Question

10th class 7th chapter first exercise all sums

Answer 1

1-(i) We know that distance between the two points is given by:
= √[(x1 – x2)2 + (y1 – y2)2]

Distance between (2, 3) and (4, 1) is:

l =√ [(2 – 4)2 + (3 – 1)2]

= √[4 + 4]= √8

= √2

(ii) Distance between (-5, 7) and (-1, 3) is:

l = √[(-5 + 1)2 + (7 – 3)2]

= √[16 + 16] = √32

= 4√2

(iii) Distance between (a, b) and (-a, -b) is:

l = √[(a+ a)2 + (b+ b)2]

= √[2a + 2b] = √(4a2 + 4b2)

= 2√(a2 + b2)



2-Distance between (0, 0) and (36, 15) is:
l = [(36 – 0)2 + (15 – 0)2]1/2
= [1296 + 225]1/2 = (1521)1/2
= 39
Yes, we can find the distance between the given towns A and B. Let us take town A at origin point (0, 0)
Hence, town B will be at point (36, 15) with respect to town A
And, as calculated above, the distance between town A and B will be39 km

3-Assume points (1, 5), (2, 3), and ( - 2, - 11) representing the vertices A, B, and C of the given triangle respectively.
Let A (1, 5), B (2, 3) and C (-2, -11)

Now,

AB = √[(1 – 2)2 + (5 – 3)2]

= √5

BC = √[(2 + 2)2 + (3 + 11)2]

= √212

CA = √[(1+ 2)2 + (5+ 11)2]

= √265

Since, AB + BC ≠ CA

4-Let us assume that points (5, - 2), (6, 4), and (7, - 2) are representing the vertices A, B, and C of the given triangle respectively
AB = [(5- 6)2 + (-2- 4)2]1/2
= √1+36
= √37
BC = [(6- 7)2 + (4+ 2)2]1/2
= √1+36
= √37
CA = [(5- 7)2 + (-2+ 2)2]1/2
= √4+0
= 2
Therefore, AB = BC
As two sides are equal in length, therefore, ABCis an isosceles triangle

5-