One day Ram went to his home town during Dussehra vacation. During his excursion, he noted
the four places Temple, TV tower, Mall and School, then he tried to locate all the places using
graph sheet by taking his position at origin. He marked A, B, C and D for School, TV Tower,
Temple and Mall respectively on the graph sheet by taking scale as 1 unit = 1 km as shown below.
(i) Find the coordinates of C.
(a) (0, -1) (b) (8, 3) (c) (6, 7) (d) (-2, 3)
(ii) Find the distance between School and TV Tower.
(a) 4 km (b) 4√5 km (c) 2√5 km (d) 3√5 km
iii) Find the distance between TV tower and Mall.
(a) 8 km (b) 10 km (c) 6 km (d) 9 km
(iv) Find the distance between School and Temple.
(a) 8 km (b) 10 km (c) 6 km (d) 9 km
(v) Name the quadrilateral ABCD so formed.
(a) Square (b) rectangle (c) rhombus (d) none of these
Answers
Solution :-
(i)
→ Coordinates of C = (6, 7) (c)
(ii)
→ Coordinates of A(School) = (0, -1)
→ Coordinates of B(Tower) = (8, 3)
So,
→ AB = √[(0 - 8)² + (-1 - 3)²]
→ AB = √(64 + 16)
→ AB = 4√5 km (b)
(iii)
→ Coordinates of B(Tower) = (8, 3)
→ Coordinates of D(Mall) = (-2, 3)
So,
→ BD = √[(8 + 2)² + (3 - 3)²]
→ BD = √(100 + 0)
→ BD = 10 km (b)
(iv)
→ Coordinates of A(School) = (0, -1)
→ Coordinates of C(Temple) = (6, 7)
So,
→ AC = √[(6 - 0)² + (7 + 1)²]
→ AC = √(36 + 64)
→ AC = 10 km (b)
(v)
→ AD = √[(0 + 2)² + (-1 - 3)²] = √(4 + 16) = 2√5 km
→ BC = √[(8 - 6)² + (3 - 7)²] = √(4 + 16) = 2√5 km
→ AB = 4√5 km
→ CD = √[(6 + 2)² + (7 - 3)²] = √(64 + 16) = 4√5 km
So, ABCD is a rectangle .
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