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

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

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15A
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15D

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.

Hope given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C are helpful to complete your math homework.

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