RD Sharma solution class 7 chapter 15 Properties of triangles Exercise 15.4

Exercise 15.4

Page-15.24

Question 1:

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

Answer 1:

(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,
$5+8 <20$

Question 2:

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) $AP....\frac{1}{2}(AB+AC+BC)$

Answer 2:

(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 <$\frac{1}{2}\left(\text{AB+AC+BC}\right)$
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 < $\frac{1}{2}\left(\text{AB+AC+BC}\right)$

Question 3:

P is a point in the interior of ∆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

Answer 3:

(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.

Page-15.25

Question 4:

O is a point in the exterior of ∆ABC. What symbol '>', '<' or '=' will you use to complete the statement OA + OB .. AB? Write two other similar statements and shown that

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

Answer 4:

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

$$\begin{gathered} OA+OB>AB ...\left(\mathrm{i}\right) \\ OB+OC>BC ...\left(\mathrm{ii}\right) \\ OA+OC>CA ...\left(\mathrm{iii}\right) \\ \mathrm{On} \mathrm{adding} \mathrm{equations} \left(\mathrm{i}\right), \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}: \\ OA+OB+OB+OC+OA+OC>AB+BC+CA \\ 2(OA+OB+OC)>AB+BC+CA \\ OA+OB+OC>\frac{AB+BC+CA}{2} \end{gathered}$$

Question 5:

In ∆ABC, ∠A = 100°, ∠B = 30°, ∠C = 50°. Name the smallest and the largest sides of the triangle.

Answer 5:

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