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.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 15 |

Chapter Name | Properties of Triangles |

Exercise | 15C |

**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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**All Subject NCERT Solutions For Class 7**

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