# Properties of Triangle: Angle Sum of a Triangle

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1. MATHEMATICS IN EVERYDAY LIFE–6
Chapter 11 : Triangles and Parallel Lines ANSWER KEYS
EXERCISE 11.1 4. (i) 12 cm, 13 cm, 15 cm
Perimeter of triangle = sum of the lengths of
1. P three sides
= 12 cm + 13 cm + 15 cm
= 40 cm
(ii) 6 cm, 6 cm, 5 cm
Perimeter of triangle = sum of the lengths of
three sides
Q R = 6 cm + 6 cm + 5 cm
(i) The side opposite to Q is PR. = 17 cm
(ii) The vertex opposite to the side PQ is R. (iii) 8 cm, 12 cm, 14 cm
(iii) The angle opposite to the side QR is P. Perimeter of triangle = sum of the lengths of
(iv) The side opposite to the vertex P is QR. three sides
2. No, we cannot draw a triangle with three collinear = 8 cm + 12 cm + 14 cm
points. In this case, sum of two sides is equal to the = 34 cm
third side. (iv) 6 cm, 5 cm, 10 cm
3. (i) 7, 8, 16 Perimeter of triangle = sum of the lengths of
For a triangle, sum of length of any two sides is three sides
always greater than the third side, i.e., = 6 cm + 5 cm + 10 cm
7 + 8 < 16 = 21 cm
8 + 16 > 7 5. (i) P
7 + 16 > 8
Since, 7 + 8 < 16 does not satisfy the condition.
So, the numbers do not denote the lengths of
side of a triangle.
(ii) 3, 2, 4
3+2>4 70° x
2+4>3 Q R
3+4>2  70° + x = 180° (Linear pair)
Hence, the numbers denote the lengths of side of  x = 180° – 70°
a triangle.  x = 110°
(iii) 5, 7, 12
A
5 + 7 = 12 (ii)
5 + 12 > 7
7 + 12 > 5
Since, 5 + 7 = 12 does not satisfy the condition.
So, the numbers do not denote the lengths of
side of a triangle.
x 130°
B C
Mathematics In Everyday Life-6 1
2.  x + 130° = 180° (Linear pair) (ii) In right angled triangle ACB, C = 90°.
 x = 180° – 130° A
 x = 50°
x
6. P
55°
B C
Q R  A + B + C = 180° (angle sum property)
S
(i) Three triangles : PSR, PQS, PQR x + 55° + 90° = 180°
(ii) Name of seven angles : QPS, SPR, QPR,  x + 145° = 180°
PQS, PRS, PSQ, PSR  x= 180° – 145°
(iii) Name of six line segments : PQ, PR, QR, PS, QS, SR
x = 35
(iv) PQS and PQR have Q as common.
(iii) In PQR,
EXERCISE 11.2 P
1. Let the two angles of a triangle be 2x and 3x. °
25
Third angle = 60°
Therefore,
x
2x + 3x + 60° = 180° (Angle sum property) 30°
Q R
 5x + 60° = 180°
 5x = 180° – 60° PQR + QRP + RPQ = 180° (Angle sum
property)
 5x = 120°
30° + x + 25° = 180°
120
 x= x + 55° = 180°
5
 x = 24° x = 180° – 55°
Henc e, other two angles of the triangle are x = 125
2 × 24° = 48° and 3 × 24° = 72°.
2. (i) P A 130°
x
(iv)
3x 2x
B C
2x 120° extA + BAC = 180° (Linear pair)
Q R  130° + BAC = 180°
extR + PRQ = 180° (Linear pair)  BAC = 180° – 130°
 120° + PRQ = 180°
BAC = 50°
 PRQ = 180° – 120° = 60°
Now, in ABC,
Now In PQR,
BAC + ACB + CBA = 180°
 Sum of the interior angles of a triangle is 180°.
(Angle sum property)
QPR + PRQ + RQP = 180°
x + 60° + 2x = 180° 50° + 2x + 3x = 180°
 3x + 60° = 180°  50° + 5x = 180°
 3x = 180° – 60°  5x = 180° – 50°
 3x = 120°  5x = 130°
120 130
 x= = 40°  x=
3 5
x = 40  x = 26
3. (v) (iii) 48°, 62°, 70°
P
110° Sum of the three angles = 48° + 62° + 70°
= 180°
Hence, a triangle with these given angles is
60° x possible.
