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2 years ago

Complete the coordinate proof of the theorem. Given: A B C D is a rectangle. Prove: The diagonals of A B C D are congruent.

Mathematics

2 answers:

Leokris [45]2 years ago

7 0

Answer:

all the side but to are eaual you must divide by two then go

Inessa05 [86]2 years ago

3 0

Step-by-step explanation:

Given coordinates of the rectangle ABCD:

A(0, 0), B(a, 0), C(x, y), D(0, b)

Lets first find coordinates of C

We can see that A and B are on same line horizontally, so C and D will be on the parallel line

The distance AB and CD are equal also distance AD and BC are equal, therefore

AB = a - 0 = a, CD = a
AD = b - 0 = b, BC = b

So the coordinates of C are:

x = a, y = b

Now the diagonals:

AC = √(a-0)² + (b-0)² = √a²+b²
BD =√(0-a)² + (b -0)² = √a² + b²

Since AC = BD, we can state they are congruent

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$\displaystyle \rm A_{ \text{c - pentagon}} =41$

$\rm \displaystyle P _{ \rm c - pentagon} = 20+ 4 \sqrt{2}$

Step-by-step explanation:

we have a square and a triangle

we want to figure out the area and the perimeter of the pentagon

to figure out the area of the pentagon we can use the given formula:

$\displaystyle \rm A_{ \text{c - pentagon}} = A_{ \text{squre}} - A_{ \text{triangle}}$

let's figure out $A_{\rm square}$ :

since the given shape is a square Every angle of its 90° Thus the triangle is a right angle triangle

therefore the height is 4

$\displaystyle A _{ \rm triangle} = \frac{1}{2} \times b \times h$

substitute h and b

$\displaystyle A _{ \rm triangle} = \frac{1}{2} \times 4 \times 4$

reduce fraction:

$\displaystyle A _{ \rm triangle} = 2 \times 4$

simplify multiplication:

$\displaystyle A _{ \rm triangle} = 8$

likewise square

$\displaystyle A _{ \rm squre} = {s}^{2}$

substitute s:

$\displaystyle A _{ \rm squre} = {7}^{2}$

simplify square

$\displaystyle A _{ \rm squre} = 49$

hence,

$\displaystyle \rm A_{ \text{c - pentagon}} =49- 8$

$\displaystyle \rm A_{ \text{c - pentagon}} =41$

to figure out perimeter

let's figure out hypotenuse first

$\displaystyle h = \sqrt{ {4}^{2} + {4}^{2} }$

$\displaystyle h = \sqrt{ 16 + 16}$

$\displaystyle h = 4 \sqrt{2}$

therefore,

$\rm \displaystyle P _{ \rm c - pentagon} = 3 + 3 + 4 \sqrt{2} + 7 + 7$

$\rm \displaystyle P _{ \rm c - pentagon} = 6 + 4 \sqrt{2} + 14$

$\rm \displaystyle P _{ \rm c - pentagon} = 20+ 4 \sqrt{2}$

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