NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties

Q.1 Find the value of the unknown exterior angle x in the following diagrams:

Answer :

(i) \(\angle \)x = 50° + 70° = 120° (Exterior angle is equal to sum of its interior opposite angles)
(ii) \(\angle \)x = 65°+ 45° = 110° (Exterior angle is equal to sum of its interior opposite angles)
(iii) \(\angle \)x = 30° + 40° = 70° (Exterior angle is equal to sum of its interior opposite angles)
(iv) \(\angle \)x = 60° + 60° = 120° (Exterior angle is equal to sum of its interior opposite angles)
(v) \(\angle \)x = 50° + 50° =100° (Exterior angle is equal to sum of its interior opposite angles)
(vi) \(\angle \)x = 30° + 60° = 90° (Exterior angle is equal to sum of its interior opposite angle)

Q.2 Find the value of the unknown interior angle x in the following figures:

Answer :

(i) \(\angle \)x + 50° = 115° (Exterior angle of a triangle)
\(\therefore \) \(\angle \)x = 115°- 50° = 65°
(ii) \(\angle \)x + 70° = 110° (Exterior angle of a triangle)
\(\therefore \) \(\angle \)x = 110° – 70° = 40°
(iii) \(\angle \) x + 90° = 125° (Exterior angle of a right triangle)
\(\therefore \) \(\angle \)x = 125° – 90° = 35°
(iv) \(\angle \)x + 60° = 120° (Exterior angle of a triangle)
\(\therefore \) \(\angle \)x = 120° – 60° = 60°
(v)\(\angle \) X + 30° = 80° (Exterior angle of a triangle)
\(\therefore \) \(\angle \)x = 80° – 30° = 50°
(vi) \(\angle \) x + 35° = 75° (Exterior angle of a triangle)
\(\therefore \) \(\angle \) x = 75° – 35° = 40°

Q.1 Find the value of the unknown x in the following diagrams:

Answer :

(i) we know that sum of all angle of a triangle = 180
\(\angle \) x + 50° + 60° = 180°
Or, \(\angle \) x + 110° = 180°
\(\therefore \) \(\angle \)x = 180° – 110° = 70°

(ii) we know that sum of all angle of a triangle = 180
\(\angle \)x + 90° + 30 = 180° [\(\triangle \) is right angled triangle]
Or, \(\angle \)x + 120° = 180°
Or, \(\angle \)x – 180° – 120° = 60°
(iii) we know that sum of all angle of a triangle = 180
\(\angle \)x + 30° + 110° =180°
Or, \(\angle \)x + 140° = 180°
\(\therefore \) \(\angle \)x = 180° – 140° = 40°

(iv ) we know that sum of all angle of a triangle = 180
\(\angle \)x + \(\angle \)x + 50° = 180°
Or, 2x + 50° = 180°
Or, 2x = 180° – 50°
Or, 2x = 130°
\(\therefore \) x=130/2=65°

(v) we know that sum of all angle of a triangle = 180
\(\angle \)x + \(\angle \)x +\(\angle \)x =180°
Or, 3 \(\angle \)x = 180°
\(\therefore \) \(\angle \)x=180°/3=60°

(vi) we know that sum of all angle of a triangle = 180
x + 2 x + 90° = 180° (\(\triangle \) is right angled triangle)
Or, 3x + 90° = 180°
Or, 3x = 180° – 90°
Or, 3x = 90°
\(\therefore \) x=90/3=30°

Q. 2 Find the values of the unknowns x and y in the following diagrams:

Answer :

(i) \(\angle \)x + 50° = 120° (Exterior angle of a triangle)
\(\therefore \) \(\angle \)x = 120°- 50° = 70°
\(\angle \)x + \(\angle \)y + 50° = 180° (Angle sum property of a triangle)
70° + \(\angle \)y + 50° = 180°
\(\angle \)y + 120° = 180°
\(\angle \)y = 180° – 120°
\(\therefore \) \(\angle \)y = 60°
Thus \(\angle \)x = 70 and \(\angle \)y – 60°

(ii) \(\angle \)y = 80° (Vertically opposite angles are same)
\(\angle \)x + \(\angle \)y + 50° = 180° (Angle sum property of a triangle)
\(\therefore \) \(\angle \)x + 80° + 50° = 180°
\(\therefore \) \(\angle \)x + 130° = 180°
\(\therefore \)\(\angle \)x = 180° – 130° = 50°
Thus, \(\angle \)x = 50° and \(\angle \)y = 80°

(iii) \(\angle \)y + 50° + 60° = 180° (Angle sum property of a triangle)
\(\angle \)y + 110° = 180°
\(\therefore \)\(\angle \)y = 180°- 110° = 70°
\(\angle \)x + \(\angle \)y = 180° (Linear pairs)
Or, \(\angle \)x + 70° = 180°
\(\therefore \) \(\angle \)x = 180° – 70° = 110°
Thus, \(\angle \)x = 110° and y = 70°

