NCERT Solutions Class 7 Mathematics Chapter 6 Triangle and its properties have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 7 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 7 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 7 Mathematics are an important part of exams for Class 7 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 7 Mathematics and also download more latest study material for all subjects. Chapter 6 Triangle and its properties is an important topic in Class 7, please refer to answers provided below to help you score better in exams
Chapter 6 Triangle and its properties Class 7 Mathematics NCERT Solutions
Class 7 Mathematics students should refer to the following NCERT questions with answers for Chapter 6 Triangle and its properties in Class 7. These NCERT Solutions with answers for Class 7 Mathematics will come in exams and help you to score good marks
Chapter 6 Triangle and its properties NCERT Solutions Class 7 Mathematics
Exercise 6.1
Q.1) In Δ PQR, D is the mid-point of QR.
PM is _______
PD is ________
Is QM = MR?
Sol.1) Given: QD = DR
∴ PM is altitude.
PD is median.
No, 𝑄𝑀 ≠ 𝑀𝑅 as D is the mid-point of QR
Q.2) Draw rough sketches for the following:
(a) In Δ 𝐴𝐵𝐶, BE is a median.
(b)In Δ 𝑃𝑄𝑅, PQ and PR are altitudes of the triangle.
(c) In Δ 𝑋𝑌𝑍, YL is an altitude in the exterior of the triangle.
Sol.2) (a) Here, BE is a median in Δ𝐴𝐵𝐶 and 𝐴𝐸 = 𝐸𝐶
Q.3) Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same
Sol.3) Isosceles triangle means any two sides are same.
Take Δ𝐴𝐵𝐶 and draw the median when 𝐴𝐵 = 𝐴𝐶.
AL is the median and altitude of the given triangle.
Exercise 6.2
Q.1) Find the value of the unknown exterior angle 𝑥 in the following diagrams:
Sol.1) Since, Exterior angle = Sum of interior opposite angles, therefore
(i) 𝑥 = 50° + 70° = 120°
(ii) 𝑥 = 65° + 45° = 110°
(iii) 𝑥 = 30° + 40° + 170°
(iv) 𝑥 = 60° + 60° = 120°
(v) 𝑥 = 50° + 50° = 100°
(vi) 𝑥 = 60° + 30° = 90°
Q.2) Find the value of the unknown interior angle𝑥in thefollowing figures:
Sol.2) Since, Exterior angle = Sum of interior opposite angles, therefore
(i) 𝑥 + 50° = 115° ⇒𝑥 = 115° − 50° = 65°
(ii) 70° + 𝑥 = 100° ⇒𝑥 = 100°−70° = 30°
(iii) 𝑥 + 90° = 125° ⇒𝑥 = 125° − 90° = 35°
(iv) 60° + 𝑥 = 120° ⇒𝑥 = 120° − 60° = 60°
(v) 30° + 𝑥 = 80° ⇒𝑥 = 80° − 30° = 50°
(vi) 𝑥 + 35° = 75° ⇒𝑥 = 75° − 35° = 40°
Exercise 6.3
