NCERT Solutions for Class 7 Maths Chapter 6: Triangle and Its Properties

NCERT Solutions for Class 7 Maths Chapter 6

Exercise 6.1

Question 1
In Δ PQR, D is the mid-point of QR

PM is _________________.
PD is _________________.
Is QM = MR?
Solution
(i) Altitude
(ii) Median
(iii) No
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.
Solution
(a)

(b)

(c)

Question 3
Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
Solution

AD is the altitude in the Isosceles triangle and we can see that BD=DC, So it is also the median of the triangle

Exercise 6.2

Question 1
Find the value of the unknown exterior angle in the below figures

Solution
We know that an exterior angle of a triangle is equal to the sum of its interior opposite
angles.
(i) $x=50+70 =120^0$
(ii) $x=45+65 =110^0$
(iii) $x= 30+40=70^0$
(iv) $x= 60+60 =120^0$
(v) $x=50+50 =100^0$
(vi) $x=30+60=90^0$
Question 2
Find the value of the unknown interior angle x in the following figures:

Solution
(i) $115= x+50$
$x=115-50=65^0$
(ii) $100 + x=70$
$x=100-70=30^0$
(iii) $x+90=125$
$x=35^0$
(iv) $x+60=120$
$x=60^0$
(v) $x+30=80$
$x=50^0$
(vi) $x+35=75$
$x=40^0$

Exercise 6.3

Question 1

Find the value of the unknown x in the following diagrams:
Solution
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°
x=65⁰
(v) x + x + x = 180°
3x = 180°
x=60⁰
(vi) x + 2x + 90° = 180°
3x = 180° – 90° = 90⁰
x=30⁰

Question 2
Find the values of the unknown’s x and y in the following diagrams:


Solution:
(i) y + 120° = 180° (Linear pair)
y = 180° — 120⁰ = 60⁰
x+ y + 50° = 180° (Angle sum property)
x+ 60° + 50° = 180°
x=70⁰
(ii) y = 80° (Vertically opposite angles)
y + x + 50° = 180° (Angle sum property)
80° + x + 50° = 180°
x+ 130⁰ = 180°
x=50⁰
(iii) y + 50° + 60° = 180° (Angle sum property}
y = 180° — 60° – 50° = 70⁰
x+ y = 180 (Linear pair)
x= 180° — y = 180° — 70° = 110°
iv) x=60⁰ ( Vertically opposite angles)
Now
x+y+30 =180
60+y+30=180
y=90
v) x=90⁰ ( Vertically opposite angles)
Now,
x+x+y=180
2x+90=180
X=45⁰
vi)
y=x ( Vertically opposite angles)
Other two angles are also angle x as vertically opposite angles
y+x+x=180
x+x+x=180
3x=180
x=60⁰

Exercise 6.4

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
Solution
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 + 9 =12 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.

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?

Solution
Joining the points the OP, OP and OR
(i) In DOPQ
Sum of two sides > third side
OP + OQ > PQ
(ii) In DOQR
Sum of two sides > third side
OR + OQ > QR
(iii) In DOPR
Sum of two sides > third side
OP + OR > PR
Question 3
AM is a median of a triangle ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles DABM and DAMC.)

Solution
In a triangle, the sum of the lengths of either two sides is always greater than the third
side.
In DABM,
AB + BM > AM (1)
Similarly, in DACM,
AC + CM > AM (ii)
Adding equation (1) and (ii),
AB + BM + MC + AC > AM + AM
AB + BC + AC > 2AM

Question 4
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?

Solution
Considering DABC,
AB + BC > CA (i)
In DBCD,
BC + CD > DB (ii)
In DCDA,
CD + DA > AC (iii)
In DDAB,
DA + AB > DB (iv)
Adding equations (I), (//), (iii), and (iv), we obtain
AB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD
2AB + 2BC + 2CD +2DA > 2AC + 2BD
AB + BC + CD + DA > AC + BD
Question 5
ABCD is quadrilateral.
Is AB + BC + CD + DA < 2 (AC + BD)?
Solution
Let O be the point of intersection of diagonals

In a triangle, the sum of the lengths of either two sides is always greater than the third
side.
In DOAB
OA +OB > AB
In DOBC
OB + OC > BC
In DOCD
OC + OD > CD
In DOAD
OA + OD > AD
Adding all these we get
OA + OB + OB +OC + OC +OD + OA +OD > AB + BC + CD + DA
2( AC + BD) > AB + BC + CD + DA

Exercise 6.5

Question 1
PQR. is a triangle right angled at P. If PQ = 10 cm and PR. = 24 cm, find QR
Solution
In DPQR , right angles at P
PQ² + PR² = QR²
100 +576 = QR²
QR=26 cm
Question 2
ABC is a triangle, right-angled at C. If AB = 25cm and AC = 7 cm, find BC.
Solution
In DABC , right angles at C
AC² + BC² = AB²
BC² = AB² -AC²
BC² = 625 -49 =576
BC=24 cm
Question 3
A 15m 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.

Solution
By applying Pythagoras theorem,
(15)² = (12)² + a²
225 = 144 + a²
a² = 225 - 144 = 81
a = 9 m
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, 2cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.
Solution
i) 2.5cm,6.5cm,6cm
We see that
(2.5)²+6²=6.25+36=42.25= (6.5)²
Therefore, the given lengths can be the sides of a right triangle. Also, the angle between the lengths, 2.5 cm and 6 cm is a right angle
ii) 2 cm, 2 cm, 5 cm.
Here 2+ 2= 4 < 5 , So sum of two sides less than third side. So this is not a triangle
iii) 1.5 cm, 2cm, 2.5 cm
We see that
(1.5)²+2²=2.25+4=6.25=(2.5)²
Therefore, the given lengths can be the sides of a right triangle. Also, the angle between the lengths, 1.5 cm and 2 cm is a right angle

Question 5
A tree is broken at a height of 5 m from the ground and its top touches the ground at 12 m from the base of the tree. Find the original height of the tree
Solution
Let the tree be broken at point A
Then tree length = AB + AC

In DABC ,right angle at B
AB² + BC² = AC²
AC² = 25 +144 =169
AC=13 m
So original length of tree= 5+13 =18 m
Question 6
Angles Q and R of a DPQR are 25º and 65º.

Write which of the following is true:
(i) PQ² + QR² = RP²
(ii) PQ² + RP²= QR²
(iii) RP² + QR² = PQ²

Solution
From angle property of triangle
Angle P = 180 -25-65=90⁰
So D PQR is right angle at P
So
PQ² + RP²= QR²
Question 7
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Solution

Let ABCD is the rectangle as shown above
Now DABC is right angle triangle, So
BC² + AB² = AC²
AB² = 1681 -1600=81
AB=9 cm
So perimeter of rectangle
= 2( AB + BC) = 98 cm
Question 8
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution
Let ABCD is the rhombus and the diagonals bisect at right angles

In triangle OAD
$AD^2 = OA^2 + OD^2 = 64 + 225 = 289$
$AD = 17 cm$
So , perimeter of rhombus = 4 X AD= 68 cm

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