NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties

NCERT
SOLUTIONS
CLASS - 7th

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Class : 7th
Subject : Maths
Chapter : 6
Chapter Name : The Triangle and its Properties

Exercise 6.1

Q1 In ∆ PQR, D is the mid-point of QR .
¯
¯¯¯
¯¯¯
¯¯

is _________________.
¯
¯¯¯
¯¯¯
¯¯
PM

PD is _________________.
Is QM = MR?

Answer. (i) Altitude
(ii) Median
(iii) No

Page : 116 , Block Name : Exercise 6.1

Q2 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.

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

Page : 116 , Block Name : Exercise 6.1

Q3 Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.

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 : 116 , Block Name : Exercise 6.1

Exercise 6.2

Q1 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 : 118 , Block Name : Exercise 6.2

Q2 Find the value of the unknown interior angle x in the following gures:

∘ ∘
(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 = 750 − 350 = 40

Page : 119 , Block Name : Exercise 6.2

Exercise 6.3

Q1 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
∘
130 ∘
x = = 65
2
∘
(v) x + x + x = 180

∘
3x = 180

180 ∘
x = = 60
3
∘ ∘
(vi) x + 2x + 90 = 180

∘ ∘ ∘
3x = 180 − 90 = 90
∘
90 ∘
x = = 30
3

Page : 121 , Block Name : Exercise 6.3

Q2 Find the values of the unknowns x and y in the following diagrams:

∘ ∘
(i) y + 120 = 180 (Linear pair)

∘ ∘ ∘
y = 180 − 120 = 60

∘ ∘
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 + 1300 = 180
∘ ∘ ∘
x = 180 − 130 = 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 = 60 (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
∘
90 ∘
x = = 45
2

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 ∘

∘
180 ∘
x = = 60
3

∘
y = x = 60

Page : 121 , Block Name : Exercise 6.3

Exercise 6.4

Q1 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 > 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 = S cm < 6 cm
Hence, this triangle is not possible.

Page : 126 , Block Name : Exercise 6.4

Q2 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 ΔOP Q, 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 ΔOP Q 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

Page : 126 , Block Name : Exercise 6.4

Q3 AM is a median of a triangle ABC.

Is AB + BC + CA > 2 AM? (Consider the sides of triangles ∆ABM and ∆AMC.)

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 : 126 , Block Name : Exercise 6.4

Q4 ABCD is a 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 + + CO + DA) > 2(AC + ED)
(AB + BC + CD + DA) > (AC + BD)
Yes, the given expression is true.

Page : 126 , Block Name : Exercise 6.4

Q5 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 + 0B > AB ---(i)
In ΔOBC ,
0B + 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 + BO) > AB + BC + CD + DA
Yes, the given expression is true.

Page : 126 , Block Name : Exercise 6.4

Q6 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 : 127 , Block Name : Exercise 6.4

Exercise 6.5

Q1 PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, nd QR.

By applying Pythagoras theorem in ΔP QR,
2 2 2
(P Q) + (P R) = (RQ)

2 2 2
(10) + (24) = RQ

2
100 + 576 = (QR)

2
676 = (QR)

QR = 26cm

Page : 130 , Block Name : Exercise 6.5

Q2 ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, nd BC.

By applying Pythagoras theorem in ΔABC ,
2 2 2
(AC) + (BC) = (AB)

2 2 2
(BC) = (AB) − (AC)

2 2 2
(BC) = (25) − (7)

2
(BC) = 625 − 49 = 576

BC = 24cm

Page : 130 , Block Name : Exercise 6.5

Q3 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,
2 2 2
(15) = (12) + a

2
225 = 144 + a

2
a = 225 − 144 = 81

a = 9m

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

Page : 130 , Block Name : Exercise 6.5

Q4 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, 2cm, 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
(2.5) = 6.25

2
(6.5) = 42.25

2
(6) = 36

It can be observed that,
36 + 6.25 = 42.25

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

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
(5) = 25

2 2 2
Here, (2) + (2) ≠ (5)

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, 2cm, 2.5 cm.
2
(1.5) = 2.25

2
(2) = 4

2
(2.5) = 6.25

Here,

2.25 + 4 = 6.25

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

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 : 130 , Block Name : Exercise 6.5

Q5 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.

In the given gure, 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 △ABC,
2 2 2
AC = BC + AB

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

2 2 2 2
AC = 25m + 144m = 169m

AC = 13m

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

Page : 130 , Block Name : Exercise 6.5

Q6 Angles Q and R of a ∆PQR are 25º and 65º.

Write which of the following is true:
(i) RP + QR = RP
2 2 2

(ii) RP + RP = QR 2 2 2

(iii) RP + QR = PQ2 2 2

Answer. The sum Of the measures Of all interior angles Of a triangle is 180
∘

∘
∠P QR + ∠P RQ + ∠QP R = 180
∘
250 + 650 + ∠QP R = 180
∘ ∘
90 + ∠QP R = 180
∘ ∘ ∘
∠QP R = 180 − 90 = 90

Therefore, Δ PQR is right -angled at point P.
Hence, (PR) + (PQ) = (QR) 2 2 2

Page : 130 , Block Name : Exercise 6.5

Q7 Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

In right-angled triangle DAB, AB 2
+ AD
2
= BD
2

2 2 2 2 2 2
⇒ 40 + AD = 41 ⇒ AD = 41 − 40 ⇒
2 2
AD = 1681 − 1600 ⇒ AD = 81 ⇒ AD = 9

Perimeter of the rectangle = 2(AB + AD) = 2(40 + 9) = 2(49) = 98cm
Hence, the perimeter of the rectangle is 98 cm.

Page : 130 , Block Name : Exercise 6.5

Q8 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 90 . It can be observed that
∘

AC 16
AO = = = 8cm
2 2

BD 30
BO = = = 15cm
2 2

By applying Pythagoras theorem in △AOB,
2 2 2
OA + OB = AB

2 2 2
8 + 15 = AB

2
64 + 225 = AB

2
289 = AB

AB = 17

Therefore, the length Of the side Of rhombus is 17 cm.
Perimeter of rhombus = 4 x Side of the rhombus = 4 x 17 = 68 cm

Page : 130 , Block Name : Exercise 6.5