Q.1) Find the value of unknown 𝑥 in the following diagrams:
Sol.1) (i) In ∠𝐴𝐵𝐶, Δ 𝐵𝐴𝐶 + ∠ 𝐴𝐶𝐵 + ∠ 𝐴𝐵𝐶 = 180°[By angle sum property of a triangle]
⇒ 𝑥 + 50° + 60° = 180°
⇒ 𝑥 + 110° = 180°
⇒ 𝑥 = 180° − 110° = 70°
(ii) In Δ 𝑃𝑄𝑅,∠ 𝑅𝑃𝑄 + ∠ 𝑃𝑄𝑅 + ∠ 𝑅𝑃𝑄 = 180° [By angle sum property of a triangle]
⇒ 90° + 30° + 𝑥 = 180°
⇒ 𝑥 + 120° = 180°
⇒ 𝑥 = 180° − 120° = 60°
(iii) In Δ 𝑋𝑌𝑍,∠ 𝑍𝑋𝑌 + ∠ 𝑋𝑌𝑍 + ∠ 𝑌𝑍𝑋 = 180° [By angle sum property of a triangle]
⇒ 30° + 110° + 𝑥 = 180°
⇒ 𝑥 + 140° = 180°
⇒ 𝑥 = 180° − 140° = 40°
(iv) In the given isosceles triangle,
𝑥 + 𝑥 + 50° = 180° [By angle sum property of a triangle]
⇒ 2𝑥 + 50° = 180°
⇒ 2𝑥 = 180° − 50°
⇒ 2𝑥 = 130°
⇒ 𝑥 = 130°/2 = 65°
(v) In the given equilateral triangle,
𝑥 + 𝑥 + 𝑥 = 180° [By angle sum property of a triangle]
⇒ 3𝑥 = 180°
⇒ 𝑥 = 180°/3 = 60° =
(vi) In the given right angled triangle,
𝑥 + 2𝑥 + 90° = 180° [By angle sum property of a triangle]
⇒ 3𝑥 + 90° = 180
⇒ 3𝑥 = 180° − 90°
⇒ 3𝑥 = 90°
⇒ 𝑥 = 90°/3 = 30°
Q.2) Find the values of the unknowns 𝑥 and 𝑦 in the following diagrams:
Sol.2) (i) 50° + 𝑥 = 120° [Exterior angle property of a Δ ]
𝑥 = 120° − 50° = 70°
Now, 50° + 𝑥 + 𝑦 = 180° [Angle sum property of a Δ ]
⇒ 50° + 70° + 𝑦 = 180°
⇒ 120° + 𝑦 = 180°
⇒ 𝑦 = 180° − 120° = 60°
(ii) 𝑦 = 80° ……….(i) [Vertically opposite angle]
Now, 50° + 𝑥 + 𝑦 = 180° [Angle sum property of a Δ ]
50° + 80° + 𝑦 = 180°[From equation (i)]
130° + 𝑦 = 180°
⇒ 𝑦 = 180° − 130° = 50°
(iii) 50° + 60° = 𝑥[Exterior angle property of a Δ ] 110° = x
⇒ 𝑥 = 110°
Now 50° + 60° + 𝑦 = 180° [Angle sum property of a Δ ]
⇒110° + 𝑦 = 180°
⇒ 𝑦 = 180° − 110°
⇒ 𝑦 = 170°
(iv) 𝑥 = 60° ……….(i) [Vertically opposite angle]
Now, 30° + 𝑥 + 𝑦 = 180° [Angle sum property of a Δ ]
⇒50° + 60° + 𝑦 = 180° [From equation (i)]
⇒90° + 𝑦 = 180°
⇒ 𝑦 = 180° − 90° = 90°
(v) 𝑦 = 90° ……….(i) [Vertically opposite angle]
Now, 𝑦 + 𝑥 + 𝑥 = 180° [Angle sum property of a Δ ]
⇒ 90° + 2𝑥 = 180° [From equation (i)]
⇒ 2𝑥 = 180° − 90°
⇒ 2𝑥 = 90°
⇒ 𝑥 = 45°/2
(vi) 𝑥 = 𝑦 ……….(i) [Vertically opposite angle]
Now, 𝑥 + 𝑥 + 𝑦 = 180° [Angle sum property of a Δ ]
2𝑥 + 𝑥 = 180° [From equation (i)]
⇒ 3𝑥 = 180°
⇒ 𝑥 = 180°/3 = 60°
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
Sol.1) Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side.
(i) 2 cm, 3 cm, 5 cm
2 + 3 > 5 No
2 + 5 > 3 Yes
3 + 5 > 2 Yes
This triangle is possible
(ii) 3 cm, 6 cm, 7 cm
3 + 6 > 7 Yes
6 + 7 > 3 Yes
3 + 7 > 6 Yes
This triangle is not possible.
