RD Sharma Solutions Class 7 Properties Of Triangles Exercise 15.4

RD Sharma Solutions Class 7 Chapter 15 Exercise 15.4

RD Sharma Class 7 Solutions Chapter 15 Ex 15.4 PDF Free Download

Exercise 15.4

Q1. In each of the following, there are three positive numbers. State if these numbers could possibly be the lengths of the sides of a triangle:

(i) 5, 7, 9

(ii) 2, 10.15

(iii) 3, 4, 5

(iv) 2, 5, 7

(v) 5, 8, 20

(i) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side. Here, 5+7>9, 5+9>7, 9+7>5

(ii) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.

(iii) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of triangle is always greater than the third side. Here, 3+4 >5, 3+5> 4, 4+5> 3

(iv) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 2 + 5 = 7

(v) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 5 + 8 <20

Q2. In Fig, P is the point on the side BC. Complete each of the following statements using symbol ‘ =’,’ > ‘or ‘ < ‘so as to make it true:

(i) AP… AB+ BP

(ii) AP… AC + PC

(iii) AP…. $\frac{1}{2}(AB+AC+BC)$

(i) In triangle APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side.

(ii) In triangle APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side.

(iii) AP < $\frac{1}{2}(AB+AC+BC)$ In triangles ABP and ACP, we can see that:

AP < AB + BP…(i) (Because the sum of any two sides of a triangle is greater than the third side)

AP < AC + PC…(ii) (Because the sum of any two sides of a triangle is greater than the third side)

On adding (i) and (ii), we have:

AP + AP < AB + BP + AC + PC

2AP < AB + AC + BC (BC = BP + PC)

AP < (AB-FAC+BC)

Q3. P is a point in the interior of $\triangle ABC$ as shown in Fig. State which of the following statements are true (T) or false (F):

(i) AP+ PB< AB

(ii) AP+ PC> AC

(iii) BP+ PC = BC

(i) False

We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.

(ii) True

We know that the sum of any two sides of a triangle is greater than the third side: it is true for the given triangle.

(iii) False

We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.

Q4. O is a point in the exterior of $\triangle ABC$. What symbol ‘>’,’<’ or ‘=’ will you see to complete the statement OA+OB….AB? Write two other similar statements and show that

OA+OB+OC>$\frac{1}{2}(AB+BC+CA)$

Because the sum of any two sides of a triangle is always greater than the third side, in triangle OAB, we have:

OA+OB> AB —(i)

OB+OC>BC —-(ii)

OA+OC > CA —–(iii)

On adding equations (i), (ii) and (iii) we get :

OA+OB+OB+OC+OA+OC> AB+BC+CA

2(OA+OB+OC) > AB+BC +CA

OA+ OB + OC > $\frac{AB+BC+CA}{2}$

Q5. In $\triangle ABC$, $\angle B=30^{\circ}$, $\angle C=50^{\circ}$. Name the smallest and the largest sides of the triangle.

Because the smallest side is always opposite to the smallest angle, which in this case is $30^{\circ}$, it is AC. Also, because the largest side is always opposite to the largest angle, which in this case is $100^{\circ}$,, it is BC.