Q R (iv) 35°, 43°, 130°
Sum of the three angles = 35° + 43° + 130°
 ext P + QPR = 180° (Linear pair) = 208°  180°
 110° + QPR = 180° Hence, it is not possible to have a triangle with
QPR = 180° – 110° these given angles.
QPR = 70° (v) 40°, 50°, 80°
In PQR,
QPR + PRQ + RQP = 180° Sum of the three angles = 40° + 50° + 80°
(Angle sum property) = 170°  180°
70° + x + 60° = 180° Hence, it is not possible to have a triangle with
these given angles.
 x = 180° – 130°
(vi) 35°, 95°, 90°
x = 50
Sum of the three angles = 35° + 95° + 90°
(vi) A
= 220°  180°
x
Hence, it is not possible to have a triangle with
these given angles.
150° x
B C 4. A
 ext B + ABC = 180° (Linear pair)
150° + ABC = 180° 3x
ABC = 180° – 150°
ABC = 30°
In ABC,
ABC + ACB + CBA = 180° 2x 120°
(Angle sum property) B
C
30° + x + x = 180°
Let the opposite interior angles of a ABC be 2x and
 30° + 2x = 180°
3x.
 2x = 180° – 30°
 ext C + ACB = 180° (Linear pair)
 2x = 150°
 120° + ACB = 180°
150
 x = = 75° ACB = 180° – 120°
2
ACB = 60°
Hence, x = 75
In ABC,
3. (i) 88°, 42°, 120°
BAC + ACB + CBA = 180°
 Sum of the angles of a triangle is 180°.
3x + 60° + 2x = 180°
 Sum of the three angles = 88° + 42° + 120°
 5x + 60° = 180°
= 250°  180°
Hence, it is not possible to have a triangle with  5x = 180° – 60°
these given angles. 120
(ii) 50°, 50°, 50°  x= = 24°
5
 Sum of the angles of a triangle is 180°. Hence,
 Sum of the three angles = 50° + 50° + 50° BAC = 3  24° = 72°
= 150°  180° CBA = 2  24° = 48°
Hence, it is not possible to have a triangle with
these given angles. ACB = 60°.
Mathematics In Everyday Life-6 3
4. 5. Let equal angles of a ABC be x. Therefore, EXERCISE 11.3
sum of interior angles of a triangle = 180° A
Q
 A + B + C = 180°
 30° + x + x = 180°
30°
 30° + 2x = 180° P
 2x = 180° – 30°
150 1. C D
 x= = 75°
2 x x No, the line segments PQ and CD do not intersect. In
Hence, each of equal angles = 75°. B C this case both line segments will intersect, when
6. (i) No, the reason a triangle can’t have 2 obtuse produced towards the left. So, they are not parallel
angles is because it would add upto more than to each other, since the distance between them is not
180°, which is not possible in a triangle as sum same throughout.
of the interior angles of a triangle is 180°.
Q S
(ii) No, the reason a triangle can’t have 2 right 2. m
angles is because the sum of the angles would
add upto more than 180°.
(iii) No, because all the angles of an equilateral
triangle are same and equal to 60° (acute).
(iv) No, an equilateral triangle has all the angles
90° 90°
equal to 70° which lead to the sum of angles of n
triangle is 70° + 70° + 70° = 210°  180°. P R
7. (i) In ABC, AB = AC = 4 cm. So, it is an isosceles Since, m || n, QP  n, SR  n and PQ = 2.5 cm.
triangle.  The distance between two parallel lines m and n
(ii) In PQR, PQ = 12 cm, QR = 5 cm, RP = 13 cm, remains same throughout.
since, all sides are of different lengths. So, it is a
So, RS = PQ = 2.5 cm
scalene triangle.
