Area - GED Math

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. . $\angle C$ is a right angle. What percent of $\Delta ACE$ has been shaded in?

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Answer

$\Delta BCD$ is a right triangle with legs ; its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 3 \times 4 = 6$

$\Delta ACE$ is a right triangle with legs

and

;

its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 5 \times 12 = 30$

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

This region is therefore

$$\frac{24}{30 }$ \times 100 % = 80 %$ of $\Delta ACE$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. . $\angle C$ is a right angle. What is the area of the shaded region?

Tap to reveal answer

Answer

$\Delta BCD$ is a right triangle with legs ; its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 8 \times 8 = 32$

$\Delta ACE$ is a right triangle with legs

and

;

its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 12 \times 24= 144$

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

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Question

The above hexagon is regular. Give its area.

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Answer

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

All segments shown are congruent, and, since the diameter shown in the original diagram is 4, each sidelength is half this, or 2.

Each equilateral triangle has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{2^{2}$$ $\sqrt{3}$}{4} = $\frac{4\sqrt{3}$}{4} = $\sqrt{3}$$ .

There are six such triangles, so the total area of the hexagon is six times this, or $6$\sqrt{3}$$ .

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Question

Give the area of a regular hexagon with perimeter 36.

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Answer

A hexagon has six sides; a regular hexagon with perimeter 36 has sidelength

$36 \div 6 = 6$ .

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

Each equilateral triangle has sidelength 6, so each has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{6^{2}$$ $\sqrt{3}$}{4} = $\frac{36 \sqrt{3}$}{4} = 9 $\sqrt{3}$$ .

The total area of the hexagon is the area of six such triangles:

$6 \cdot 9 $\sqrt{3}$ = 54 $\sqrt{3}$$

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Question

Determine the area of a square with a side length of .

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Answer

Write the area of a square.

Substitute the side into the formula.

The answer is:

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Question

Figure NOT drawn to scale.

Refer to the above figure. Every angle shown is a right angle.

Give its area.

Tap to reveal answer

Answer

Examine the bottom figure, in which the bottom two sides have been connected. Note that the figure is now a rectangle cut out of a rectangle, and, since the opposite sides of a rectangle have the same length, we can fill in some of the side lengths as shown:

The figure is a 60-by-40 rectangle cut from a 100-by-100 square, so, to get the area of the figure, subtract the area of the former from that of the latter. The area of a rectangle is equal to the product of its dimensions, so the areas of the rectangle and the square are, respectively,

$40 \times 60 = 2,400$

and

$100 \times 100 = 10,000$ ,

making the area of the figure

.

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Question

A circle is inscribed in square that has a side length of , as shown by the figure below.

Find the area of the shaded region. Use $\pi=3.14$ .

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Answer

Since the circle is inscribed in the square, the diameter of the circle is the same length as the length of a square.

Start by finding the area of the square.

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

For the given square,

$$\text{Area}$=144$

Now, because the diameter of the circle is the same as the length of a side of the square, we now also know that the radius of the circle must be . Next recall how to find the area of a circle.

$\text{Area of Circle}$=\pi $r^2$

Plug in the found radius to find the area of the circle.

$$\text{Area of $Circle}$=\pi(6)^2$=36\pi=113.04$

Now, the shaded area is the area left over when the area of the circle is subtracted from the area of the square. Thus, we can write the following equation to find the area of the shaded region.

$$\text{Area of Shaded Region}$=144-113.04=30.96$

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Question

Josh wants to build a circular pool in his square yard that measures feet on each side. He wants to build the pool as big as possible, then pave the rest of his yard in tile. In square feet, what is the area of the yard that will be tiled? Round your answer to the nearest tenths place.

Tap to reveal answer

Answer

Start by drawing out the square yard and the circular pool in a way that maximizes the area of the pool.

Notice that the diameter of the pool will be the same length as the side of the square.

Since the question asks about the area that is left over after the pool is built, we can find that area by subtracting the area of the pool from the area of the square.

Start by finding the area of the square.

$$\text{Area of $Square}$=50^2$=2500$

Next, find the area of the circular pool.

Since the diameter of the pool is , the radius of the pool must be . Recall how to find the area of a circle:

$\text{Area}$=\pi $r^2$

Plug in the radius of the circle.

$$\text{Area}$=\pi $(25)^2$=625\pi$

Subtract the area of the circle from that of the square.

$$\text{Tiled Area}$=2500-625\pi=536.5$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. . $\angle C$ is a right angle. What percent of $\Delta ACE$ has been shaded in?

Tap to reveal answer

Answer

$\Delta BCD$ is a right triangle with legs ; its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 3 \times 4 = 6$

$\Delta ACE$ is a right triangle with legs

and

;

its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 5 \times 12 = 30$

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

This region is therefore

$$\frac{24}{30 }$ \times 100 % = 80 %$ of $\Delta ACE$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. . $\angle C$ is a right angle. What is the area of the shaded region?

Tap to reveal answer

Answer

$\Delta BCD$ is a right triangle with legs ; its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 8 \times 8 = 32$

$\Delta ACE$ is a right triangle with legs

and

;

its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 12 \times 24= 144$

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

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Question

The above hexagon is regular. Give its area.

Tap to reveal answer

Answer

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

All segments shown are congruent, and, since the diameter shown in the original diagram is 4, each sidelength is half this, or 2.