(iii) 6 cm, 3 cm, 2 cm
6 + 3 > 2 Yes
6 + 2 > 3 Yes
2 + 3 > 6 No
This 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 ?
Sol.2) Join OR, OQ and OP.
(i) Is OP + OQ > PQ ?
Yes, POQ form a triangle.
(ii) Is OQ + OR > QR ?
Yes, RQO form a triangle.
(iii) Is OR + OP > RP ? P
Yes, ROP form a triangle
Q.3) AM is a median of a triangle ABC. Is AB + BC + CA > 2AM? (Consider the sides of triangles Δ 𝐴𝐵𝑀 and Δ 𝐴𝑀𝐶.)
Sol.3) Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore, In Δ 𝐴𝐵𝑀, 𝐴𝐵 + 𝐵𝑀 > 𝐴𝑀 ... (i)
In Δ 𝐴𝑀𝐶, 𝐴𝐶 + 𝑀𝐶 > 𝐴𝑀 ... (ii)
Adding eq. (i) and (ii),
𝐴𝐵 + 𝐵𝑀 + 𝐴𝐶 + 𝑀𝐶 > 𝐴𝑀 + 𝐴𝑀
⇒ 𝐴𝐵 + 𝐴𝐶 + (𝐵𝑀 + 𝑀𝐶) > 2𝐴𝑀
⇒ 𝐴𝐵 + 𝐴𝐶 + 𝐵𝐶 > 2𝐴𝑀
Hence, it is true.
Q.4) ABCD is a quadrilateral. Is 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 > 𝐴𝐶 + 𝐵𝐷?
Sol.4) Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore, In Δ 𝐴𝐵𝐶, 𝐴𝐵 + 𝐵𝐶 > 𝐴𝐶 ……….(i)
In Δ 𝐴𝐷𝐶, 𝐴𝐷 + 𝐷𝐶 > 𝐴𝐶 ……….(ii)
In Δ 𝐷𝐶𝐵, 𝐷𝐶 + 𝐶𝐵 > 𝐷𝐵 ……….(iii)
In Δ 𝐴𝐷𝐵, 𝐴𝐷 + 𝐴𝐵 > 𝐷𝐵 ……….(iv)
Adding equations (i), (ii), (iii) and (iv), we get
𝐴𝐵 + 𝐵𝐶 + 𝐴𝐷 + 𝐷𝐶 + 𝐷𝐶 + 𝐶𝐵 + 𝐴𝐷 + 𝐴𝐵 > 𝐴𝐶 + 𝐴𝐶 + 𝐷𝐵 + 𝐷𝐵
⇒ (𝐴𝐵 + 𝐴𝐵) + (𝐵𝐶 + 𝐵𝐶) + (𝐴𝐷 + 𝐴𝐷) + (𝐷𝐶 + 𝐷𝐶) > 2𝐴𝐶 + 2𝐷𝐵
⇒ 2𝐴𝐵 + 2𝐵𝐶 + 2𝐴𝐷 + 2𝐷𝐶 > 2(𝐴𝐶 + 𝐷𝐵)
⇒ 2(𝐴𝐵 + 𝐵𝐶 + 𝐴𝐷 + 𝐷𝐶) > 2(𝐴𝐶 + 𝐷𝐵)
⇒ 𝐴𝐵 + 𝐵𝐶 + 𝐴𝐷 + 𝐷𝐶 > 𝐴𝐶 + 𝐷𝐵
⇒ 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 > 𝐴𝐶 + 𝐷𝐵
Hence, it is true.
Q.5)` ABCD is quadrilateral. Is 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 < 2 (𝐴𝐶 + 𝐵𝐷) ?