(iv) \(\angle \)x = 60° (Vertically opposite angles)
\(\angle \)x + \(\angle \)y + 30° = 180° (Angle sum Or, 60° + \(\angle \)y + 30° = 180°
Or, \(\angle \)y + 90° = 180°
Or, \(\angle \)y = 180° – 90° = 90°
Thus, \(\angle \)x = 60° and \(\angle \)y = 90°

(v) \(\angle \)y = 90° (Vertically opposite angles)
\(\angle \)x + \(\angle \)x + \(\angle \)y = 180° (Angle sum property of a triangle)
Or, 2 \(\angle \)x + 90° = 180°
Or, 2\(\angle \)x = 180° – 90°
Or, 2\(\angle \)x = 90°
\(\therefore \) \(\angle \)x=90°/2=45°
Thus, \(\angle \)x = 45° and \(\angle \)y = 90

(vi) From the given figure, we have
\(\angle \)y = \(\angle \)x
\(\angle \)1 = \(\angle \)x
\(\angle \)2 = \(\angle \)x
Adding both sides, we have Adding both sides, we have Adding both sides, we have \(\angle \)y + \(\angle \)1 + \(\angle \)2 = 3\(\angle \)x
Or, 180° = 3\(\angle \)x [Angle sum property of a triangle]
\(\therefore \) \(\angle \)x=180°/3=60°
\(\angle \)x = 60°, \(\angle \)y = 60°

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

Answer :

We know that for a triangle, the sum of any two sides must be greater than the third side.
(i) Given sides are 2 cm, 3 cm, 5 cm
(2 + 3) = 5
5 = 5
Hence, the triangle is not possible.
(i) Given sides are 3 cm, 6 cm, 7 cm
(3 + 6) = 9 > 7
(6 + 7) = 13 > 3
(7 + 3) = 10 > 6
Hence, the triangle is possible.
(iii) Given sides are 6 cm, 3 cm, 2 cm
(3 + 2) = 5 < 6
Hence, the triangle is not possible.

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?

Answer :

(i) Yes, \(\triangle \)OPQ has sides OP, OQ and PQ.
So, OP + OQ > PQ
(ii) Yes, \(\triangle \)OQR has sides OR, OQ and QR.
So, OQ + OR > QR
(iii) Yes, \(\triangle \)ORP has sides OR, OP and PR.
So, OR + OP > RP

Q.3 AM is a median of a \(\triangle \) ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles \(\triangle \)ABM and \(\triangle \)AMC.)

Answer :

We know that,
The sum of the length of any two sides is always greater than the third side.
Now In \(\triangle \)ABM,
Here, AB + BM > AM … [equation i]
Then, In \(\triangle \)ACM
Here, AC + CM > AM … [equation ii]
By adding equation [i] and [ii] we get,
AB + BM + AC + CM > AM + AM
From the figure we have, BC = BM + CM
AB + BC + AC > 2 AM
Hence, the given expression is true.

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

Answer :

In \(\triangle \)ABC, we have
AB + BC > AC …(i) [Sum of any two sides is greater than the third side]
In \(\triangle \)BDC, we have
BC + CD > BD …(ii)
In \(\triangle \)ADC, we have CD + DA > AC .... (iii)
In \(\triangle \)DAB, we have DA + AB > BD …(iv)
Adding eq. (i), (ii), (iii) and (iv), we get
2AB + 2BC + 2CD + 2DA > 2AC + 2BD
Or, AB + BC + CD + DA > AC + BD [Dividing both sides by 2]
Hence, proved

Q.5 ABCD is a quadrilateral.
Is AB + BC + CD + DA < 2(AC + BD)?

Answer :

\(\triangle \)AOB, we have AB < AO + BO …(i) [Any side of a triangle is less than the sum of other two sides]
In \(\triangle \)BOC, we have
BC < BO + CO …(ii)
In \(\triangle \)COD, we have
CD < CO + DO …(iii)
In \(\triangle \)AOD, we have
DA < DO + AO …(iv)
Adding eq. (i), (ii), (iii) and (iv), we have
AB + BC + CD + DA < 2AO + 2BO + 2CO + 2DO
Or, AB + BC + CD + DA < 2(AO + BO + CO + DO)
Or, AB + BC + CD + DA < 2 [(AO + CO) + (BO + DO)]
Or, AB + BC + CD + DA < 2(AC + BD)
Thus, AB + BC + CD + DA < 2(AC + BD)
Hence, proved.

Q.6 The length of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Answer :

Sum of two sides
= 12 cm + 15 cm = 27 cm
Difference of the two sides
= 15 cm – 12 cm = 3 cm
\(\therefore \) The measure of third side should fall between 3 cm and 27 cm.

