NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Mathematics (Maths) Class 7

Created by: Praveen Kumar

Class 7 : NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

The document NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev is a part of the Class 7 Course Mathematics (Maths) Class 7.
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Exercise 6.4 

Question 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 1: 

(i) 2 cm, 3 cm, 5 cm
2 + 3 > 5 No
2 + 5 > 3 Yes
3 + 5 > 2 Yes
This triangle is not possible.

(ii) 3 cm, 6 cm, 7 cm
3 + 6 > 7 Yes
6 + 7 > 3 Yes
3 + 7 > 6 Yes
This triangle is 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.

Question 2:

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

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

(i) OP + OQ > PQ ?
 (ii) OQ + OR > QR?
 (iii) OR + OP > RP ?

Answer 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 ?
Yes, ROP form a triangle.

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Question 3: 

AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM? (Consider the sides of triangles ΔABM and ΔAMC.)

 

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Answer 3: 

Since, the sum of lengths of any two sides in a triangle should be greater titan the length of third side.

Therefore, In ΔABM, AB + BM > AM ... (i)
In ΔAMC, 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.

Question 4:

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

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Answer 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 ΔDCB, DC + CB > DB  (iii)
In ΔADB, 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.

Question 5: 

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

Answer 5: 

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Therefore, In A AOB, AB < OA + OB ......... (i)
In A BOC, BC < OB + OC  (ii)
In A COD, CD<OC + OD  (iii)
InAAOD, 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 + 20B + 2OC + 2OD
⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]
⇒ AB + BC + CD + DA < 2(AC + BD)
Hence, it is proved.

Question 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? 

Answer 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 

Question 1: 

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

Answer 1: 

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

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 
⇒ x2 = 100 + 576 = 676
NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Thus, the length of QR is 26 cm.

Question 2: 

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

Answer 2: 

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

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+x2
 ⇒ 625 = 49 + x2
 ⇒ x2 = 625 - 49 = 576
 ⇒ 
NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Thus, the length of BC is 24 cm.

Question 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 Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Answer 3:

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

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
⇒ (15)2 + (a)2 = (12)2
⇒ 225 = a2 + 144
⇒ a2 = 225 - 144 = 81 
⇒ NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Thus, the distance of the foot of the ladder from the wall is 9 m.

Question 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 

In the case of right angled triangles, identify the right angles. 

Answer 4: 

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

(i) 2.5 cm, 6.5 cm, 6 cm

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

In AABC, (AC)2 = (AB)2 + (BC)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 ΔPQR, (PR)2 = (PQ)3 + (RQ)2

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

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.

Question 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 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.

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

AB = 12 m and BC = 5 m
Using Pythagoras theorem, In ΔABC
(AC)2 =(AB)2 + (BC)2 
⇒ (AC)2 = (12)2 + (5)2
⇒ (AC)2 = 144 + 25
⇒ (AC)2 = 169
⇒ AC = 13 m

Hence, the total height o f the tree = AC + CB 13 + 5 = 18 m.

Question 6: 

Angles Q and R of a ΔPQR are 25° and 65°
 Write which of the following is true:
 (i) PQ2 + QR2 = RP2
 (ii) PQ2 + RP2 = QR
 (iii) RP2 + QR2 = PQ2

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev  

Answer 6: 

In ΔPQR, ∠PQR + ∠QRP + ∠RPQ = 180°      [By Angle sum property of a Δ] 
⇒ 25° + 65° + ∠RPQ = 180°
⇒ 90° + ∠RPQ = 180°
⇒ ∠RPQ = 180°- 90° = 90°
Thus, ΔPQR 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

Question 7: 

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

Answer 7: 

Given diagonal (PR) = 41 cm, length (PQ) = 40 cm

Let breadth (QR) be x cm. 

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Now, in right angled triangle PQR,

(PR)2 = (RQ)2 +(PQ)2      [By Pythagoras theorem]
⇒ (41)2 = x2 + (40)2
⇒ 1681 = x2 + 1600
⇒ x2 = 1681 - 1600
⇒ x2 = 81
⇒  NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Therefore the breadth of the rectangle is 9 cm,
Perimeter of rectangle - 2(length + breadth)
= 2 [9 + 49)
= 2 x 49 = 98 cm
Hence, the perimeter of the rectangle is 98 cm.

Question 8: 

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

Answer 8: 

Given: Diagonals AC = 30 cm and DB = 16 cm.

Since the diagonals of the rhombus bisect at right angle to each other.

Therefore,  NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

And  NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Now , In right angle triangle DOC,
(DC)2 =(OD)2 + (OC)2    [By Pythagoras dieorem]
⇒  (DC)2 = (8)2+(15)2
⇒ (DC)2 = 64 + 225 = 289
⇒  NCERT Solution(Part - 2) - Triangle and Its Properties Class 7 Notes | EduRev

Perimeter of rhombus = 4 x side = 4 x 17 = 68 cm
Thus, die perimeter of rhombus is 68 cm.

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