Sol.5) Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore, 𝐼𝑛 Δ 𝐴𝑂𝐵, 𝐴𝐵 < 𝑂𝐴 + 𝑂𝐵 ……….(i)
In Δ 𝐵𝑂𝐶, 𝐵𝐶 < 𝑂𝐵 + 𝑂𝐶 ……….(ii)
In Δ 𝐶𝑂𝐷, 𝐶𝐷 < 𝑂𝐶 + 𝑂𝐷 ……….(iii)
In Δ 𝐴𝑂𝐷, 𝐷𝐴 < 𝑂𝐷 + 𝑂𝐴 ……….(iv)
Adding equations (i), (ii), (iii) and (iv), we get
𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 < 𝑂𝐴 + 𝑂𝐵 + 𝑂𝐵 + 𝑂𝐶 + 𝑂𝐶 + 𝑂𝐷 + 𝑂𝐷 + 𝑂𝐴
⇒ 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 < 2𝑂𝐴 + 2𝑂𝐵 + 2𝑂𝐶 + 2𝑂𝐷
⇒ 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 < 2[(𝐴𝑂 + 𝑂𝐶) + (𝐷𝑂 + 𝑂𝐵)]
⇒ 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 < 2(𝐴𝐶 + 𝐵𝐷)
Hence, it is proved.
Q.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?
Sol.6) Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
It is given that two sides of triangle are 12 cm and 15 cm.
Therefore, the third side should be less than 12 + 15 = 27 𝑐𝑚.
And also the third side cannot be less than the difference of the two sides.
Therefore, the third side has to be more than 15 – 12 = 3 𝑐𝑚.
Hence, the third side could be the length more than 3 cm and less than 27 cm.
Exercise 6.5
Q.1) PQR is a triangle right angled at P. If 𝑃𝑄 = 10 𝑐𝑚 and 𝑃𝑅 = 24 𝑐𝑚, find QR.
Sol.1) Given: PQ = 10 cm, PR = 24 cm
Let QR be x cm.
In right angled triangle QPR,
(𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2 = (𝐵𝑎𝑠𝑒)2 + (𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟)2 [By Pythagoras theorem]
⇒ (𝑄𝑅)2 = (𝑃𝑄)2 + (𝑃𝑅)2
⇒ 𝑥2 = (10)2 + (24)2
⇒ 2𝑥 = 100 + 576 = 676
⇒ 𝑥 = √676 = 26 𝑐𝑚
Thus, the length of QR is 26 𝑐𝑚
Q.2) ABC is a triangle right angled at C. If 𝐴𝐵 = 25 𝑐𝑚 and 𝐴𝐶 = 7 𝑐𝑚, find 𝐵𝐶.
Sol.2) Given: 𝐴𝐵 = 25 𝑐𝑚, 𝐴𝐶 = 7 𝑐𝑚
Let BC be 𝑥 𝑐𝑚.
In right angled triangle ACB,
(𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2 = (𝐵𝑎𝑠𝑒)2 + (𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟)2 [By Pythagoras theorem]
⇒ (𝐴𝐵)2 = (𝐴𝐶)2 + (𝐵𝐶)2
⇒ (25)2 = (7)2 + (𝑥)2
⇒ 625 = 49 + 2𝑥
⇒ 2𝑥 = 625 – 49 = 576
⇒ 𝑥 = √576 = 24 𝑐𝑚
Thus, the length of BC is 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.
Sol.3) Let AC be the ladder and A be the window.
Given: AC = 15 m, AB = 12 m, CB = a m
In right angled triangle ACB,
(𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2 = (𝐵𝑎𝑠𝑒)2 + (𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟)2 [By Pythagoras theorem]
⇒ (𝐴𝐶)2 = (𝐶𝐵)2 + (𝐴𝐵)2
⇒ 152 = 𝑎2 + 122
⇒ 225 = 2 𝑎 + 144
⇒ 2 𝑎 = 225 – 144 = 81
⇒ 𝑎 = √81 = 9 𝑐𝑚
Thus, the distance of the foot of the ladder from the wall is 9 m.
