Quadrilaterals

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Geometry › Quadrilaterals

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Find the area of a square if it has a diagonal of $\sqrt5$ .

$\frac{5}{2}$

$\frac{15}{2}$

$5\sqrt2$

Explanation

The diagonal of a square is also the hypotenuse of a triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Substitute in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(\sqrt5)^2}{2}$

Simplify.

$\text{Area}=\frac{5}{2}$

Two congruent equilateral triangles with sides of length are connected so that they share a side. Each triangle has a height of . Express the area of the shape in terms of .

$\frac{h}{2}$

Explanation

The shape being described is a rhombus with side lengths 1. Since they are equilateral triangles connected by one side, that side becomes the lesser diagonal, so .

The greater diagonal is twice the height of the equaliteral triangles, .

The area of a rhombus is half the product of the diagonals, so:

$A=\frac{pq}{2}=\frac{1 \cdot 2h}{2}=h$

If the diagonal of a square is $6\sqrt2$ , what is the area of the square?

Explanation

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{6\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=6$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}=6^2$

Solve.

$\text{Area}=36$

In the figure, the area of the parallelogram is . Find the length of the base.

Cannot be determined

Explanation

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

Expand this equation.

Now factor this equation.

Solve for .

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Find the area of a kite with diagonal lengths of and .

Explanation

Write the formula for the area of a kite.

$A= \frac{d1\cdot d2}{2}$

Plug in the given diagonals.

$A= \frac{(a+b)\cdot (2a-2b)}{2}$

Pull out a common factor of two in and simplify.

$A= \frac{(a+b)\cdot2(a-b)}{2}$

Use the FOIL method to simplify.

The sides of a square garden are 10 feet long. What is the area of the garden?

$140 ft^{2}$

$20ft^{2}$

$40ft^{2}$

$110ft^{2}$

Explanation

The formula for the area of a square is

$Area = s^{2}$

where is the length of the sides. So the solution can be found by

$Area=(10)^{2}=100 ft^{2}$

Find the area of a square if it has a diagonal of $\sqrt5$ .

$\frac{5}{2}$

$\frac{15}{2}$

$5\sqrt2$

Explanation

The diagonal of a square is also the hypotenuse of a triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Substitute in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(\sqrt5)^2}{2}$

Simplify.

$\text{Area}=\frac{5}{2}$

Find the length of a side of a rhombus that has diagonal lengths of and .

Explanation

Recall that in a rhombus, the diagonals are not only perpendicular to each other, but also bisect one another.

Thus, we can find the lengths of half of each diagonal and use that in the Pythagorean Theorem to find the length of the side of the rhombus.

First, find the lengths of half of each diagonal.

$\text{Half Diagonal 1}=\frac{12}{2}=6$

$\text{Half Diagonal 2}=\frac{14}{2}=7$

Now, use these half diagonals as the legs of a right triangle that has the side of the rhombus as its hypotenuse.

$\text{Side of Rhombus}=\sqrt{(\text{Half Diagonal 1})^2+(\text{Half Diagonal 2})^2}$

Plug in the lengths of the half diagonals to find the length of the rhombus.

$\text{Side of Rhombus}=\sqrt{6^2+7^2}=\sqrt{85}=9.22$

Make sure to round to places after the decimal.

Know that in a Major League Baseball infield the distance between home plate and first base is 90 feet and the infield is a perfect square.

What is the area of a Major League Baseball infield?

$8100\ ft^2$

$90\ ft^2$

$900\ ft^2$

$360\ ft^2$

$180\ ft^2$

Explanation

Because the infield is a square, the distance between each set of bases is 90 feet.

To find the area of a square you multiply the length by the width.

In this case

$90\times90=8100\ ft^2$ .

Find the area of a rhombus if the both diagonals have a length of $\sqrt{x+2}$ .

$\frac{x+2}{2}$

$\frac{\sqrt{2x+4}}{2}$

$\sqrt{2x+4}$

Explanation

Write the formula for the area of a rhombus.

$A=\frac{d_1 \cdot d_2}{2}$

Since both diagonals are equal, . Plug in the diagonals and reduce.

$A=\frac{(\sqrt x+2)(\sqrt x+2)}{2}=\frac{\sqrt{x+2}^2}{2}= \frac{x+2}{2}$

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