Rhombuses - Geometry

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Question

A rhombus has a side length of 5. Which of the following is NOT a possible value for its area?

Answer

The area of a rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

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Question

Which of the following shapes is a rhombus?

Answer

A rhombus is a four-sided figure where all sides are straight and equal in length. All opposite sides are parallel. A square is considered to be a rhombus.

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Question

Assume quadrilateral $\small ABCD$ is a rhombus. The perimeter of $\small ABCD$ is , and the length of one of its diagonals is $\small 12$ . What is the area of $\small ABCD$ ?

Answer

To solve for the area of the rhombus $\small ABCD$ , we must use the equation $\small A=\frac{pq}{2}$ , where $\small p$ and $\small q$ are the diagonals of the rhombus. Since the perimeter of the rhombus is $\small 40$ , and by definition all 4 sides of a rhombus have the same length, we know that the length of each side is $\small 10$ . We can find the length of the other diagonal if we recognize that the two diagonals combined with a side edge form a right triangle. The length of the hypotenuse is $\small 10$ , and each leg of the triangle is equal to one-half of each diagonal. We can therefore set up an equation involving Pythagorean's Theorem as follows:

$\small 10^2=x^2+6^2$ , where $\small x$ is equal to one-half the length of the unknown diagonal.

We can therefore solve for $\small x$ as follows:

$\small x^2=10^2-6^2=100-36=64$

$\small x$ is therefore equal to 8, and our other diagonal is 16. We can now use both diagonals to solve for the area of the rhombus:

$\small A=\frac{(16)(12)}{2}=\frac{192}{2}=96$

The area of rhombus $\small ABCD$ is therefore equal to $\small 96$

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Question

Assume quadrilateral is a rhombus. If diagonal $\overline{MO}=29$ and diagonal $\overline{NP}=14$ , what is the area of rhombus

Answer

Solving for the area of rhombus requires knowledge of the equation for finding the area of a rhombus. The equation is $\small A=\frac{pq}{2}$ , where and $\small q$ are the two diagonals of the rhombus. Since both of these values are given to us in the original problem, we merely need to substitute these values into the equation to obtain:

$\small A=\frac{(29)(14)}{2}=\frac{406}{2}=203$

The area of rhombus $\small MNOP$ is therefore $\small 203$ square units.

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Question

What is the area of the rhombus above?

Answer

The formula for the area of a rhombus from the diagonals is half the product of the diagonals, or in mathematical terms:

$A=\frac{p\cdot q}{2}$ where and are the lengths of the diagonals.

Substituting our values yields,

$A=\frac{8*12}{2}=\frac{96}{2}=48$

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Question

Above is a rhombus imposed on a rectangle. What is the area of the rhombus?

Answer

One of the formulas for a rhombus is base times height,

$A=b \cdot h$

Since a rhombas has equal sides, the base is 5 and the height of the rhombus is the same as the height of the rectangle, 4.

Substituting in these values we get the following:

$A=5 \cdot 4 =20$

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Question

The above figure shows a rhombus . Give its area.

Answer

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

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

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

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Question

Find the area of a rhombus if the diagonal lengths are and .

Answer

The area of the rhombus is given below. Substitute the diagonals into the formula.

$A=\frac{d1\cdot d2}{2} = \frac{(1)(2)}{2}=1$

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Question

What is the area of a rhombus with diagonal lengths of $\sqrt2$ and $\sqrt3$ ?

Answer

The area of a rhombus is given below. Plug in the diagonals and solve for the area.

$A=\frac{d1 \cdot d2}{2} = \frac{\sqrt2 \cdot \sqrt3}{2}=\frac{\sqrt6}{2}$

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Question

Rhombus has perimeter 48; . What is the area of Rhombus ?

Answer

Each side of a rhombus is congruent, so if it has perimeter 48, it has sidelength 12. Also, the diagonals of a rhombus are each other's perpendicular bisectors, so if they are both constructed, and their point of intersection is called , then . The following figure is formed by the rhombus and its diagonals.

$\bigtriangleup AMB$ is a right triangle with its short leg half the length of its hypotenuse, so it is a 30-60-90 triangle, and its long leg measures $6\sqrt{3}$ by the 30-60-90 Theorem. Therefore, $BD = 12\sqrt{3}$ . The area of a rhombus is half the product of the lengths of its diagonals:

$\frac{1}{2} \times 12 \times 12 \sqrt{3} = 72 \sqrt{3}$

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Question

A rhombus contains diagonals with the length and . Find the area of the rhombus.

Answer

The equation for the area of a rhombus is given by:

$A= \frac{p \cdot q}{2}$

where and are the two diagonal lengths.

