How to find the area of a polygon - PSAT Math

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Question

In the square above, the radius of each half-circle is 6 inches. What is the area of the shaded region?

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Answer

We can find the area of the shaded region by subtracting the area of the semicircles, which is much easier to find. Two semi-circles are equivalent to one full circle. Thus we can just use the area formula, where r = 6:

π(62) → 36π

Now we must subtract the area of the semi-circles from the total area of the square. Since we know that the radius also covers half of a side, 6(2) = 12 is the full length of a side of the square. Squaring this, 122 = 144. Subtracting the area of the circles, we get our final terms,

= 144 – 36π

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Question

If square A has a side of length 5 inches, how many times bigger is the area of square B if it has a side of length 25 inches?

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Answer

First find the area of both squares using the formula .

For square A, s = 5.

For square B, s = 25.

The question is asking for the ratio of these two areas, which will tell us how many times bigger square B is. Divide the area of square B by the area of square A to find the answer.

$\frac{A_B}{A_A}=\frac{625}{25}=25$

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Question

If Bailey paints the wall shaped like above and uses one bucket per 5 square units, how many buckets does Bailey need?

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Answer

To solve, we will need to find the area of the wall. We can do this by finding the areas of each section and adding them together. Break the area into a rectange and two triangles.

The area of the rectangle will be equal to the base times the height. The area of each triangle will be one half its given base times its height.

$A_{rec}=bh$

$A_{tri}=\frac{1}{2}bh$

For the rectangle, the base is 12 and the height is 4 (both given in the figure).

$A_{rec}=(12)(4)=48$

The triangle to the right has a given base of 6, but we need to solve for its height. The height will be equal to the difference between the total height (6) and the height of the rectangle (4).

$h_{tri}=6-4=2$

We now have the base and height of the triangle to the right, allowing us to calculate its area.

$A_{tri_r}=\frac{1}{2}(6)(2)=6$

Now we need to solve the triangle to the left. We solved for its height (2), but we still need to solve for its base. The total base of the rectangle is 12. Subtract the base of the right-side triangle (6) and the small segment at the top of the rectangle (3) from this total length to solve for the base of the left triangle.

The left-side triangle has a base of 3 and a height of 2, allowing us to calculate its area.

$A_{tri_l}=\frac{1}{2}(3)(2)=3$

Add together the two triangles and the rectangle to find the total area.

We know that each bucket of paint will cover 5 square units, and we have 57 square units total. Divide to find how many buckets are required.

$\frac{57}{5}=11.4$

We will need 11 full buckets and part of a twelfth bucket to cover the wall, meaning that we will need 12 buckets total.

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Question

A square is inscribed within a circle with a radius $\small 3\sqrt{2}$ . Find the area of the circle that is not covered by the square.

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Answer

First, find the area of the circle.

$A=\pi(3\sqrt{2})^2$

$\small A=18\pi$

Next, find the length of 1 side of the square using the Pythagorean Theorem. Two radii from the center of the circle to adjacent corners of the square will create a right angle at the center of the circle. The radii will be the legs of the triangle and the side of the square will be the hypotenuse.

$\small (3\sqrt{2})^2 +(3\sqrt{2})^2 =c^2$

$\small 18+18=36=c^2$

$\small c=6$

Find the area of the square.

$\small A=c^2=6^2=36$

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

$\small 18\pi-36$

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Question

Refer to the above figure. Quadrilateral is a square. Give the area of Polygon in terms of .

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Answer

$\overline{BE }$ is both one side of Square and the hypotenuse of $\Delta ABE$ ; its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

$BE = \sqrt{(AB)^{2}+ \left ( AE\right )^{2}}$

$BE = \sqrt{x^{2}+ \left ( 2x\right )^{2}} = \sqrt{x^{2}+ 4x ^{2}} = \sqrt{5x ^{2}} = x\sqrt{5}$

Polygon is the composite of $\Delta ABE$ and Square , so its area is the sum of those of the two figures.

The area of $\Delta ABE$ is half the product of its legs:

$A= \frac{1}{2} \cdot x \cdot 2x = x^{2}$

The area of Square is the square of the length of a side:

$A = \left ( x\sqrt{5} \right )^{2} = 5x^{2}$

Add the areas:

$x^{2}+ 5x^{2} = 6x^{2}$

This is the area of the polygon.

