

3x = 180°

y = x = 60°

Video Solution for The Triangle and its Properties (Page: 121 , Q.No.: 2)

NCERT Solution for Class 7 math - The Triangle and its Properties 121 , Question 2



Page No 126:

Question 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

Answer:

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

(i) Given that, the sides of the triangle are 2 cm, 3 cm, 5 cm.

It can be observed that,

2 + 3 = 5 cm

However, 5 cm = 5 cm

Hence, this triangle is not possible.

(ii) Given that, the sides of the triangle are 3 cm, 6 cm, 7 cm.

Here, 3 + 6 = 9 cm > 7 cm

6 + 7 = 13 cm > 3 cm

3 + 7 = 10 cm > 6 cm

Hence, this triangle is possible.

(iii) Given that, the sides of the triangle are 6 cm, 3 cm, 2 cm.

Here, 6 + 3 = 9 cm > 2 cm

However, 3 + 2 = 5 cm < 6 cm

Hence, this triangle is not possible.

Page No 126:

Question 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?

Answer:

If O is a point in the interior of a given triangle, then three triangles ΔOPQ, ΔOQR, and ΔORP can be constructed. In a triangle, the sum of the lengths of either two sides is always greater than the third side.

(i) Yes, as ΔOPQ is a triangle with sides OP, OQ, and PQ.

OP + OQ > PQ

(ii) Yes, as ΔOQR is a triangle with sides OR, OQ, and QR.

OQ + OR > QR

(iii) Yes, as ΔORP is a triangle with sides OR, OP, and PR.

OR + OP > PR

Video Solution for The Triangle and its Properties (Page: 126 , Q.No.: 2)

NCERT Solution for Class 7 math - The Triangle and its Properties 126 , Question 2

Page No 126:

Question 3:

AM is a median of a triangle ABC.

Is AB + BC + CA > 2 AM?

(Consider the sides of triangles ΔABM and ΔAMC.)

Answer:

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

In ΔABM,

AB + BM > AM (i)

Similarly, in ΔACM,

AC + CM > AM (ii)

Adding equation (i) and (ii),

AB + BM + MC + AC > AM + AM

AB + BC + AC > 2AM

Yes, the given expression is true.

Page No 126:

Question 4:

ABCD is quadrilateral.

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

Answer:

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

Considering ΔABC,

AB + BC > CA (i)

In ΔBCD,

BC + CD > DB (ii)

In ΔCDA,

CD + DA > AC (iii)

In ΔDAB,

DA + AB > DB (iv)

Adding equations (i), (ii), (iii), and (iv), we obtain

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)

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

Yes, the given expression is true.

Page No 126:

Question 5:

ABCD is quadrilateral.

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

Answer:

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

Considering ΔOAB,

OA + OB > AB (i)

In ΔOBC,

OB + OC > BC (ii)

In ΔOCD,

OC + OD > CD (iii)

In ΔODA,

OD + OA > DA (iv)

Adding equations (i), (ii), (iii), and (iv), we obtain

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

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

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

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

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

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

Yes, the given expression is true.



Page No 127:

Question 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?

Answer:

In a triangle, the sum of the lengths of either two sides is always greater than the third side and also, the difference of the lengths of either two sides is always lesser than the third side. Here, the third side will be lesser than the sum of these two (i.e., 12 + 15 = 27) and also, it will be greater than the difference of these two (i.e., 15 − 12 = 3). Therefore, those two measures are 27cm and 3 cm.



Page No 130:

Question 1:

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

Answer:

By applying Pythagoras theorem in ΔPQR,

(PQ)2 + (PR)2 = (RQ)2

(10)2 + (24)2 = RQ2

100 + 576 = (QR)2

676 = (QR)2

QR = 26 cm

Page No 130:

Question 2:

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

Answer:

By applying Pythagoras theorem in ΔABC,

(AC)2 + (BC)2 = (AB)2

(BC)2 = (AB)2 − (AC)2

(BC)2 = (25)2 − (7)2

(BC)2 = 625 − 49 = 576

BC = 24 cm

Page No 130:

Question 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.

