S.chand publication New Learning Composite mathematics solution of class 7 Chapter 10 Properties of Triangles Exercise 10C

 Exercise 10C

Question 1

Determine whether a triangle can have sides with the given lengths. Explain.

(a) 4, 6, 8

Sol :

4+6=10>8

6+8=14>4

4+8=12>6

Two number are greater than third number. So, they can represent the lengths of the sides of a triangle

(b) 5, 7.5, 14

Sol :

5+7.5=12.5<14

So there numbers can not be the measure of the sides of triangle

(c) 7.2, 8, 10

Sol :

72+8=15.2>14

7.2+10=17.2>8

8+10=18>10

Since, length of any two sides > length of third side so a triangle with the above measure is possible

(d) 9, 14, 23

Sol :

9+14=23

Sum of two side=length of the third side

So, a triangle with a above measure is not possible

(e) 4.5, 4.5, 7

Sol :

4.5+4.5=9>7

4.5+7=11.5>4.5

Since, length of two side> length of the third side

So, a triangle with the above measure is possible

(f) 11.9, 5.8, 5.8

Sol :

5.8+5.8=11.6<11.9

Since, the length of any two side< length of third side, a triangle with there measure is not possible

(g) 3x, 2x + 1, x² when x = 5

Sol :

2x+1=(2×5)+1=11

x²=5²=25

15+11=26>25

11+25=36>15

15+25=40>11

Since, the length of any two side> length of third side, a triangle with there measure is possible

Question 2

A is any point within a △PQR whose sides are 5 cm, 7 cm, 10 cm respectively. Prove that, AP+AQ+AR>11 cm.

Sol :

Fig to be added

Sum of two sides is greater than three sides

AP+AQ>5 cm

AP+AR>7 cm

AQ+AR>10 cm

Add all equations

2AP+2AR+2AQ>5+7+10

2AP+2AR+2AQ>22

2(AP+AR+AQ)>2×11

∴AP+AR+AQ>11 cm