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,2cm,2.5 cm
Sol.4) Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem
(𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2 = (𝐵𝑎𝑠𝑒)2 + (𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟)2 [By Pythagoras theorem]
(i) 2.5 cm, 6.5 cm, 6 cm
In Δ𝐴𝐵𝐶,
(𝐴𝐶)2 = (𝐶𝐵)2 + (𝐴𝐵)2
L.H.S. = (6.5)2 = 42.25 𝑐𝑚
R.H.S. = (6)2 + (2.5)2 = 36 + 6.25 = 42.25 𝑐𝑚
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 6.5 𝑐𝑚, i.e., at B.
(ii) 2 cm, 2 cm, 5 cm
In the given triangle, (5)2 = (2)2 + (2)2
L.H.S. = (5)2 = 25
R.H.S. = (2)2 + (2)2 = 4 + 4 = 8
Since, 𝐿. 𝐻. 𝑆. ≠ 𝑅. 𝐻. 𝑆.
Therefore, the given sides are not of the right angled triangle.
(iii) 1.5 cm, 2 cm, 2.5 cm
In Δ PQR, (𝑃𝑅)2 = (𝑃𝑄)2 + (𝑅𝑄)2
L.H.S. = (2.5)2 = 6.25 𝑐𝑚
R.H.S. = (1.5)2 + (2)2 = 2.25 + 4 = 6.25 𝑐𝑚
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 2.5 𝑐𝑚, i.e., at Q.
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.
Sol.5) Let A’CB represents the tree before it broken at the point C and let the top A’ touches the ground at A after it broke.
Then Δ𝐴𝐵𝐶 is a right angled triangle, right angled at B.
Using Pythagoras theorem, In Δ𝐴𝐵𝐶
(𝐴𝐶)2 = (𝐶𝐵)2 + (𝐴𝐵)2
⇒ (𝐴𝐶)2 = (5)2 + (12)2
⇒ (𝐴𝐶)2 = 25 + 144
⇒ (𝐴𝐶)2 = 169
⇒ 𝐴𝐶 = 13 𝑚
Hence, the total height of the tree = 𝐴𝐶 + 𝐶𝐵 = 13 + 5 = 18 𝑚.
Q.6) Angles Q and R of a Δ 𝑃𝑄𝑅 are 25° and 65°. Write which of the following is true :
i) 𝑃𝑅2 + 𝑄𝑅2 = 𝑅𝑃2 ii) 𝑃𝑄2 + 𝑅𝑃2 = 𝑄𝑅2 iii) 𝑅𝑃2 + 𝑄𝑅2 = 𝑃𝑄2
Sol.6) In Δ 𝑃𝑄𝑅,
∠ 𝑃𝑄𝑅 + ∠ 𝑄𝑅𝑃 + ∠ 𝑅𝑃𝑄 = 180° [By Angle sum property of a Δ]
⇒ 25° + 65° + ∠𝑅𝑃𝑄 = 180°
⇒ 90° + ∠ 𝑅𝑃𝑄 = 180°
⇒ ∠ 𝑅𝑃𝑄 = 180° − 90° = 90°
Thus, Δ PQR is a right angled triangle, right angled at P.
(𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2 = (𝐵𝑎𝑠𝑒)2 + (𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟)2 [By Pythagoras theorem]
⇒ (𝑄𝑅)2 = (𝑃𝑅)2 + (𝑄𝑃)2
Hence, Option (ii) is correct
Q.7) Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Sol.7) Given diagonal (𝑃𝑅) = 41 𝑐𝑚, length (𝑃𝑄) = 40 𝑐𝑚
Let breadth (QR) be x cm.
Now, in right angled triangle PQR,
(𝑃𝑅)2 = (𝑅𝑄)2 + (𝑄𝑃)2 [By Pythagoras theorem]
⇒ 412 = 𝑥2 + 402
⇒ 1681 = 𝑥2 + 1600
⇒ 𝑥2 = 1681 – 1600
⇒ 𝑥2 = 81
⇒ 𝑥 = √ 81 = 9 𝑐𝑚
Therefore, the breadth of the rectangle is 9 𝑐𝑚.
