In this chapter, we provide RD Sharma Class 7 ex 15.4 Solutions Chapter 15 Properties of Triangles for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 15.4 Solutions Chapter 15 Properties of Triangles Maths pdf, 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 | Ex 15.4 |
RD Sharma Solutions for Class 7 Chapter 15 Properties of Triangles Ex 15.4 Download PDF
Chapter 15: Properties of Triangles Exercise – 15.4
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
Solution:
(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…. 1/2(AB + AC + BC)

Solution:
(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 < 12(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)
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
Solution:
(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.
Question: 4
O is a point in the exterior of △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 > 1/2(AB + BC +CA)
Solution:
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 > (AB + BC + CA)/2
Question: 5
In ∆ABC, ∠B = 30°, ∠C = 50°. Name the smallest and the largest sides of the triangle.
Solution:
Because the smallest side is always opposite to the smallest angle, which in this case is 30°, it is AC. Also, because the largest side is always opposite to the largest angle, which in this case is 100°, it is BC.
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