NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and Its Properties
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The NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and Its Properties are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and Its Properties
Students can access the NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and Its Properties. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 6.1
In Δ PQR, D is the mid-point of 
(i) 
(ii) PD is ___.
(iii) Is QM = MR?
(i) Altitude
An altitude has one endpoint at a vertex of the triangle and another on the line containing the opposite side.
(ii) Median
A median connects a vertex of a triangle to the midpoint of the opposite side.
(iii) No, QM ≠ MR because D is the midpoint of QR.
Verify by drawing a diagram if the median and altitude of an isosceles triangle can be the same.
Draw a line segment PS ⊥ BC. It is an altitude for this triangle. Here, we observe that the length of QS and SR is also the same. So PS is also a median of this triangle.
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) A median connects a vertex of a triangle to the midpoint of the opposite side.
(b)
An altitude has one endpoint at a vertex of the triangle and another on the line containing the opposite side.
(c)
In the figure, we may observe that for ΔXYZ, YL is an altitude drawn exteriorly to side XZ which is extended up to point L.
Exercise 6.2
Find the value of the unknown exterior angle x in the following diagram.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 50o + 70o
= x = 120o
(ii) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 65o + 45o
= x = 110o
(iii) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 30o + 40o
= x = 70o
(iv) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 60o + 60o
= x = 120o
(v) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 50o + 50o
= x = 100o
(vi) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 30o + 60o
= x = 90o
Find the value of the unknown interior angle x in the following figures.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x + 50o = 115o
By transposing 50o from LHS to RHS, it becomes – 50o
= x = 115o – 50o
= x = 65o
(ii) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= 70o + x = 100o
By transposing 70o from LHS to RHS, it becomes – 70o
= x = 100o – 70o
= x = 30o
(iii) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right-angled triangle. So, the angle opposite to the x is 90o.
= x + 90o = 125o
By transposing 90o from LHS to RHS, it becomes – 90o
= x = 125o – 90o
= x = 35o
(iv) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x + 60o = 120o
By transposing 60o from LHS to RHS, it becomes – 60o
= x = 120o – 60o
= x = 60o
(v) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right-angled triangle. So, the angle opposite to the x is 90o.
= x + 30o = 80o
By transposing 30o from LHS to RHS, it becomes – 30o
= x = 80o – 30o
= x = 50o
(vi) We know that,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right-angled triangle. So, the angle opposite to the x is 90o.
= x + 35o = 75o
By transposing 35o from LHS to RHS, it becomes – 35o
= x = 75o – 35o
= x = 40o
Exercise 6.3
Find the value of the unknown x in the following diagrams:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i)
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= ∠BAC + ∠ABC + ∠BCA = 180o
= x + 50o + 60o = 180o
= x + 110o = 180o
By transposing 110o from LHS to RHS it becomes – 110o
= x = 180o – 110o
= x = 70o
(ii)
We know that,
The sum of all the interior angles of a triangle is 180o.
The given triangle is a right angled triangle. So the ∠QPR is 90o.
Then,
= ∠QPR + ∠PQR + ∠PRQ = 180o
= 90o + 30o + x = 180o
= 120o + x = 180o
By transposing 110o from LHS to RHS it becomes – 110o
= x = 180o – 120o
= x = 60o
(iii)
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= ∠XYZ + ∠YXZ + ∠XZY = 180o
= 110o + 30o + x = 180o
= 140o + x = 180o
By transposing 140o from LHS to RHS it becomes – 140o
= x = 180o – 140o
= x = 40o
(iv)
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= 50o + x + x = 180o
= 50o + 2x = 180o
By transposing 50o from LHS to RHS it becomes – 50o
= 2x = 180o – 50o
= 2x = 130o
= x = 130o/2
= x = 65o
(v)
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= x + x + x = 180o
= 3x = 180o
= x = 180o/3
= x = 60o
∴The given triangle is an equiangular triangle.
(vi)
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= 90o + 2x + x = 180o
= 90o + 3x = 180o
By transposing 90o from LHS to RHS it becomes – 90o
= 3x = 180o – 90o
= 3x = 90o
= x = 90o/3
= x = 30o
Then,
= 2x = 2 × 30o = 60o
Find the values of the unknowns x and y in the following diagrams:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i)
We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
Then,
= 50o + x = 120o
By transposing 50o from LHS to RHS it becomes – 50o
= x = 120o – 50o
= x = 70o
We also know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= 50o + x + y = 180o
= 50o + 70o + y = 180o
= 120o + y = 180o
By transposing 120o from LHS to RHS it becomes – 120o
= y = 180o – 120o
= y = 60o
(ii)
From the rule of vertically opposite angles,
= y = 80o
Then,
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= 50o + 80o + x = 180o
= 130o + x = 180o
By transposing 130o from LHS to RHS it becomes – 130o
= x = 180o – 130o
= x = 50o
(iii)
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= 50o + 60o + y = 180o
= 110o + y = 180o
By transposing 110o from LHS to RHS it becomes – 110o
= y = 180o – 110o
= y = 70o
Now,
From the rule of linear pair,
= x + y = 180o
= x + 70o = 180o
By transposing 70o from LHS to RHS it becomes – 70o
= x = 180o – 70
= x = 110o
(iv)
From the rule of vertically opposite angles,
= x = 60o
Then,
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= 30o + x + y = 180o
= 30o + 60o + y = 180o
= 90o + y = 180o
By transposing 90o from LHS to RHS it becomes – 90o
= y = 180o – 90o
= y = 90o
(v)
From the rule of vertically opposite angles,
= y = 90o
Then,
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= x + x + y = 180o
= 2x + 90o = 180o
By transposing 90o from LHS to RHS it becomes – 90o
= 2x = 180o – 90o
= 2x = 90o
= x = 90o/2
= x = 45o
(vi)
From the rule of vertically opposite angles,
= x = y
Then,
We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= x + x + x = 180o
= 3x = 180o
= x = 180o/3
= x = 60o
Exercise 6.4
ABCD is a quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?
