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THE TRIANGLE AND ITS
PROPERTIES
Exercise 6.1
Q.1. In ΔPQR, D is the mid-point of QR .
PM is ___________.
PD is ____________.
Is QM = MR?
Ans. PM is altitude.
PD is median.
Hence, no, QM ≠ MR.
Q.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.
Ans.
Q.3. Verify by drawing a diagram if the median and altitude
of an isosceles triangle can be same.
Ans. Draw an isosceles triangle ABC.
Find out the mid-point of BC and mark
it by D. Most the vertex A to this
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mid-point D. AD is the median of
the triangle ABC.
Now, measure the angle ADC, which comes out 90°. It
means AD is perpendicular to the side BC. AD is also the
altitude of ΔABC.
Exercise 6.2
Q.1. Find the value of the unknown exterior angle x in the
following diagrams :
Ans. (i) x is an exterior angle, an exterior angle is the sum of its
two interior opposite angles.
So,
x = 50° + 70°
or
x = 120°
(by property)
Hence, exterior angle is 120°.
(ii) x is an exterior angle, an exterior angle is the sum of its
two interior opposite angles.
So,
x = 65° + 45°
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or
x = 110°
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(by property)
Hence, exterior angle is 110°.
(iii) x is an exterior angle, an exterior angle is the sum of its
two interior opposite angles.
So,
x = 30° + 40°
or
x = 70°
(by property)
Hence, exterior angle is 110°.
(iv) x is an exterior angle.
∴
x = 60° + 60°
or
x = 120°
(by property)
Hence, exterior angle is 120°.
(v) x is an exterior angle.
∴
x = 50° + 50°
or
x = 100°
(by property)
Hence, exterior angle is 100°.
(vi) x is an exterior angle.
∴
x = 60° + 30°
or
x = 90°
(by property)
Hence, exterior angle is 90°.
Q.2. Find the value of the unknown interior angle x in the
following figures :
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Ans. Exterior angle = Sum of interior opposite angles
(i)
115° = x + 50°
⇒ 115° – 50° = x
⇒
65° = x
So,
(ii)
x = 65°
70° + x = 100°
⇒
x = 100° – 70°
So,
x = 30°
(iii)
x + 90° = 125°
⇒
x = 125° – 90°
So,
x = 35°
(iv)
60° + x = 120°
⇒
x = 120° – 60°
So,
x = 60°
(v)
30° + x = 80°
⇒
x = 80° – 30°
So,
x = 50°
(vi)
x + 35° = 75°
⇒
x = 75° – 35°
So,
x = 40°
Exercise 6.3
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Q.1. Find the value of the unknown x in the following
diagrams :
Ans. (i)
(ii)
In ΔABC, ∠A + ∠B + ∠C = 180° (by angle sum
property of a triangle)
∴
x + 50° + 60° = 180°
or
x + 110° = 180°
or
x = 180° – 110°
or
x = 70°
In ΔPQR, ∠P + ∠Q + ∠R = 180° (by angle sum
property of a triangle)
∴
90° + 30° + x = 180°
or
120° + x = 180°
or
(iii)
x = 180° – 120° = 60°
In ΔXYZ, ∠X + ∠Y + ∠Z = 180° (by angle sum
property of a triangle)
∴
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30° + 110° + x = 180°
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or
140° + x = 180°
or
x = 180° – 140° = 40°
(iv) ∴
x + x + 50° = 180° (by property)
or
2x + 50° = 180°
or
(v)
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2x = 180° – 50° = 130°
130°
= 65°
2
or
x =
∴
x + x + x = 180°
or
(by property)
3x = 180°
180
x =
= 60°
3
x + 2x + 90° = 180° (by property)
or
(vi)
∴
3x = 180° – 90° = 90°
or
x =
90°
= 30°
3
Q.2. Find the values of the unknowns x and y in the following
diagrams :
Ans. (i) x + y + 50° = 180°
(by the angle sum property
of a triangle)
x + y = 180° – 50°
x + y = 130°
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120° = x + 50°
⇒
∴
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(by property of an
exterior angle)
x = 120° – 50° = 70°
70° + y = 130°
⇒
y = 130° – 70° = 60°
(ii)
y = 80°
x + y + 50° = 180°
or
x + 80° + 50° = 180°
or
x + 130° = 180°
(vertically opposite angles)
(by the angle sum property
of a triangle)
or
x = 180° – 130°
Hence,
x = 50° and y = 80°.
