uDeties of a square and use the distancefm EXERCISE 1 In the

uDeties of a square and use the distancefm EXERCISE #1 In the coordinate plane, A (4,-2),B (5,5), C(-2,6) and D (-3,-1) are the vertices of a parallelogram a) Find AC and b) Which theorem proves that ABCD is a rectangle? Find AB and BC c) d) What definition proves that ABCD is a square?

Solution

Ans:

Step 1: Given a parallelogram with coordinates A(4, -2), B(5, 5), C(-2, 6) and D(-3, -1)

a) To find AC and BD

A(4, -2) and C(-2, 6)

Using distance formula

Two points with coordinates (x₁, y₁) and (x₂, y₂)

distance = [(x₂ - x₁)² + (y₂ - y₁)²]^0.5

AC = [(-2 - 4)² + (6 - {-2})²]^0.5

AC = [(-6)² + (8²)]^0.5

AC = [36 + 64]^0.5

AC = 100^0.5 = 10

B(5, 5) and D(-3, -1)

using distance formula

BD = [(-3 - 5)² + (-1 - 5)²]^0.5

BD = [(-8)² + (-6)²]^0.5

BD = [64 + 36]^0.5

BD = 100^0.5 = 10

b) Now AC and BD are equal each having length 10

Theorem: A parallelogram whose diagonals are congruent is a rectangle

Since, AC and BD are the diagonals of parallelogram ABCD, and, AC = BD =10

Hence parallelogram ABCD is a rectangle

c) To find AB and BC

A(4, -2) and B(5, 5)

using distance formula

AB = [(5 - 4)² + (5 - {-2})²]^0.5

AB = [1² + 7²]^0.5

AB = [1 + 49]^0.5

AB = 50^0.5 = 7.07

B(5, 5) and C(-2, 6)

using distance formula

BC = [(-2 - 5)² + (6 - 5)²]^0.5

BC = [(-7)² + 1²]^0.5

BC = [49 + 1]^0.5

BC = 50^0.5

BC = 7.07

d) Since ABCD is a parallelogram, opposite sides are equal by definition

AB = CD

BC = AD

from part c, we know that AB = BC = 7.07

Therefore, AB = CD = 7.07

Also, BC = CD = 7.07

Hence, all four sides are equal

We have already proved in part b that ABCD is a rectangle.

Theorem: A rectangle is a square if all the sides are equal

Since ABCD is also a rectangle and all sides are equal to 7.07, hence it is a square