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Question

Note: Figure NOT drawn to scale.

The above figure shows Rectangle ; is the midpoint of $\overline{CA}$ ; $EC = 2 \cdot RE$ ; $RC = 2 \cdot RN$ .

What percent of Rectangle is shaded?

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Answer

The area of , the shaded region in question, is that of the rectangle minus those of $\Delta ERN$ and $\Delta ECT$ . We look at both.

The answer is independent of the sidelengths of the rectangle, so to ease calculations - this will become more apparent later - we will arbitrarily assign to the rectangle the dimensions

and, subsequently,

and, since is the midpoint of $\overline{CA}$ ,

.

The area of Rectangle is equal to $RN \cdot RC = 6 \cdot 12 = 72$ .

Since $EC = 2 \cdot RE$ , and ,

and

The area of $\Delta ERN$ is equal to $\frac{1}{2}(RE)(RN)= \frac{1}{2} \cdot 4 \cdot 6 = 12$ .

The area of $\Delta ECT$ is equal to $\frac{1}{2}(EC)(CT)= \frac{1}{2} \cdot 8 \cdot 3 = 12$

The area of the shaded region is therefore , which is

$\frac{48}{72} \times 100 % = 66 \frac{2}{3} %$ of the rectangle.

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Question

A square has an area of 36 cm2. A circle is inscribed and cut out. What is the area of the remaining shape? Use 3.14 to approximate π.

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Answer

We need to find the area of both the square and the circle and then subtract the two. Inscribed means draw within a figure so as to touch in as many places as possible. So the circle is drawn inside the square. The opposite is circumscribed, meaning drawn outside.

Asquare = s2 = 36 cm2 so the side is 6 cm

6 cm is also the diameter of the circle and thus the radius is 3 cm

A circle = πr2 = 3.14 * 32 = 28.28 cm2

The resulting difference is 7.74 cm2

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Question

In the square above, the radius of each half-circle is 6 inches. What is the area of the shaded region?

Tap to reveal answer

Answer

We can find the area of the shaded region by subtracting the area of the semicircles, which is much easier to find. Two semi-circles are equivalent to one full circle. Thus we can just use the area formula, where r = 6:

π(62) → 36π

Now we must subtract the area of the semi-circles from the total area of the square. Since we know that the radius also covers half of a side, 6(2) = 12 is the full length of a side of the square. Squaring this, 122 = 144. Subtracting the area of the circles, we get our final terms,

= 144 – 36π

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Question

If square A has a side of length 5 inches, how many times bigger is the area of square B if it has a side of length 25 inches?

Tap to reveal answer

Answer

First find the area of both squares using the formula .

For square A, s = 5.

For square B, s = 25.

The question is asking for the ratio of these two areas, which will tell us how many times bigger square B is. Divide the area of square B by the area of square A to find the answer.

$\frac{A_B}{A_A}=\frac{625}{25}=25$

← Didn't Know|Knew It →

Question

If Bailey paints the wall shaped like above and uses one bucket per 5 square units, how many buckets does Bailey need?

Tap to reveal answer

Answer

To solve, we will need to find the area of the wall. We can do this by finding the areas of each section and adding them together. Break the area into a rectange and two triangles.

The area of the rectangle will be equal to the base times the height. The area of each triangle will be one half its given base times its height.

$A_{rec}=bh$

$A_{tri}=\frac{1}{2}bh$

For the rectangle, the base is 12 and the height is 4 (both given in the figure).

$A_{rec}=(12)(4)=48$

The triangle to the right has a given base of 6, but we need to solve for its height. The height will be equal to the difference between the total height (6) and the height of the rectangle (4).

$h_{tri}=6-4=2$

We now have the base and height of the triangle to the right, allowing us to calculate its area.

$A_{tri_r}=\frac{1}{2}(6)(2)=6$

Now we need to solve the triangle to the left. We solved for its height (2), but we still need to solve for its base. The total base of the rectangle is 12. Subtract the base of the right-side triangle (6) and the small segment at the top of the rectangle (3) from this total length to solve for the base of the left triangle.

The left-side triangle has a base of 3 and a height of 2, allowing us to calculate its area.

$A_{tri_l}=\frac{1}{2}(3)(2)=3$

Add together the two triangles and the rectangle to find the total area.

