## 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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Varinder says

Very good and nice