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Advanced Geometry : Plane Geometry

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6 Diagnostic Tests 57 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #12 : Rhombuses

A rhombus contains diagonals with the length $12 \:cm$ and $6 \:cm$ . Find the area of the rhombus.

Possible Answers:

$72\:cm^2$

$36\:cm^2$

$54\:cm^2$

$88\:cm^2$

$96\:cm^2$

Correct answer:

$36\:cm^2$

Explanation:

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:

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

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

$A= 36\:cm^2$

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Example Question #11 : How To Find The Area Of A Rhombus

Find the area of a rhombus if the diagonals lengths are $20\:cm$ and $40\:cm$ .

Possible Answers:

$600\:cm^2$

$200\:cm^2$

$400\:cm^2$

$800\:cm^2$

$900\:cm^2$

Correct answer:

$400\:cm^2$

Explanation:

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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Example Question #11 : How To Find The Area Of A Rhombus

Find the area of a rhombus if the diagonals lengths are and 5a^2 .

Possible Answers:

10a^3

5a^3

5a^2

2a+5a^2

7a^3

Correct answer:

5a^3

Explanation:

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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Example Question #1 : How To Find The Area Of A Rhombus

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

Possible Answers:

Correct answer:

Explanation:

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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Example Question #13 : Rhombuses

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

Rhombus area

The rhombus shown has the following coordinates:

(-2,5)(4,5)(3,-3) (-3,-3)

Round to the nearest hundredth.

Possible Answers:

24.18

48.36

Correct answer:

24.18

Explanation:

Finding the area of a rhombus follows the formula:

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

In this rhombus, you will find that p=4-(-2) , which are the two x coordinates.

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

1^2+8^2=65

$\sqrt{65}=8.06$ .

Therefore, the area is

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

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Example Question #16 : Rhombuses

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

Varsity4

Possible Answers:

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

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

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

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

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

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

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

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

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

$\frac{1}{4}d_{1}\cdot \frac{1}{4}d_{2}$

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

$\frac{1}{2}\cdot (d_{1}\cdot d_{2})$

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

Correct answer:

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

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

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

Explanation:

1) The given rhombus is divded into two congruent isosceles triangles.

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

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

4) The areas of the two isosceles triangles are added together,

A + A

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

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

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

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Example Question #17 : Rhombuses

Find the area of the rhombus below.

Possible Answers:

120

100

Correct answer:

120

Explanation:

Recall that the diagonals of the rhombus are perpendicular bisectors. From the given side and the given diagonal, we can find the length of the second diagonal by using the Pythagorean Theorem.

$\text{side}^2=(\frac{\text{diagonal 1}}{2})^2-(\frac{\text{diagonal 2}}{2})^2$

Let the given diagonal be diagonal 1, and rearrange the equation to solve for diagonal 2.

$(\frac{\text{diagonal 2}}{2})^2=\text{side}^2-(\frac{\text{diagonal 1}}{2})^2$

$(\frac{\text{diagonal 2}}{2})=\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}$

$\text{diagonal 2}=2\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}$

Plug in the given side and diagonal to find the length of diagonal 2.

$\text{diagonal 2}=2\sqrt{13^2-(\frac{10}{2})^2}=2\sqrt{144}=24$

Now, recall how to find the area of a rhombus:

$\text{Area of Rhombus}=\frac{(\text{diagonal 1})(\text{diagonal 2})}{2}$

Plug in the two diagonals to find the area.

$\text{Area of Rhombus}=\frac{(10)(24)}{2}=120$

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Example Question #102 : Plane Geometry

Find the area of the rhombus below.

Possible Answers:

101.51

187.51

109.51

203.02

Correct answer:

203.02

Explanation:

Recall that the diagonals of the rhombus are perpendicular bisectors. From the given side and the given diagonal, we can find the length of the second diagonal by using the Pythagorean Theorem.

$\text{side}^2=(\frac{\text{diagonal 1}}{2})^2-(\frac{\text{diagonal 2}}{2})^2$

Let the given diagonal be diagonal 1, and rearrange the equation to solve for diagonal 2.

