NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4
NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4
These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties 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:
(i) 2 cm, 3 cm, 5 cm
2 cm + 3 cm = 5 cm
third side = 5 cm
∵ Sum of the measures of length of two sides = length of the third side
∴ It is not possible with these sides to form a triangle.
(ii) 3 cm, 6 cm, 7 cm
3 cm + 6 cm = 9 cm
9 cm > 7 cm
3 cm + 7 cm = 10 cm and 10 cm > 6 cm.
6 cm + 7 cm = 13 cm and 13 cm > 3 cm.
Thus, a triangle can be possible with these sides.
(iii) 6 cm, 3 cm and 2 cm.
6 cm + 3 cm = 9 cm and 9 cm > 2 cm
3 cm + 2 cm = 5 cm and 5 cm < 6 cm 2 cm + 6 cm = 8 cm and 8 cm > 5 cm
Thus, a triangle cannot be possible with these sides.
Question 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 OP + QQ> PQ
( The sum of lengths of any two sides of a triangle is greater than the length of the third side)
(ii) Yes, OQ + OR > QR
(iii) Yes, OR + OP > RP
Question 3.
AM is a median of a triangle ABC. IsAB + BC + CA > 2 AM?
(Consider the sides of triangles ΔABM and ΔAMC.)
Answer:
Since, the sum of the length of any two sides of a triangle is greater than the length
of the third side
∴ In ΔABM, we have
(AB + BM) > AM …………. (i)
Similarly in ΔACM
(CA + CM) > AM ………. (ii)
Adding (1) and (2), we have
[ (AB + BM) + (CA + CM)] > AM + AM
[ AB + (BM + CM) + CA] > 2 AM
[ AB + BC + CA]> 2AM
Thus, (AB + BC + CA) > 2 AM
Question 4.
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?
Answer:
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
In ABC, we have
AB + BC > AC …………(i)
Similarly, in ΔACD, we have
CD + DA > AC …………(ii)
Adding (i) and (ii), we get
[(AB + BC) + (CD + DA)] > 2 AC …………(iii)
Again In ΔABD, we have
AB + DA > BD …………….(iv)
In ΔBCD, we have
BC + CD > BD …(v)
Adding (iv) and (v), we get
[(AB + DA) + (BC + CD)] > 2 BD …………..(vi)
Now Adding (iii) and (vi), we get
2[(AB + BC) + (CD+ DA)] > 2 (AC + BD) (AB + BC + CD + DA) > (AC + BD)
Question 5.
ABCD is a quadrilateral.
Is AB + BC + CD + DA < 2(AC + BD)?
Answer:
Since the sum of the length of any two sides of a triangle is greater than the length of the third side.
∴ In ΔAOB, we have (OA + OB) > AB …………(i)
Similarly,
In ΔOBC, we have
(OB + OC) > BC …………..(ii)
In ΔOCD, we have
(OC + OD) > CD ………….(iii)
In ΔOAD we have
(OA + OD) > AD …………..(iv)
Adding (i), (ii), (iii) and (iv), we have
2] OA + OB + OC + OD] > (AB + BC + CD + DA)]
⇒ AB + BC + CD + DA < 2 (OA + OB + OC + OD)
⇒ AB + BC + CD + DA < 2 [(OA + OC) + (OB + OD)]
⇒ AB + BC + CD + DA < 2 (AC + BD)
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:
Since, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Sum of 12 cm and 15 cm is greater than the length of the third side,
i.e. (12 cm + 15 cm) > Third side 27 cm > Third side (or) Third side < 27 cm.
Also, the difference of the lengths of any two sides is less than the length of the third side.
(15 cm – 12 cm) < Third side 3 cm < Third side
Thus, we have 3 cm < Third side < 27 cm.
∴ The third side should be of any length between 3 cm and 27 cm.
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