(iii)
y + 50° + 60° = 180° (by the angle sum property
of a triangle)
or
y + 110° = 180°
or
y = 180° – 110°
or
y = 70°
x + y = 180°
⇒
(Linear pair)
x + 70° = 180°
or
x = 180° – 70°
or
x = 110°
(iv)
x = 60°
x + y + 30° = 180°
(vertically opposite angles)
(by the angle sum property
of a triangle)
or 60° + y + 30° = 180°
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y + 90° = 180°
or
y = 180° – 90°
or
y = 90°
Hence,
x = 60° and y = 90°
(v)
y = 90°
∴
x + y + z = 180°
⇒
x + 90° + x = 180°
or
2x + 90° = 180°
(vertically opposite angles)
(by the angle sum property
of a triangle)
or
2x = 90°
or
x = 45°
Hence,
x = 45° and y = 90°.
(vi)
y = x
(vertically opposite angles)
∴
x + x + y = 180°
⇒
x + x + x = 180°
or
3x = 180°
or
x = 60°
So,
y = 60°
Hence,
x = 60° and y = 60°.
(by angle sum property of
a triangle)
Exercise 6.4
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
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Ans. We know that the sum of the lengths of any two sides of a
triangle is greater than the length of the third side.
(i)
2 cm + 3 cm = 5 cm = 5 cm
3 cm + 5 cm = 8 cm > 2 cm
Hence, it is not possible to have a triangle.
(ii)
3 cm + 6 cm = 9 cm > 7 cm
3 cm + 7 cm = 10 cm > 6 cm
6 cm + 7 cm = 13 cm > 3 cm
Hence, it is possible to have a triangle.
(iii)
6 cm + 3 cm = 9 cm > 2 cm
6 cm + 2 cm = 8 cm > 3 cm
2 cm + 3 cm = 5 cm < 6 cm
Hence, it is not possible to have a triangle.
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?
Ans. The sum of the lengths of any two sides of a triangle is
greater than the length of the third side.
(i) In ΔOPQ, yes, OP + OQ > PQ
(ii) In ΔORQ, yes, OR + OQ > QR
(iii) In ΔPOR, yes, OR + OP > PR
Q.3. AM is a median of a triangle ABC.
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Is AB + BC + CA > 2 AM?
(Consider the sides of triangles ΔABM and ΔAMC).
Ans. The sum of the lengths of any two sides of a triangle is
greater than the length of the third side.
In ΔAMB,
AB + BM > AM ----- (i)
In ΔAMC
AC + CM > AM ----- (ii)
Adding (i) and (ii), we have
AB + BM + AC + CM > AM + AM
or
AB + AC + (BM + CM) > 2 AM
or
AB + AC + BC > 2 AM
or
AB + BC + CA > 2AM
Q.4. ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?
Ans. Given : A quadrilateral ABCD in which AC and BD are
diagonals.
To prove : AB + BC + CD + DA > AC + BD
Proof : In ΔABD,
AB + DA > BD ----- (i)
In ΔBDC,
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BC + CD > BD ----- (ii)
In ΔADC,
DA + CD > AC -----(iii)
In ΔABC,
AB + BC > AC ----- (iv)
Adding (i), (ii), (iii) and (iv), we have
AB + DA + BC + CD + DA + CD + AB + BC
> BD + AC + AC + BD
or
2 [AB + BC + CD + DA] > 2[BD + AC]
or
AB + BC + CD + DA > BD + AC
Q.5. ABCD is a quadrilateral.
Is AB + BC + CD + DA < 2 (AC + BD)?
Ans. Given : A quadrilateral ABCD in which AC and BD are
diagonals.
To prove : AB + BC + CD + DA < 2(AC + BD)
Proof : In ΔAOB,
OB + OA > AB ------ (i)
In ΔBOC,
OB + OC > BC ------ (ii)
In ΔCOD,
OD + OC > CD ------ (iii)
In ΔAOD,
OA + OD > DA ------ (iv)
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Adding (i), (ii), (iii) and (iv), we have
OB + OA + OB + OC + OD + OC + OA + OD >
AB + BC + CD + DA
or
2OB + 2OD + 2OC + 2OA > AB + BC + CD
+ DA
or
2 (OB + OD) + 2 (OC + OA) > AB + BC + CD
+ DA
or
2 (BD + AC) > AB + BC + CD + DA
or
AB + BC + CD + AD < 2 (BD + AC)
Q.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?
Ans. Let x cm be the length of the third side. We know that the
sum of the lengths of any two sides of a triangle is greater
than the length of the third side.
∴ 12 + 15 > x, so, 27 > x
x + 12 > 15, so, x > 3
x + 15 > 12, so, x > – 3
The numbers between 3 and 27 satisfy these.
∴ The length of the third side could be any length between
3 cm and 27 cm.
Exercise 6.5
Q.1. PQR is a triangle, right-angled at P. If PQ = 10 cm and
PR = 24 cm, find QR.