Let us consider ABCD is quadrilateral and P is the point where the diagonals intersect. As shown in the figure below,
We know that,
The sum of the length of any two sides is always greater than the third side.
Now consider the ΔPAB,
Here, PA + PB > AB … [equation i]
Then, consider the ΔPBC
Here, PB + PC > BC … [equation ii]
Consider the ΔPCD
Here, PC + PD > CD … [equation iii]
Consider the ΔPDA
Here, PD + PA > DA … [equation iv]
By adding equation [i], [ii], [iii] and [iv] we get,
PA + PB + PB + PC + PC + PD + PD + PA > AB + BC + CD + DA
2PA + 2PB + 2PC + 2PD > AB + BC + CD + DA
2PA + 2PC + 2PB + 2PD > AB + BC + CD + DA
2(PA + PC) + 2(PB + PD) > AB + BC + CD + DA
From the figure we have, AC = PA + PC and BD = PB + PD
Then,
2AC + 2BD > AB + BC + CD + DA
2(AC + BD) > AB + BC + CD + DA
Hence, the given expression is true.
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?
We know that,
The sum of the length of any two sides is always greater than the third side.
From the question, it is given that two sides of the triangle are 12 cm and 15 cm.
So, the third side’s length should be less than the sum of other two sides,
12 + 15 = 27 cm.
Then, it is given that the third side cannot not be less than the difference of the two sides, 15 – 12 = 3 cm
So, the length of the third side falls between 3 cm and 27 cm.
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) Clearly, we have:
(2 + 3) = 5
5 = 5
Thus, the sum of any two of these numbers is not greater than the third.
Hence, it is not possible to draw a triangle whose sides are 2 cm, 3 cm and 5 cm.
(ii) Clearly, we have:
(3 + 6) = 9 > 7
(6 + 7) = 13 > 3
(7 + 3) = 10 > 6
Thus, the sum of any two of these numbers is greater than the third.
Hence, it is possible to draw a triangle whose sides are 3 cm, 6 cm and 7 cm.
(iii) Clearly, we have:
(3 + 2) = 5 < 6
Thus, the sum of any two of these numbers is less than the third.
Hence, it is not possible to draw a triangle whose sides are 6 cm, 3 cm and 2 cm.
AM is a median of a triangle ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles ΔABM and ΔAMC.)
We know that,
The sum of the length of any two sides is always greater than the third side.
Now consider the ΔABM,
Here, AB + BM > AM … [equation i]
Then, consider the ΔACM
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.
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?
We know that,
The sum of the length of any two sides is always greater than the third side.
Now consider the ΔABC,
Here, AB + BC > CA … [equation i]
Then, consider the ΔBCD
Here, BC + CD > DB … [equation ii]
Consider the ΔCDA
Here, CD + DA > AC … [equation iii]
Consider the ΔDAB
Here, DA + AB > DB … [equation iv]
By adding equation [i], [ii], [iii] and [iv] we get,
AB + BC + BC + CD + CD + DA + DA + AB > CA + DB + AC + DB
2AB + 2BC + 2CD + 2DA > 2CA + 2DB
Take out 2 on both the side,
2(AB + BC + CA + DA) > 2(CA + DB)
AB + BC + CA + DA > CA + DB
Hence, the given expression is true.
Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?
If we take any point O in the interior of a triangle PQR and join OR, OP, OQ.
Then, we get three triangles ΔOPQ, ΔOQR and ΔORP is shown in the figure below.
We know that,
The sum of the length of any two sides is always greater than the third side.
(i) Yes, ΔOPQ has sides OP, OQ and PQ.
So, OP + OQ > PQ
(ii) Yes, ΔOQR has sides OR, OQ and QR.
So, OQ + OR > QR
(iii) Yes, ΔORP has sides OR, OP and PR.
So, OR + OP > RP
Exercise 6.5
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) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm
Let us assume the largest value is the hypotenuse side i.e. b = 6.5 cm.