Q.1 PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Answer :

In right angled triangle PQR, we have
\(QR^2 = PQ^2 + PR^2 \) [From Pythagoras property]
\(QR^2 = 10^2 + 24^2 \)
\(QR^2 = 100 + 576 = 676 \)
\(\therefore QR = \sqrt{676} = 26 cm \) The, the required length of QR = 26 cm.

Q.2 ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Answer :

In right angled?ABC, we have
\( BC^2 + 7^2 = 25^2 \)(By Pythagoras property)
Or, \( BC^2 + 49 = 625 \)
Or, \( BC^2 = 625 – 49 \)
Or, \( BC^2 = 576 \)
\(\therefore BC = \sqrt{576} = 24 cm \)
Thus, the required length of 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.

Answer :

Here, the ladder forms a right angled triangle.
\(\therefore \) \(a^2 + 12^2 = 15^2 \)(By Pythagoras property)
Or, \(a^2+ 144 = 225\)
Or, \(a^2 = 225 – 144 \)
Or, \(a^2 = 81 \)
\(\therefore \) a = \(\sqrt{81}= 9 \) m
Thus, the distance of the foot from the ladder = 9m

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

Answer :

(i) Given sides are 2.5 cm, 6.5 cm, 6 cm.
Square of the longer side = \((6.5)^2 = 42.25 \) cm.
Sum of the square of other two sides = \( (2.5)^2 + (6)^2 = 6.25 + 36 \) = 42.25 cm.
Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.
\(\therefore \)The given sides form a right triangle.

(ii) Given sides are 2 cm, 2 cm, 5 cm .
Square of the longer side =\( (5)^2 = 25 \)cm
Sum of the square of other two sides =\( (2)^2 + (2)^2 =4 + 4 = 8 cm \) Since 25 cm \(\neq \)8 cm
\(\therefore \)The given sides do not form a right triangle.

(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm
Square of the longer side = \( (2.5)^2 = 6.25 \)cm
Sum of the square of other two sides = \((1.5)^2 + (2)^2 = 2.25 + 4 = 6.25cm = 6.25 cm = 6.25 cm \)
Since the square of longer side in a triangle is equal to the sum of square of other two sides.
\(\therefore \)The given sides form a right triangle.

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.

Answer :

Let AB be the original height of the tree and broken at C touching the ground at D such that
AC = 5 m
and AD = 12 m
In right triangle \(\triangle \)CAD,
\(AD^2 + AC^2 = CD^2 \) (By Pythagoras property)
Or, \( (12)^2 + (5)^2 = CD^2 \)
Or, 144 + 25 =\( CD^2 \) Or, 169 = \( CD^2\)
\(\therefore \)CD =\(\sqrt{169} \) = 13 m
But CD = BC
AC + CB = AB
5 m + 13 m = AB
\(\therefore \) AB = 18 m .
Thus, the original height of the tree = 18 m.

Q.6 Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.
(i) \( PQ^2 + QR^2 = RP^2 \)
(ii)\( PQ^2 + RP^2 = QR^2 \)
(iii)\( RP^2 + QR^2 = PQ^2 \)

Answer :

We know that
\(\angle P + \angle Q + \angle R = 180° \)(Angle sum property)
\(\angle \)P + 25° + 65° = 180°
\(\angle \)P + 90° = 180°
\(\angle \)P = 180° – 90° – 90°
\(\triangle \)PQR is a right triangle, right angled at P
(i) Not True
\(\therefore PQ^2 + QR^2 \neq RP^2 \) (By Pythagoras property)
(ii) True
\(\therefore \) \(PQ^2 + RP^2 = QP^2 \) (By Pythagoras property)
(iii) Not True
\(\therefore RP^2 + QR^2 \neq PQ^2 \)(By Pythagoras property)

Q.7 Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Answer :

Length of rectangle = 40cm
Diagonal = 41 cm
\(Breath^2 = 41^2 - 40^2 \)
Or, \( breath^2= 1681 – 1600 \)
Or, breath = \(\sqrt{81} \) cm
Breath = 9 cm
Perimeter of rectangle = 2×(40+9) = 2×49 = 98 cm

Q.8 The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer :

Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm
Since, the diagonals of a rhombus bisect each other at 90°.
\(\therefore \) OA = OC = 8 cm and OB = OD = 15 cm
In right \(\triangle \)OAB,
\(AB^2 = OA^2 + OB^2 \) (By Pythagoras property)
= \((8)^2+ (15)^2 = 64 + 225 \) = 289
\(\therefore \)AB =\(\sqrt{289} \)= 17 cm
Since AB = BC = CD = DA (Property of rhombus)
\(\therefore \)Required perimeter of rhombus = 4 × side = 4 × 17 = 68 cm.