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




In this page we have NCERT Solutions for Class 7 Maths Chapter 6: Triangle and Its Properties . Hope you like them and do not forget to like , social share and comment at the end of the page.

Exercise 6.1

Question 1
In Δ PQR, D is the mid-point of QR
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
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)
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
(b)
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
(c)
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
Question 3
Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
Solution
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
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
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
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
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
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:
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
NCERT Solutions for Class 7 Maths  Chapter 6: Triangle and Its Properties
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
Important question
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?
Solution
There are two things for the triangle
(i) sum of the lengths of any two sides of a triangle is greater than the length of the third side.
(ii) difference between the length of any two sides of a triangle is smaller than the length of the third side.
Let x be the third side
So, from first point
12+15 > x
Or x < 27 cm
Now from second point
15-12 < x
Or x > 3
So x lies between 3 cm and 27 cm

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

Similar type Practice questions

a) classify triangles based on the (i) sides (ii) angles.
b) What is a median
c) What can you say about each of the interior opposite angles, when the exterior angle is
(i) a right angle? (ii) an obtuse angle? (iii) an acute angle?
d) Two angles of a triangle are 45° and 80°. Find the third angle.
e) One of the angles of a triangle is 60° and the other two angles are equal. Find the measure of each of the equal angles.
f) The three angles of a triangle are in the ratio 1:2:3. Find all the angles of the triangle.
g) Can these be sides of the triangle
i) 2, 4, 7
ii) 3,4,5
iii) 10.2 , 5.8 , 4.5
h) What is the angle sum property of a triangle


Download NCERT Solutions Triangle and its properties PDF link to this page by copying the following text
Reference Books for class 7 Science

Given below are the links of some of the reference books for class 7 math.

  1. Science for Class 7 by Lakhmir Singh
  2. Science Foundation Course For JEE/NEET/NSO/Olympiad - Class 7
  3. CBSE All In One Science Class 7 by Arihant Experts (Author)
  4. IIT Foundation Physics, Chemistry & Maths for Class 7

You can use above books for extra knowledge and practicing different questions.


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