Hence, RS = 2.5 cm
(iii) In MNP, MN = NP = PM = 7.5 cm, all sides are
of equal lengths. So, it is an equilateral triangle. 3. (i) Two lines perpendicular to the same line are
8. Let the angles of a triangle be x, 3x and 5x. parallel. (True)
 Sum of the three angles of a triangle = 180° (ii) Distance between two parallel lines is not same
everywhere. (False)
 x + 3x + 5x = 180°
(iii) No parallel line segments intersect. (True)
 9x = 180°
(iv) If l  m, n  m and l  n, then l || n. (True)
180
 x= = 20° (v) Railway tracks are parallel to each other. (True)
9
4. (i)
Hence, angles of the triangle are 20°, 60° and 100°. A
9. (i) P
D E
Q R
D F
Altitude = AD; Median = PF
( F is mid point of QR) B C
(ii) A DE || BC
(ii) S R
C
E B D
Altitude = AE; Median = AD P Q
( D is mid point of BC)
SR || PQ and SP || RQ
5. (iii) S R MENTAL MATHS CORNER
Fill in the blanks:
1. All the sides of a scalene triangle are of different
lengths.
2. The angles opposite to equal sides of an isosceles
P Q triangle are equal.
SR || PQ 3. The region inside the triangle is called its interior.
4. Parallel lines always lie in the same plane.
MULTIPLE CHOICE QUESTIONS 5. A triangle can have one right angle.
1. A perpendicular drawn from the vertex to the 6. An equiangular triangle is also called an equilateral
opposite side of a triangle is known as an altitude. triangle.
Hence, option (a) is correct. 7. A triangle has 3 altitudes and 3 medians.
2. The sum of the angles of a triangle is 180°. 8. A line intersecting two or more given lines at
Hence, option (d) is correct. different points is called a transversal.
3.  Sum of the angles of a triangle is 180°.
 Third angle = 180° – (55° + 35°) REVIEW EXERCISE
= 180° – 90° = 90° 1. Sides of a triangle are 7 cm, 5.5 cm and 6.5 cm.
Third angle = 90°  Perimeter of triangle = sum of the lengths of sides
Hence, option (b) is correct.  Perimeter of triangle = 7 cm + 5.5 cm + 6.5 cm
4. In PQR, P = 40°, Q = 50° = 19 cm
 P + Q + R = 180°
Hence, the perimeter of triangle is 19 cm.
(Angle sum property)
2. (i) 8.4, 16.1, 4.5
 40° + 50° + R = 180°
8.4 + 16.1 > 4.5
 R = 180° – 90° = 90°
8.4 + 4.5 < 16.1
Since, R is 90°. So, PQR is a right angled triangle.
Hence, option (c) is correct. 16.1 + 4.5 > 8.4
5. Total number of parts of a triangle is 6. (3 sides, 3 angles) Since, 8.4 + 4.5 < 16.1
Hence, option (a) is correct. So, the given numbers do not denote the sides of
6. A triangle has three altitudes. a triangle.
Hence, option (c) is correct. (ii) 7, 16, 8
7. The sum of the lengths of the sides of a triangle is 7 + 16 > 8
known as its perimeter. 7 + 8 < 16
Hence, option (b) is correct. 8 + 16 > 7
8. Line segments joining the vertices to the mid-points Since 7 + 8 < 16, so the given numbers do not
of opposite sides of a triangle are known as medians. denote the sides of a triangle.
Hence, option (c) is correct.
(iii) 7, 5, 8
9. A triangle whose two sides are equal is known as
isosceles. 7+5>8
Hence, option (c) is correct. 7+8>5
10. The angle formed 8+5>7
N
between North and  Sum of the lengths of any two sides is
Sourth-East directions is greater than the third side.
an obtuse angle. Hence, the given numbers denote the sides of a
Henc e, option (a) is triangle.
W E
correct.
S-E
S
Mathematics In Everyday Life-6 5
6. 3. S R In PQR, PQ = QR = 2.5 cm, PR = 4.2 cm.
Since, two sides PQ and QR are of equal length.
So, it is an isosceles triangle.
5. (i) A
P Q 58°
In PQS,
SPQ + PQS + QSP = 180° ...(i)
In QRS,
SQR + QRS + RSQ = 180° ...(ii)
Adding equations (i) and (ii), we get 58° 64°
B C
SPQ + PQS + QSP + SRQ
In ABC, A = B = 58°and each angle is less
+ SQR + RSQ = 360°
than 90°. Also, BC = AC.
or,
 LMN is an acute angled or an isosceles
SPQ + PQS + SQR + SRQ triangle.
