Question

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.1) Find the coordinates of A, B, C and D 2)Find the distance between School and TV Tower. 3) Find the distance between TV tower and Mall. 4)Find the distance between School and Temple. 5)Name the quadrilateral ABCD so formed.

Answer 1

Answer:

mam your voice of the year

Answer 2

Ram's vacation visit to his Town:

1) Coordinates of A = (0,-1)

  Coordinates of B = (-2,3)

  Coordinates of C = (6,7)

  Coordinates of D = (8,3)

2) The distance between School and TV Tower = 4$\sqrt[2]{5}$ Km

3) The distance between TV tower and Mall = 10 Km

4) The distance between School and Temple =  10 km

5) The quadrilateral ABCD so formed is Rectangle.

Step-by-step explanation:

1) The coordinates of A,B,C,D is found by taking x-axis and y-axis value in graph into account where

A = School

B = TV Tower

C = Temple

D = Mall.

2) AB = $\sqrt[2]{(8-0)^{2} + (3+1)^{2} }$

          = $\sqrt[2]{64 + 16}$

          = $\sqrt[2]{80}$

          = 4$\sqrt[2]{5}$ Km

3) The distance between TV tower and Mall is calculated by counting number of units between B = TV tower and D = Mall.ween

4) The distance between School and Temple is calculated by :

AC = $\sqrt[2]{(6-0)^{2} + (7+1)^{2} }$

     = $\sqrt[2]{(36 + 64)}$

     = $\sqrt[2]{100}$

     = 10 Km

5) The quadrilateral ABCD so formed is :

AB = 4$\sqrt[2]{5}$

DC =$\sqrt[2]{(6 + 2)^{2} + (7 - 3)^{2} }$ = 4$\sqrt[2]{5}$

AD =$\sqrt[2]{(-2 - 0)^{2} + (3 + 1)^{2} }$ = 2$\sqrt[2]{5}$

BC =$\sqrt[2]{(6-8)^{2} + (7 - 3)^{2} }$ = 2$\sqrt[2]{5}$

:: AB = DC and AD = BC

Thus, opposite sides are equal.

:: BD = 10 and AC = 10, diagonals are equal.

Therefore ABCD is Rectangle.