Plane Geometry

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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:

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

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.

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:

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

Find the area of the rhombus.

Explanation

Recall that one of the ways to find the area of a rhombus is with the following formula:

$\text{Area}=\text{base}\times\text{height}$

Now, since all four sides in the rhombus are the same, we know from the given side value what the length of the base will be. In order to find the length of the height, we will need to use sine.

$\sin x=\frac{\text{height}}{\text{side}}$ , where is the given angle.

$\text{height}=(\text{side})(\sin x)$

Now, plug this into the equation for the area to get the following equation:

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

Plug in the given side length and angle values to find the area.

$\text{Area}=(5)^2\sin 33=13.62$

Make sure to round to places after the decimal.

A kite has an area of square units, and one diagonal is units longer than the other. In unites, what is the length of the shorter diagonal?

Explanation

Let be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by .

Recall how to find the area of a kite:

$\text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for .

$119=\frac{x(x+3)}{2}$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, .

The length of the shorter diagonal is units long.

Which of the following shapes is a trapezoid?

Shapes

Explanation

A trapezoid is a four-sided shape with straight sides that has a pair of opposite parallel sides. The other sides may or may not be parallel. A square and a rectangle are both considered trapezoids.

Find the area of the rhombus.

Explanation

Recall that one of the ways to find the area of a rhombus is with the following formula:

$\text{Area}=\text{base}\times\text{height}$

Now, since all four sides in the rhombus are the same, we know from the given side value what the length of the base will be. In order to find the length of the height, we will need to use sine.

$\sin x=\frac{\text{height}}{\text{side}}$ , where is the given angle.

$\text{height}=(\text{side})(\sin x)$

Now, plug this into the equation for the area to get the following equation:

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

Plug in the given side length and angle values to find the area.

$\text{Area}=(5)^2\sin 33=13.62$

Make sure to round to places after the decimal.

Rhombus_1

The above figure shows a rhombus . Give its area.

$40 \sqrt{11}$

$40\sqrt{61}$

Explanation

Construct the other diagonal of the rhombus, which, along with the first one, form a pair of mutual perpendicular bisectors.

Rhombus_1

By the Pythagorean Theorem,

$n = \sqrt{12^{2}- 10^{2}} =\sqrt{144- 100} = \sqrt{44}= \sqrt{4}\cdot \sqrt{11} = 2 \sqrt{11}$

The rhombus can be seen as the composite of four congruent right triangles, each with legs 10 and $2 \sqrt{11}$ , so the area of the rhombus is

$A = 4 \cdot \frac{1}{2}\cdot 10 \cdot 2\sqrt{11} =40 \sqrt{11}$ .

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:

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

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