RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15C

In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 15 Properties of Triangles Ex 15C for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 15 Properties of Triangles Ex 15C Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C Maths book pdf download. Now you will get step by step solution to each question.

RS Aggarwal Solutions for Class 7 Chapter 15 Properties of Triangles Ex 15C Download PDF

Question 1.
Solution:
We know that in a triangle, sum of any two sides is greater than the third side. Therefore :
(i) 1cm, 1cm, 1cm
It is possible to draw a triangle
(1 + 1) cm > 1cm (sum of two sides is greater than the third)
(ii) 2cm, 3cm, 4cm
It is also possible to draw the triangle
(2 + 3) cm > 4cm (sum of two sides is greater than third side)
(iii) 7cm, 8cm, 15cm
It is not possible to draw the triangle
(7 + 8)cm not > 15cm
But (7 + 8) cm = 15 cm
(iv) 3.4 cm, 2.1 cm, 5.3 cm
It is possible to draw the triangle
(3.4 + 2.1) cm > 5.3 cm
⇒ 5.5cm > 5.3 cm
(v) 6cm, 7cm, 14cm
It is not possible to draw
(6 + 7) cm not > 14cm
i.e. 13cm not > 14cm (13cm < 14cm)

Question 2.
Solution:
Two sides of a triangle are 5 cm and 9 cm long
Then the third side will be less then (5 + 9) or less than 14 cm

Question 3.
Solution:
(i) In ∆APB,
PA + PB > AB (sum of two sides is greater than its third side)
(ii) In ∆PBC,
PB + PC > BC (sum of two sides is greater than its third side)
(iii) In ∆PAC,
AC < PA + PC (PA + PC > AC)

Question 4.
Solution:
Proof: AM is the median of ∆ABC
M is mid-point of BC

In ∆ABM,
AB + BM > AM ….(i)
(Sum of any two sides of a triangle is greater than its third side)
Similarly in ∆ACM,
AC + MC > AM ….(ii)
Adding (i) and (ii)
AB + BM + AC + MC > 2 AM
⇒ AB + AC + BM + MC > 2AM
⇒ AB + AC + BC > 2AM
Hence proved.

Question 5.
Solution:
Given: In ∆ABC, P is a point on BC.
AP is joined.
To prove :
(AB + BC + AC) > 2AP
Proof : In ∆ABP,
AB + BP > AP …(i) (Sum of two sides is greater than third)
Similarly in ∆ACP,

AC + PC > AP …(ii)
Adding (i) and (ii)
AB + BP + AC + PC > AP + AP
⇒ AB + BP + PC + CA > 2AP
⇒ AB + BC + CA > 2AP
Hence proved.

Question 6.
Solution:
ABCD is a quadrilateral AC and BD are joined.
Proof: Now in ∆ABC
AB + BC > AC ….(i)

(Sum of any two sides of a triangle is greater than its third side)
Similarly in ∆ADC,
AD + CD > AC ….(ii)
In ∆ABD,
AB + AD > BD ….(iii)
and in ∆BCD,
BC + CD > BD ……..(iv)
Adding (i), (ii), (iii) and (iv)
AB + BC + CD + AD + AB + AD + BC + CD > AC + AC + BD + BD
⇒ 2 (AB + BC + CD + AD) > 2(AC + BD)
⇒ AB + BC + CD + AD > AC + BD
Hence proved.

Question 7.
Solution:
Given : O is any point outside of the ∆ABC
To prove : 2(OA + OB + OC) > (AB + BC + CA)
Construction : Join OA, OB and DC.
Proof: In ∆AOB,

OA + OB > AB ….(i) (Sum of two sides of a triangle is greater than its third side)
Similarly in ∆BOC,
OB + OC > BC …(ii)
and in ∆COA
OC + OA > CA …(iii)
Adding (i), (ii) and (iii), we get:
OA + OB + OB + OC + OC + OA > AB + BC + CA
2 (OA + OB + OC) > (AB + BC + CA)
Hence proved.

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