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.

(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 kmiii) 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





Can any send me the answers for this please​

Answer 1

Answer:

1) 6,7

2) 4km

3) 10km

4) 8km

5) Rectangle

Answer 2

Concept

An address that aids in locating a spot in space is a coordinate. The coordinates of a point in a two-dimensional space are (x, y).

Abscissa: The distance from the origin along the x-axis is represented by the x value at the point (x, y).

Ordinate: It is the y value at the coordinates (x, y) and the angle at which the point lies in relation to the x-axis, which runs parallel to the y-axis.

A point's coordinates can be used for a variety of tasks, including calculating distance, midpoint, a line's slope, and its equation.

Given

the places are marked as A,B,C and D on the graph.

Find

we need to find:

• the coordinates of C.
• the distance between School and TV Tower.
• the distance between TV tower and Mall.
• the distance between School and Temple.
• the name of the quadrilateral ABCD so formed.

Solution

(i) the coordinate of C consists of abscissa and ordinate, the point is situated on the x and y axis respectively.

hence we get the abscissa as 6 on x axis and ordinate as 7 on y axis.

(ii) the distance between School and TV Tower, means the distance between A and B.

distance between two points = √(x₂₋x₁)² ₊ (y₂₋y₁)²

two points are: A = (0,₋1) as (x₁,y₁) and B = (8,3) as (x₂,y₂)

d = √(8₋0)² ₊ (3₋(₋1))²

= √8² ₊ 4²

= 4√5 km

(iii) the distance between TV Tower and mall, means the distance between B and D.

two points are: B = (8,3) as (x₁,y₁) and D = (₋2,3) as (x₂,y₂)

d = √(₋2₋8)² ₊ (3₋3)²

= √(₋10²) ₊ 0

= 10 km

(iv) the distance between school and temple, means the distance between A and C.

two points are: A = (0,₋1) as (x₁,y₁) and C = (6,7) as (x₂,y₂)

d = √(6₋0)² ₊ (7₋(₋1)²

= √6² ₊8²

= 10 km

(v) the quadrilateral ABCD formed is a rectangle as the diagonals are equal.

hence we determined the coordinates and distances between the given points.

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