Each equilateral triangle has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{2^{2}$$ $\sqrt{3}$}{4} = $\frac{4\sqrt{3}$}{4} = $\sqrt{3}$$ .

There are six such triangles, so the total area of the hexagon is six times this, or $6$\sqrt{3}$$ .

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Question

Give the area of a regular hexagon with perimeter 36.

Tap to reveal answer

Answer

A hexagon has six sides; a regular hexagon with perimeter 36 has sidelength

$36 \div 6 = 6$ .

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

Each equilateral triangle has sidelength 6, so each has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{6^{2}$$ $\sqrt{3}$}{4} = $\frac{36 \sqrt{3}$}{4} = 9 $\sqrt{3}$$ .

The total area of the hexagon is the area of six such triangles:

$6 \cdot 9 $\sqrt{3}$ = 54 $\sqrt{3}$$

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Question

Determine the area of a square with a side length of .

Tap to reveal answer

Answer

Write the area of a square.

Substitute the side into the formula.

The answer is:

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Question

Figure NOT drawn to scale.

Refer to the above figure. Every angle shown is a right angle.

Give its area.

Tap to reveal answer

Answer

Examine the bottom figure, in which the bottom two sides have been connected. Note that the figure is now a rectangle cut out of a rectangle, and, since the opposite sides of a rectangle have the same length, we can fill in some of the side lengths as shown:

The figure is a 60-by-40 rectangle cut from a 100-by-100 square, so, to get the area of the figure, subtract the area of the former from that of the latter. The area of a rectangle is equal to the product of its dimensions, so the areas of the rectangle and the square are, respectively,

$40 \times 60 = 2,400$

and

$100 \times 100 = 10,000$ ,

making the area of the figure

.

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Question

A circle is inscribed in square that has a side length of , as shown by the figure below.

Find the area of the shaded region. Use $\pi=3.14$ .

Tap to reveal answer

Answer

Since the circle is inscribed in the square, the diameter of the circle is the same length as the length of a square.

Start by finding the area of the square.

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

For the given square,

$$\text{Area}$=144$

Now, because the diameter of the circle is the same as the length of a side of the square, we now also know that the radius of the circle must be . Next recall how to find the area of a circle.

$\text{Area of Circle}$=\pi $r^2$

Plug in the found radius to find the area of the circle.

$$\text{Area of $Circle}$=\pi(6)^2$=36\pi=113.04$

Now, the shaded area is the area left over when the area of the circle is subtracted from the area of the square. Thus, we can write the following equation to find the area of the shaded region.

$$\text{Area of Shaded Region}$=144-113.04=30.96$

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Question

Josh wants to build a circular pool in his square yard that measures feet on each side. He wants to build the pool as big as possible, then pave the rest of his yard in tile. In square feet, what is the area of the yard that will be tiled? Round your answer to the nearest tenths place.

Tap to reveal answer

Answer

Start by drawing out the square yard and the circular pool in a way that maximizes the area of the pool.

Notice that the diameter of the pool will be the same length as the side of the square.

Since the question asks about the area that is left over after the pool is built, we can find that area by subtracting the area of the pool from the area of the square.

Start by finding the area of the square.

$$\text{Area of $Square}$=50^2$=2500$

Next, find the area of the circular pool.

Since the diameter of the pool is , the radius of the pool must be . Recall how to find the area of a circle:

$\text{Area}$=\pi $r^2$

Plug in the radius of the circle.

$$\text{Area}$=\pi $(25)^2$=625\pi$

Subtract the area of the circle from that of the square.

$$\text{Tiled Area}$=2500-625\pi=536.5$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. . $\angle C$ is a right angle. What percent of $\Delta ACE$ has been shaded in?

Tap to reveal answer

Answer

$\Delta BCD$ is a right triangle with legs ; its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 3 \times 4 = 6$

$\Delta ACE$ is a right triangle with legs

and

;

its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 5 \times 12 = 30$

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

This region is therefore

$$\frac{24}{30 }$ \times 100 % = 80 %$ of $\Delta ACE$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. . $\angle C$ is a right angle. What is the area of the shaded region?

Tap to reveal answer

Answer

$\Delta BCD$ is a right triangle with legs ; its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 8 \times 8 = 32$

$\Delta ACE$ is a right triangle with legs

and

;

its area is half the product of its legs, which is

$Area= $\frac{1}{2}$ \times 12 \times 24= 144$

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

← Didn't Know|Knew It →

Question

The above hexagon is regular. Give its area.

Tap to reveal answer

Answer

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

All segments shown are congruent, and, since the diameter shown in the original diagram is 4, each sidelength is half this, or 2.

Each equilateral triangle has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{2^{2}$$ $\sqrt{3}$}{4} = $\frac{4\sqrt{3}$}{4} = $\sqrt{3}$$ .

There are six such triangles, so the total area of the hexagon is six times this, or $6$\sqrt{3}$$ .

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Question

Give the area of a regular hexagon with perimeter 36.

Tap to reveal answer

Answer

A hexagon has six sides; a regular hexagon with perimeter 36 has sidelength

$36 \div 6 = 6$ .

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

Each equilateral triangle has sidelength 6, so each has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{6^{2}$$ $\sqrt{3}$}{4} = $\frac{36 \sqrt{3}$}{4} = 9 $\sqrt{3}$$ .

The total area of the hexagon is the area of six such triangles:

$6 \cdot 9 $\sqrt{3}$ = 54 $\sqrt{3}$$

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