Answer:

By applying Pythagoras theorem,

(15)2 = (12)2 + a2

225 = 144 + a2

a2 = 225 − 144 = 81

a = 9 m

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

Video Solution for The Triangle and its Properties (Page: 130 , Q.No.: 3)

NCERT Solution for Class 7 math - The Triangle and its Properties 130 , Question 3

Page No 130:

Question 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

In the case of right-angled triangles, identify the right angles.

Answer:

(i) 2.5 cm, 6.5 cm, 6 cm

(2.5)2 = 6.25

(6.5)2 = 42.25

(6)2 = 36

It can be observed that,

36 + 6.25 = 42.25

(6)2 + (2.5)2 = (6.5)2

The square of the length of one side is the sum of the squares of the lengths of the remaining two sides. Hence, these are the sides of a right-angled triangle. Right angle will be in front of the side of 6.5 cm measure.

(ii) 2 cm, 2 cm, 5 cm

(2)2 = 4

(2)2 = 4

(5)2 = 25

Here, (2)2 + (2)2 ≠ (5)2

The square of the length of one side is not equal to the sum of the squares of the lengths of the remaining two sides. Hence, these sides are not of a right-angled triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm

(1.5)2 = 2.25

(2)2 = 4

(2.5)2 = 6.25

Here,

2.25 + 4 = 6.25

(1.5)2 + (2)2 = (2.5)2

The square of the length of one side is the sum of the squares of the lengths of the remaining two sides. Hence, these are the sides of a right-angled triangle.

Right angle will be in front of the side of 2.5 cm measure.

Page No 130:

Question 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.

Answer:

In the given figure, BC represents the unbroken part of the tree. Point C represents the point where the tree broke and CA represents the broken part of the tree. Triangle ABC, thus formed, is right-angled at B.

Applying Pythagoras theorem in ΔABC,

AC2 = BC2 + AB2

AC2 = (5 m)2 + (12 m)2

AC2 = 25 m2 + 144 m2 = 169 m2

AC = 13 m

Thus, original height of the tree = AC + CB = 13 m + 5 m = 18 m

Video Solution for The Triangle and its Properties (Page: 130 , Q.No.: 5)

NCERT Solution for Class 7 math - The Triangle and its Properties 130 , Question 5

Page No 130:

Question 6:

Angles Q and R of a ΔPQR are 25° and 65°.

Write which of the following is true:

(i) PQ2 + QR2= RP2

(ii) PQ2 + RP2= QR2

(iii) RP2 + QR2= PQ2

Answer:

The sum of the measures of all interior angles of a triangle is 180º.

∠PQR + ∠PRQ + ∠QPR = 180º

25º + 65º + ∠QPR = 180º

90º + ∠QPR = 180º

∠QPR = 180º − 90º = 90º

Therefore, Δ PQR is right-angled at point P.

Hence, (PR)2 + (PQ)2 = (QR)2

Thus, (ii) is true.

Page No 130:

Question 7:

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

Answer:

In a rectangle, all interior angles are of 90º measure. Therefore, Pythagoras theorem can be applied here.

(41)2 = (40)2 + x2

1681 = 1600 + x2

x2 = 1681 − 1600 = 81

x = 9 cm

Perimeter = 2(Length + Breadth)

= 2(x + 40)

= 2 (9 + 40)

= 98 cm

Page No 130:

Question 8:

The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer:

Let ABCD be a rhombus (all sides are of equal length) and its diagonals, AC and BD, are intersecting each other at point O. Diagonals in a rhombus bisect each other at 90º. It can be observed that

By applying Pythagoras theorem in ΔAOB,

OA2 + OB2 = AB2

82 + 152 = AB2

64 + 225 = AB2

289 = AB2

AB = 17

Therefore, the length of the side of rhombus is 17 cm.

Perimeter of rhombus = 4 × Side of the rhombus = 4 × 17 = 68 cm

Video Solution for The Triangle and its Properties (Page: 130 , Q.No.: 8)

NCERT Solution for Class 7 math - The Triangle and its Properties 130 , Question 8



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