Perimeter of rectangle = 2(𝑙𝑒𝑛𝑔𝑡ℎ + 𝑏𝑟𝑒𝑎𝑑𝑡ℎ) = 2 (9 + 49) = 2 × 49 = 98 𝑐𝑚
Hence, the perimeter of the rectangle is 98 cm.
Q.8) The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Sol.8) Given: Diagonals 𝐴𝐶 = 30 𝑐𝑚 and 𝐷𝐵 = 16 𝑐𝑚.
Since the diagonals of the rhombus bisect at right angle to each other.
Therefore, 𝑂𝐷 = 𝐷𝐵/2 = 16/2 = 8 𝑐𝑚
And 𝑂𝐶 = 𝐴𝐶/2 = 30/2 = 15 𝑐𝑚
Now, In right angle triangle 𝐷𝑂𝐶,
(𝐷𝐶)2 = (𝑂𝐶)2 + (𝑂𝐷)2 [By Pythagoras theorem]
⇒ (𝐷𝐶)2 = (8)2 + (15)2
⇒ (𝐷𝐶)2 = 64 + 225
⇒ 𝐷𝐶 = √289 = 17 𝑐𝑚
Perimeter of rhombus = 4 × 𝑠𝑖𝑑𝑒 = 4 × 17 = 68 𝑐𝑚
Thus, the perimeter of rhombus is 68 𝑐𝑚.
| NCERT Solutions Class 7 Mathematics Chapter 1 Integers |
| NCERT Solutions Class 7 Mathematics Chapter 2 Fractions and Decimals |
| NCERT Solutions Class 7 Mathematics Chapter 3 Data Handling |
| NCERT Solutions Class 7 Mathematics Chapter 4 Simple Equations |
| NCERT Solutions Class 7 Mathematics Chapter 5 Lines and angles |
| NCERT Solutions Class 7 Mathematics Chapter 6 Triangle and its properties |
| NCERT Solutions Class 7 Mathematics Chapter 7 Congruence of Triangle |
| NCERT Solutions Class 7 Mathematics Chapter 8 Comparing Quantities |
| NCERT Solutions Class 7 Mathematics Chapter 9 Rational Numbers |
| NCERT Solutions Class 7 Mathematics Chapter 10 Practical Geometry |
| NCERT Solutions Class 7 Mathematics Chapter 11 Perimeter and Area |
| NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions |
| NCERT Solutions Class 7 Mathematics Chapter 13 Exponents and Power |
| NCERT Solutions Class 7 Mathematics Chapter 14 Symmetry |
| NCERT Solutions Class 7 Mathematics Chapter 15 Visualizing Solid Shapes |
NCERT Solutions Class 7 Mathematics Chapter 6 Triangle and its properties
The above provided NCERT Solutions Class 7 Mathematics Chapter 6 Triangle and its properties is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 7 Mathematics textbook online or you can easily download them in pdf. The answers to each question in Chapter 6 Triangle and its properties of Mathematics Class 7 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 6 Triangle and its properties Class 7 chapter of Mathematics so that it can be easier for students to understand all answers. These solutions of Chapter 6 Triangle and its properties NCERT Questions given in your textbook for Class 7 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 7.
You can download the NCERT Solutions for Class 7 Mathematics Chapter 6 Triangle and its properties for latest session from StudiesToday.com
Yes, the NCERT Solutions issued for Class 7 Mathematics Chapter 6 Triangle and its properties have been made available here for latest academic session
Regular revision of NCERT Solutions given on studiestoday for Class 7 subject Mathematics Chapter 6 Triangle and its properties can help you to score better marks in exams
Yes, studiestoday.com provides all latest NCERT Chapter 6 Triangle and its properties Class 7 Mathematics solutions based on the latest books for the current academic session
Yes, NCERT solutions for Class 7 Chapter 6 Triangle and its properties Mathematics are available in multiple languages, including English, Hindi
All questions given in the end of the chapter Chapter 6 Triangle and its properties have been answered by our teachers