Then, by Pythagoras theorem,
b2 = a2 + c2
6.52 = 2.52 + 62
42.25 = 6.25 + 36
42.25 = 42.25
The sum of square of two side of triangle is equal to the square of third side,
∴The given triangle is right-angled triangle.
Right angle lies on the opposite of the greater side 6.5 cm.
(ii) Let a = 2 cm, b = 2 cm, c = 5 cm
Let us assume the largest value is the hypotenuse side i.e. c = 5 cm.
Then, by Pythagoras theorem,
c2 = a2 + b2
52 = 22 + 22
25 = 4 + 4
25 ≠ 8
The sum of square of two side of triangle is not equal to the square of third side,
∴The given triangle is not right-angled triangle.
(iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm
Let us assume the largest value is the hypotenuse side i.e. b = 2.5 cm.
Then, by Pythagoras theorem,
b2 = a2 + c2
2.52 = 1.52 + 22
6.25 = 2.25 + 4
6.25 = 6.25
The sum of square of two side of triangle is equal to the square of third side,
∴The given triangle is right-angled triangle.
Right angle lies on the opposite of the greater side 2.5 cm.
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.
Let ABC is the triangle and B is the point where tree is broken at the height 5 m from the ground.
Tree top touches the ground at a distance of AC = 12 m from the base of the tree,
By observing the figure we came to conclude that right angle triangle is formed at A.
From the rule of Pythagoras theorem,
BC2 = AB2 + AC2
BC2 = 52 + 122
BC2 = 25 + 144
BC2 = 169
BC = √169
BC = 13 m
Then, the original height of the tree = AB + BC
= 5 + 13
= 18 m
Angles Q and R of a ΔPQR are 25o and 65o.
Write which of the following is true:
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2
Given that ∠Q = 25o, ∠R = 65o
Then, ∠P =?
We know that sum of the three interior angles of triangle is equal to 180o.
∠PQR + ∠QRP + ∠RPQ = 180o
25o + 65o + ∠RPQ = 180o
90o + ∠RPQ = 180o
∠RPQ = 180 – 90
∠RPQ = 90o
Also, we know that side opposite to the right angle is the hypotenuse.
∴ QR2 = PQ2 + PR2
Hence, (ii) is true
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Let ABCD be the rectangular plot.
Then, AB = 40 cm and AC = 41 cm
BC =?
According to Pythagoras theorem,
From right angle triangle ABC, we have:
= AC2 = AB2 + BC2
= 412 = 402 + BC2
= BC2 = 412 – 402
= BC2 = 1681 – 1600
= BC2 = 81
= BC = √81
= BC = 9 cm
Hence, the perimeter of the rectangle plot = 2 (length + breadth)
Where, length = 40 cm, breadth = 9 cm
Then,
= 2(40 + 9)
= 2 × 49
= 98 cm
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Let PQRS be a rhombus, all sides of rhombus has equal length and its diagonal PR and SQ are intersecting each other at a point O. Diagonals in rhombus bisect each other at 90o.
So, PO = (PR/2)
= 16/2
= 8 cm
And, SO = (SQ/2)
= 30/2
= 15 cm
Then, consider the triangle POS and apply the Pythagoras theorem,
PS2 = PO2 + SO2
PS2 = 82 + 152
PS2 = 64 + 225
PS2 = 289
PS = √289
PS = 17 cm
Hence, the length of side of rhombus is 17 cm
Now,
Perimeter of rhombus = 4 × side of the rhombus
= 4 × 17
= 68 cm
∴ Perimeter of rhombus is 68 cm.
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Let us draw a rough sketch of right-angled triangle
By the rule of Pythagoras Theorem,
Pythagoras theorem states that for any right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of square on the legs.
In the above figure RQ is the hypotenuse,
QR2 = PQ2 + PR2
QR2 = 102 + 242
QR2 = 100 + 576
QR2 = 676
QR = √676
QR = 26 cm
Hence, the length of the hypotenuse QR = 26 cm.
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Let us draw a rough sketch of right-angled triangle
By the rule of Pythagoras Theorem,
Pythagoras theorem states that for any right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of square on the legs.
In the above figure RQ is the hypotenuse,
AB2 = AC2 + BC2
252 = 72 + BC2
625 = 49 + BC2
By transposing 49 from RHS to LHS it becomes – 49
BC2 = 625 – 49
BC2 = 576
BC = √576
BC = 24 cm
Hence, the length of the BC = 24 cm.
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.
By the rule of Pythagoras Theorem,
Pythagoras theorem states that for any right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of square on the legs.
In the above figure RQ is the hypotenuse,
152 = 122 + a2
225 = 144 + a2
By transposing 144 from RHS to LHS it becomes – 144
a2 = 225 – 144
a2 = 81
a = √81
a = 9 m
Hence, the length of a = 9 m.
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The NCERT solution for Class 7 Chapter 6: The Triangle and its Properties is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 7 Chapter 6: The Triangle and its Properties is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. The Triangle and its Propertiesally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 7 Maths Chapter 6 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 7 Maths Chapter 6: The Triangle and its Properties as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 7 Maths Chapter 6 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.