This problem very quickly becomes one of the "plug and chug" type, where the given values just need to be substituted into the equation and the equation then solved. By plugging in the values given, we get:

$A = \frac{12:cm \cdot 6:cm}{2}$

$A = \frac{72:cm^2}2{}$

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Question

Find the area of a rhombus if its diagonal lengths are $\sqrt{2x}:cm$ and $\sqrt{4x}:cm$ .

Answer

Write the equation for the area of a rhombus.

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

Substitute the diagonals and evaluate the area.

$A= \frac{\sqrt{2x}:cm\sqrt{4x}:cm}{2} = \frac{\sqrt{8x^2}}{2}:cm^2 = \frac{2x\sqrt2}{2}:cm^2=x\sqrt2:cm^2$

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Question

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

Answer

Write the formula for the area of a rhombus.

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

Substitute the given diagonal lengths:

$A=\frac{(2x+2)(4x+4)}{2}$

Use FOIL to multiply the two parentheticals in the numerator:

First: $2x\cdot4x=8x^2$

Outer: $2x\cdot4=8x$

Inner: $2\cdot4x=8x$

Last: $2\cdot 4=8$

Add your results together:

$A=\frac{8x^2+16x+8}{2}$

Divide all elements in the numerator by two to arrive at the correct answer:

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Question

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

Answer

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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Question

What is the area of a rhombus if the diagonals are $\frac{6}{x}$ and $\frac{6}{x^2}$ ?

Answer

Write the formula for an area of a rhombus.

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

Substitute the diagonal lengths provided into the formula.

$A=\frac{d1\cdot d2}{2} = \frac{\frac{6}{x}\cdot\frac{6}{x^2}}{2}$

Multiply the two terms in the numerator.

$A= \frac{\frac{36}{x^3}}{2}$

You can consider the outermost division by two as multiplying everything in the numerator by $\frac{1}{2}$ .

$A= \frac{1}{2}(\frac{36}{x^3})$

Multiply across and reduce to arrive at the correct answer.

$A= \frac{1}{2}(\frac{36}{x^3})=\frac{36}{2x^3}=\frac{18}{x^3}$

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Question

Find the area of a rhombus if the diagonals lengths are and .

Answer

Write the formula for the area of a rhombus:

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

Substitute the given lengths of the diagonals and solve:

$A=\frac{d1 \cdot d2}{2} = \frac{20:cm \cdot 40:cm}{2} =\frac{800:cm^2}{2} = 400:cm^2$

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Question

Find the area of a rhombus if the diagonals lengths are and .

Answer

Write the formula for finding the area of a rhombus. Substitute the diagonals and evaluate.

$A=\frac{d1\cdot d2}{2}= \frac{2a \cdot 5a^2}{2}= 5a^3$

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Question

Show algebraically how the formula for the area of a rhombus is developed.

Answer

The given rhombus is divded into two congruent isosceles triangles.

Each isosceles triangle has a height $h = \frac{1}{2}d_{1}$ and a base $b = d_{2}$ .

The area of each isosceles triangle is $A = \frac{1}{2}bh$ .

The areas of the two isosceles triangles are added together,

$= \frac{1}{2}(\frac{1}{2}d_{1})(d_{2}) + \frac{1}{2}(\frac{1}{2}d_{1})(d_{2})$

$= \frac{1}{4}d_{1}d_{2} +\frac{1}{4}d_{1}d_{2}$

$=\frac{1}{2}d_{1}d_{2}$

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Question

A rhombus is $d_{2} = 10$ meters across. $d_{1} = \frac{1}{2}d_{2}$ . Find the area.

Answer

Substitute $d_{2} = 10$ into the expression $d_{1} = \frac{1}{2}d_{2}$ , $d_{1} = \frac{1}{2}\cdot 10 = 5$ .

Substitute $d_{2} = 10$ and $d_{1} = 5$ into the rhombus area formula,

$A = \frac{1}{2}d_{1}d_{2}$

$A = \frac{1}{2}\cdot 5 \cdot 10$

$= \frac{1}{2}\cdot 50$

$= 25\textup{ m}^2$

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Question

Find the area of the rhombus shown below. You will have to find the lengths of the sides as well.

The rhombus shown has the following coordinates:

Round to the nearest hundredth.

Answer

Finding the area of a rhombus follows the formula:

$\frac{p\times q}{2}$

In this rhombus, you will find that , which are the two x coordinates.

The length of q is a more involved process. You can find q by using the Pythagorean Theorem.

$\sqrt{65}=8.06$ .

Therefore, the area is

$\frac{6\times 8.06}{2}=24.18$ .

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