+ PSQ + QSR = 360° (ii)
P
4. (i) A
60°
6 cm 7 cm
60° 60°
B C Q R
7 cm
In PQR, P = Q = R = 60° or QR = PR = PQ .
In ABC, AB = 6 cm, BC = AC = 7 cm.
Since, each angles less than 90°.
Since, two sides BC and AC are of equal length.
 PQR is an acute angled or an equilateral
So, it is an isosceles triangle. triangle.
(ii) P (iii)
X
m
11
52°
c
cm
11
Q R
11 cm
In PQR, PQ = PR = QR = 11 cm. Since, all the
sides are of equal length. 60° 68°
Y Z
So, it is an equilateral triangle.
In XYZ, X = 52°, Y = 60°, Z = 68° all are
(iii) P acute angles. Also, XY  YZ  ZX
So, XYZ is an acute angled or scalene triangle.
6. (i) 45°, 80°, 60°
4.2 cm  The sum of the angles of a triangle is 180°.
2.5 cm Sum of the given angles = 45° + 80° + 60°
= 185°  180°.
R So, it is not possible to have a triangle with these
Q 2.5 cm angles.
7. (ii) 30°, 50°, 90° 9.  l || m, QP  l and SR  l.
The sum of the given angles = 30° + 50° + 90°
Q S
= 170°  180° m
So, it is not possible to have a triangle with these
angles.
3.5 cm
(iii) 52°, 52°, 76°
The sum of the given angles = 52° + 52° + 76°
= 180° l
P R
Hence, it is possible to have a triangle with these
 The distance between two parallel lines always
given angles.
remains same throughout.
7. In XYZ, XY = YZ = ZX = 5 cm  SR = QP = 3.5 cm
It is an equilateral triangle. Hence, SR = 3.5 cm.
 X = Y = Z = x(let) 10.
C
In XYZ,
 X + Y + Z = 180°
x + x + x = 180° E
 3x = 180°
180
 x= = 60° A
3 B F
               X = Y = Z = 60°.
8. P
D
60°
Hence, AD, BE and CF are three altitudes of ABC.
HOTS QUESTIONS
1. If the three sides of a triangle are equal, it is
necessary that if any two of the sides are chosen,
50°
Q R they are also of the same length. Thus equilateral
S triangles are isosceles.
In PQR, This does not imply, that all isosceles triangles have
to be equilateral.
PQR + QRP + RPQ = 180° (Angle sum property)
2. (a) Line segment is a part of the line that has two
       PQS + 50° + 60° = 180° ( PQR = PQS) end points.
 PQS + 110° = 180° (b) Line is a straight path that goes on indefinitely
 PQS = 180° – 110° in two opposite directions.
(c) Ray is a straight path that goes on indefinitely in
 PQS = 70°
one direction.
In RPS, 
(d) AB means, a ray AB.
 RPS + PSR + SRP = 180°
(e) Parallel lines are lines that are same distance
   RPS + 90° + 50° = 180° ( SRP = QRP) apart and never meet.
   RPS + 140° = 180° Hence, (a)  (iv), (b)  (iii), (c)  (i), (d)  (ii),
(e)  (v)
 RPS = 180° – 140° = 40°
Hence, PQS= 70° and RPS= 40°.
Mathematics In Everyday Life-6 7
8. First of all, count all the triangles whose vertices are vertices of pentagon. There are exactly 10 of these,
one for every different sets of three vertices.
ACD, BDE, CEA, DAB, EBC, BCD, ABC, CDE, DEA, A
EAB.
Now, count triangles using exactly two adjacent vertices of the
pentagon and one inside the pentagon. Each pair of adjacent vertices
has three such triangles. So, there are 15 more. F G
E B
JCD, ICD, HCD, IBC, HBC, GBC, HAB, GAB, FAB,
GEA, FEA, JEA, FDE, JDE, IDE.
J H
Now, count triangle which are formed using two non-adjacent vertices I
of the pentagon and on inside vertex.
There is one triangle for each such pair, and there are 5 pairs. So we
get 5 more triangles. D C
FBD, GCE, HDA, IEB, JAC
Now, count the triangles using one pentagon vertex and two inside vertices.
There is only one for each vertex. So, 5 more.
DJI, EFJ, AGF, BHG, CIH
There are no triangle using only inside vertices.
Total number of triangles = 10 + 15 + 5 + 5 = 35.