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In Δ PQR, D is the mid-point of QR

PM is _________________.

PD is _________________.

Is QM = MR?

(i) Altitude

(ii) Median

(iii) No

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.

(a)

(b)

(c)

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

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

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

(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) $x=30+60=90^0$

Find the value of the unknown interior angle x in the following figures:

(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) $x+35=75$

$x=40^0$

Find the value of the unknown x in the following diagrams:

(i) x + 50° + 60° = 180°

(ii) x + 90° + 30° = 180°

x +120° = 180°

(iii) x+30° +110° = 180°

(iv) 50° +

50° +2x = 180°

2x = 180°- 50° = 130°

x=65

(v) x + x +

3x = 180°

x=60

(vi)

3x = 180° – 90° = 90

x=30

Find the values of the unknown’s

(i) y + 120° = 180° (Linear pair)

y = 180° — 120

x+ y + 50° = 180° (Angle sum property)

x+ 60° + 50° = 180°

x=70

(ii) y = 80° (Vertically opposite angles)

y +

80° + x + 50° = 180°

x+ 130

x=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

Now

x+y+30 =180

60+y+30=180

y=90

v) x=90

Now,

x+x+y=180

2x+90=180

X=45

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=60

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

(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.

Take any point O in the interior of a triangle PQR. Is

(i) OP + OQ > PQ?

(ii) OQ + OR > QR?

(iii) OR + OP > RP?

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

AM is a median of a triangle ABC.

Is AB + BC + CA > 2 AM?

(Consider the sides of triangles DABM and DAMC.)

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

ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD?

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

ABCD is quadrilateral.

Is AB + BC + CD + DA < 2 (AC + BD)?

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.

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

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?

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

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

In DPQR , right angles at P

PQ

100 +576 = QR

QR=26 cm

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

In DABC , right angles at C

AC

BC

BC

BC=24 cm

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.

By applying Pythagoras theorem,

(15)

225 = 144 + a

a

a = 9 m

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.

i) 2.5cm,6.5cm,6cm

We see that

(2.5)

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)

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

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

Let the tree be broken at point A

Then tree length = AB + AC

In DABC ,right angle at B

AB

AC

AC=13 m

So original length of tree= 5+13 =18 m

Angles Q and R of a DPQR are 25º and 65º.

Write which of the following is true:

(i) PQ

(ii) PQ

(iii) RP

From angle property of triangle

Angle P = 180 -25-65=90

So D PQR is right angle at P

So

PQ

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

Let ABCD is the rectangle as shown above

Now DABC is right angle triangle, So

BC

AB

AB=9 cm

So perimeter of rectangle

= 2( AB + BC) = 98 cm

The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

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

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

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

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

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

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