Chapter 12.3, Problem 11PSB

a.

To determine

To calculate: The area of KITE .

Answer to Problem 11PSB

The area of KITE is $64$ .

Explanation of Solution

Given information:

The coordinates of point K are $(0,6)$ .

The coordinates of point I are $(-4,2)$ .

The coordinates of point T are $(0,-10)$ .

The coordinates of point E are $(4,2)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoras theorem.

D $=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Area of rhombus $=\frac{1}{2}(d_1\times d_2)$

d₁ and d₂ are diagonals of rhombus.

Calculation:

  

The coordinates of point K are $(0,6)$ .

The coordinates of point I are $(-4,2)$ .

The coordinates of point T are $(0,-10)$ .

The coordinates of point E are $(4,2)$ .

The distance KT and IE can be calculated by applying Distance Formula.

  $\begin{array}{l}KT=\sqrt{{\left(0-0\right)}^2+{\left(-10-6\right)}^2}\\ KT=\sqrt{{\left(0\right)}^2+{\left(-16\right)}^2}\\ KT=\sqrt{256}\\ KT=16\end{array}$

  $\begin{array}{l}IE=\sqrt{{\left(4-\left(-4\right)\right)}^2+{\left(2-2\right)}^2}\\ IE=\sqrt{{\left(8\right)}^2+{\left(0\right)}^2}\\ IE=\sqrt{64}\\ IE=8\end{array}$

Area of KITE $=\frac{1}{2}(d_1\times d_2)$

Area of KITE $=\frac{1}{2}(16\times 8)$

Area of KITE $=64$

b.

To determine

To find: The area of rectangle ABCD .

Answer to Problem 11PSB

The area of rectangle ABCD is $32$ .

Explanation of Solution

Given information:

The coordinates of point K are $(0,6)$ .

The coordinates of point I are $(-4,2)$ .

The coordinates of point T are $(0,-10)$ .

The coordinates of point E are $(4,2)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoras theorem.

D $=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle

Calculation:

  

From midpoint form,

Coordinates of point D = mid $\overline{KE}$ = $(2,4)$ ,

Coordinates of point A = mid $\overline{KI}$ = $(-2,4)$ ,

Coordinates of point C = mid $\overline{ET}$ = $(2,-4)$ .

The distance AD and DC can be calculated by applying Distance Formula.

  $\begin{array}{l}AD=\sqrt{{\left(2-\left(-2\right)\right)}^2+{\left(4-4\right)}^2}\\ AD=\sqrt{{\left(4\right)}^2+{\left(0\right)}^2}\\ AD=\sqrt{16}\\ AD=4\end{array}$

  $\begin{array}{l}DC=\sqrt{{\left(2-2\right)}^2+{\left(-4-4\right)}^2}\\ DC=\sqrt{{\left(0\right)}^2+{\left(-8\right)}^2}\\ DC=\sqrt{64}\\ DC=8\end{array}$

Area of rectangle ABCD :

  $A=DC\times AD=8\times 4=32$