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

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Updated at 2021-02-11


NCERT solutions for class 7 Maths Chapter 6 The Triangle And Its Properties Exercise 6.1

Q.1 In ?PQR, D is the mid-point of QR¯ ,
(i) PM¯ is..........
(ii) PD¯ is...........
(iii) Is QM = MR?



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


Answer :

(i) PM¯ is altitude.
(ii) PD¯ is median.
(iii) No, QM ? MR.

Q.2 Draw rough sketches for the following:
(a) In ?ABC, BE is a median.
(ib) In ?PQR, PQ and PR are altitudes of the triangle.
(c) In ?XYZ, YL is an altitude in the exterior of the triangle.



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


Answer :

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



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


Answer :


Draw a Line segment AD ?BC. It is an altitude for this triangle. Here we observe that length of QS and SR is also same. So PS is also a median of this triangle.

NCERT solutions for class 7 Maths Chapter 6 The Triangle And Its Properties Exercise 6.2

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



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


Answer :

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



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


Answer :

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

NCERT solutions for class 7 Maths Chapter 6 The Triangle And Its Properties Exercise 6.3

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



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


Answer :

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

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

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

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

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

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



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


Answer :

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

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

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

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

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

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

NCERT solutions for class 7 Maths Chapter 6 The Triangle And Its Properties 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



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


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?



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


Answer :

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

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



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


Answer :

We know that,
The sum of the length of any two sides is always greater than the third side.
Now In ?ABM,
Here, AB + BM > AM … [equation i]
Then, In ?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?



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


Answer :

In ?ABC, we have
AB + BC > AC …(i) [Sum of any two sides is greater than the third side]
In ?BDC, we have
BC + CD > BD …(ii)
In ?ADC, we have CD + DA > AC .... (iii)
In ?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)?



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


Answer :


?AOB, we have AB < AO + BO …(i) [Any side of a triangle is less than the sum of other two sides]
In ?BOC, we have
BC < BO + CO …(ii)
In ?COD, we have
CD < CO + DO …(iii)
In ?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?



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


Answer :

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

NCERT solutions for class 7 Maths Chapter 6 The Triangle And Its Properties Exercise 6.5

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



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


Answer :


In right angled triangle PQR, we have
QR2=PQ2+PR2 [From Pythagoras property]
QR2=102+242
QR2=100+576=676
?QR=676=26cm 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.



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


Answer :


In right angled?ABC, we have
BC2+72=252(By Pythagoras property)
Or, BC2+49=625
Or, BC2=62549
Or, BC2=576
?BC=576=24cm
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.



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


Answer :

Here, the ladder forms a right angled triangle.
? a2+122=152(By Pythagoras property)
Or, a2+144=225
Or, a2=225144
Or, a2=81
? a = 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



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


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.
?The given sides form a right triangle.

(ii) Given sides are 2 cm, 2 cm, 5 cm .
Square of the longer side =(5)2=25cm
Sum of the square of other two sides =(2)2+(2)2=4+4=8cm Since 25 cm ?8 cm
?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=6.25cm
Sum of the square of other two sides = (1.5)2+(2)2=2.25+4=6.25cm=6.25cm=6.25cm
Since the square of longer side in a triangle is equal to the sum of square of other two sides.
?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.



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


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 ?CAD,
AD2+AC2=CD2 (By Pythagoras property)
Or, (12)2+(5)2=CD2
Or, 144 + 25 =CD2 Or, 169 = CD2
?CD =169 = 13 m
But CD = BC
AC + CB = AB
5 m + 13 m = AB
? 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) PQ2+QR2=RP2
(ii)PQ2+RP2=QR2
(iii)RP2+QR2=PQ2



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


Answer :

We know that
?P+?Q+?R=180°(Angle sum property)
?P + 25° + 65° = 180°
?P + 90° = 180°
?P = 180° – 90° – 90°
?PQR is a right triangle, right angled at P
(i) Not True
?PQ2+QR2?RP2 (By Pythagoras property)
(ii) True
? PQ2+RP2=QP2 (By Pythagoras property)
(iii) Not True
?RP2+QR2?PQ2(By Pythagoras property)

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



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


Answer :

Length of rectangle = 40cm
Diagonal = 41 cm
Breath2=412?402
Or, breath2=16811600
Or, breath = 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.



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


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°.
? OA = OC = 8 cm and OB = OD = 15 cm
In right ?OAB,
AB2=OA2+OB2 (By Pythagoras property)
= (8)2+(15)2=64+225 = 289
?AB =289= 17 cm
Since AB = BC = CD = DA (Property of rhombus)
?Required perimeter of rhombus = 4 × side = 4 × 17 = 68 cm.



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