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Ans. In ΔPQR, by Pythagoras Property,
(QR)2 = (PR)2 + (PQ)2
⇒
(QR)2 = (24)2 + (10)2
⇒
(QR)2 = 576 + 100
⇒
(QR)2 = 676
⇒
QR =
676 = 26 cm
Q.2. ABC is a triangle, right-angled at C. If AB = 25 cm and
AC = 7 cm, find BC.
Ans. In ΔABC, according to Pythagoras Property,
(AB)2 = (AC)2 + (BC)2
⇒
(25)2 = 72 + (BC)2
⇒
625 – 49 = (BC)2
⇒
576 = (BC)2
⇒
576 = BC
⇒
24 = BC
∴
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.
Ans. Let the distance of the ladder from the wall is a m.
Now, in given right triangle,
(15)2 = (12)2 + a2 (By Pythagoras property)
⇒
225 – 144 = a2
⇒
81 = a2
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or
81 = a
or
9 = a
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or a = 9 m
Hence, the distance of the foot of the ladder from the
wall is 9 m.
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
In the case of right-angled triangles, identify the
right angles.
Ans. (i)
Let a = 2.5 cm, c = 6.5 cm, b = 6 cm
c2 = a2 + b2
(by Pythagoras property)
⇒
(6.5)2 = (2.5)2 + (6)2
⇒
42.25 = 6.25 + 36 = 42.25
Hence, Pythagoras property holds, we can say it will be
right triangle and right angle will be the angle between side
a and b.
(ii) 2 cm, 2 cm, 5 cm
Let a = 2 cm,
c2 = a2 + b2
⇒
b = 2 cm,
c = 5 cm
(By Pythagoras property)
(5)2 = (2)2 + (2)2
⇒
25 ≠ 4 + 4
⇒
25 ≠ 8
Hence, Pythagoras property does not hold, we can say it is
not right triangle.
(iii)
Let a = 1.5 cm,
b = 2 cm,
c2 = a2 + b2
⇒
c = 2.5 cm
(By Pythagoras property)
(2.5)2 = (1.5)2 + (2)2
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⇒
6.25 = 2.25 + 4
⇒
6.25 = 6.25
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Hence, Pythagoras property holds, we can say it is right
triangle and right angle is formed between sides a and b.
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.
Ans.
AC = CD
(Given)
In right angled triangle DBC,
DC2 = BC2 + BD2 (by Pythagoras Property)
= 52 + 122
= 25 + 144 = 169
⇒
DC =
⇒
AC = 13 m
∴
AB = AC + BC
⇒
AB = 13 m + 5 m = 18 m
169 = 13 m
Hence, the original height of the tree = 18 m
Q.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
Ans.
In ΔPQR,
∠P + ∠Q + ∠R = 180°
(Sum of the angles of a triangle is 180°)
⇒
∠P + 25° + 65° = 180°
⇒
∠P + 90° = 180°
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⇒
∴
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∠P = 180° – 90° = 90°
Triangle PQR is a right angled triangle.
Now, in right angled triangle PQR, by Pythagoras Property,
QR2 = PQ2 + PR2
Hence, (ii) is true.
Q.7. Find the perimeter of the rectangle whose length is
40 cm and a diagonal is 41 cm.
Ans.
Let ABCD be a rectangle
Length (l) = 40 cm
(Given)
Diagonal = 41 cm
(Given)
We know that perimeter of rectangle = 2 (l + b) -------- (i)
Where, l = length and b = breadth
In ΔABC, according to Pythagoras property
(AC)2 = (AB)2 + (BC)2
⇒
(41)2 = (40)2 + (BC)2
⇒
(BC)2 = (41)2 – (40)2
⇒
(BC)2 = 1681 – 1600
⇒
(BC)2 = 81
So,
Thus,
BC = 9 cm (where b = BC)
b = breadth = 9 cm
Putting the values of l and b in equation (i), we have
= 2[40 + 9] = 2 × 49 = 98 cm
Hence, Perimeter of rectangle is 98 cm.
Q.8. The diagonals of a rhombus measure 16 cm and 30 cm.
Find its perimeter.
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Ans. Since diagonals of a rhombus bisect each other
perpendicularly.
∴
and
1
1
BD =
× 16 = 8 cm
2
2
1
1
AC =
× 30 = 15 cm
OC =
2
2
OB =
Now, in right ΔBOC,
BC2 = OB2 + OC2
= 82 + 152 = 64 + 225
= 289
∴
BC =
∴
All sides of rhombus are equal
∴
AB = BC = CD = AD = 17 cm
289 = 17 cm
Now, perimeter = 4 × 17 cm = 68 cm
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