$(\frac{\text{diagonal 2}}{2})^2=\text{side}^2-(\frac{\text{diagonal 1}}{2})^2$

$(\frac{\text{diagonal 2}}{2})=\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}$

$\text{diagonal 2}=2\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}$

Plug in the given side and diagonal to find the length of diagonal 2.

$\text{diagonal 2}=2\sqrt{15^2-(\frac{16}{2})^2}=2\sqrt{161}$

Now, recall how to find the area of a rhombus:

$\text{Area of Rhombus}=\frac{(\text{diagonal 1})(\text{diagonal 2})}{2}$

Plug in the two diagonals to find the area.

$\text{Area of Rhombus}=\frac{(16)(2\sqrt{161})}{2}=203.02$

Make sure to round to places after the decimal.

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Example Question #101 : Plane Geometry

Find the area of the rhombus below.

Possible Answers:

26.83

21.03

29.58

25.48

Correct answer:

26.83

Explanation:

Recall that the diagonals of the rhombus are perpendicular bisectors. From the given side and the given diagonal, we can find the length of the second diagonal by using the Pythagorean Theorem.

$\text{side}^2=(\frac{\text{diagonal 1}}{2})^2-(\frac{\text{diagonal 2}}{2})^2$

Let the given diagonal be diagonal 1, and rearrange the equation to solve for diagonal 2.

$(\frac{\text{diagonal 2}}{2})^2=\text{side}^2-(\frac{\text{diagonal 1}}{2})^2$

$(\frac{\text{diagonal 2}}{2})=\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}$

$\text{diagonal 2}=2\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}$

Plug in the given side and diagonal to find the length of diagonal 2.

$\text{diagonal 2}=2\sqrt{7^2-(\frac{4}{2})^2}=2\sqrt{45}$

Now, recall how to find the area of a rhombus:

$\text{Area of Rhombus}=\frac{(\text{diagonal 1})(\text{diagonal 2})}{2}$

Plug in the two diagonals to find the area.

$\text{Area of Rhombus}=\frac{(4)(2\sqrt{45})}{2}=26.83$

Make sure to round to places after the decimal.

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Example Question #103 : Plane Geometry

Find the area of the rhombus below.

Possible Answers:

21.20

17.43

32.98

31.11

Correct answer:

17.43

Explanation:

Recall that the diagonals of the rhombus are perpendicular bisectors. From the given side and the given diagonal, we can find the length of the second diagonal by using the Pythagorean Theorem.

$\text{side}^2=(\frac{\text{diagonal 1}}{2})^2-(\frac{\text{diagonal 2}}{2})^2$

Let the given diagonal be diagonal 1, and rearrange the equation to solve for diagonal 2.

$(\frac{\text{diagonal 2}}{2})^2=\text{side}^2-(\frac{\text{diagonal 1}}{2})^2$

$(\frac{\text{diagonal 2}}{2})=\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}$

$\text{diagonal 2}=2\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}$

Plug in the given side and diagonal to find the length of diagonal 2.

$\text{diagonal 2}=2\sqrt{6^2-(\frac{3}{2})^2}=2\sqrt{\frac{135}{4}}$

Now, recall how to find the area of a rhombus:

$\text{Area of Rhombus}=\frac{(\text{diagonal 1})(\text{diagonal 2})}{2}$

Plug in the two diagonals to find the area.

$\text{Area of Rhombus}=\frac{(3)(2\sqrt{\frac{135}{4}})}{2}=17.43$

Make sure to round to places after the decimal.

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Advanced Geometry : Plane Geometry

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #12 : Rhombuses

Example Question #11 : How To Find The Area Of A Rhombus

Example Question #11 : How To Find The Area Of A Rhombus

Example Question #1 : How To Find The Area Of A Rhombus

Example Question #13 : Rhombuses

Example Question #16 : Rhombuses

Example Question #17 : Rhombuses

Example Question #102 : Plane Geometry

Example Question #101 : Plane Geometry

Example Question #103 : Plane Geometry

All Advanced Geometry Resources

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