We know that each bucket of paint will cover 5 square units, and we have 57 square units total. Divide to find how many buckets are required.

$\frac{57}{5}=11.4$

We will need 11 full buckets and part of a twelfth bucket to cover the wall, meaning that we will need 12 buckets total.

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Question

A square is inscribed within a circle with a radius $\small 3\sqrt{2}$ . Find the area of the circle that is not covered by the square.

Tap to reveal answer

Answer

First, find the area of the circle.

$A=\pi(3\sqrt{2})^2$

$\small A=18\pi$

Next, find the length of 1 side of the square using the Pythagorean Theorem. Two radii from the center of the circle to adjacent corners of the square will create a right angle at the center of the circle. The radii will be the legs of the triangle and the side of the square will be the hypotenuse.

$\small (3\sqrt{2})^2 +(3\sqrt{2})^2 =c^2$

$\small 18+18=36=c^2$

$\small c=6$

Find the area of the square.

$\small A=c^2=6^2=36$

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

$\small 18\pi-36$

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Question

Refer to the above figure. Quadrilateral is a square. Give the area of Polygon in terms of .

Tap to reveal answer

Answer

$\overline{BE }$ is both one side of Square and the hypotenuse of $\Delta ABE$ ; its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

$BE = \sqrt{(AB)^{2}+ \left ( AE\right )^{2}}$

$BE = \sqrt{x^{2}+ \left ( 2x\right )^{2}} = \sqrt{x^{2}+ 4x ^{2}} = \sqrt{5x ^{2}} = x\sqrt{5}$

Polygon is the composite of $\Delta ABE$ and Square , so its area is the sum of those of the two figures.

The area of $\Delta ABE$ is half the product of its legs:

$A= \frac{1}{2} \cdot x \cdot 2x = x^{2}$

The area of Square is the square of the length of a side:

$A = \left ( x\sqrt{5} \right )^{2} = 5x^{2}$

Add the areas:

$x^{2}+ 5x^{2} = 6x^{2}$

This is the area of the polygon.

← Didn't Know|Knew It →

Question

Note: Figure NOT drawn to scale.

The above figure shows Rectangle ; is the midpoint of $\overline{CA}$ ; $EC = 2 \cdot RE$ ; $RC = 2 \cdot RN$ .

What percent of Rectangle is shaded?

Tap to reveal answer

Answer

The area of , the shaded region in question, is that of the rectangle minus those of $\Delta ERN$ and $\Delta ECT$ . We look at both.

The answer is independent of the sidelengths of the rectangle, so to ease calculations - this will become more apparent later - we will arbitrarily assign to the rectangle the dimensions

and, subsequently,

and, since is the midpoint of $\overline{CA}$ ,

.

The area of Rectangle is equal to $RN \cdot RC = 6 \cdot 12 = 72$ .

Since $EC = 2 \cdot RE$ , and ,

and

The area of $\Delta ERN$ is equal to $\frac{1}{2}(RE)(RN)= \frac{1}{2} \cdot 4 \cdot 6 = 12$ .

The area of $\Delta ECT$ is equal to $\frac{1}{2}(EC)(CT)= \frac{1}{2} \cdot 8 \cdot 3 = 12$

The area of the shaded region is therefore , which is

$\frac{48}{72} \times 100 % = 66 \frac{2}{3} %$ of the rectangle.

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Question

If square A has a side of length 5 inches, how many times bigger is the area of square B if it has a side of length 25 inches?

Tap to reveal answer

Answer

First find the area of both squares using the formula .

For square A, s = 5.

For square B, s = 25.

The question is asking for the ratio of these two areas, which will tell us how many times bigger square B is. Divide the area of square B by the area of square A to find the answer.

$\frac{A_B}{A_A}=\frac{625}{25}=25$

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Question

A square has an area of 36 cm2. A circle is inscribed and cut out. What is the area of the remaining shape? Use 3.14 to approximate π.

Tap to reveal answer

Answer

We need to find the area of both the square and the circle and then subtract the two. Inscribed means draw within a figure so as to touch in as many places as possible. So the circle is drawn inside the square. The opposite is circumscribed, meaning drawn outside.

Asquare = s2 = 36 cm2 so the side is 6 cm

6 cm is also the diameter of the circle and thus the radius is 3 cm

A circle = πr2 = 3.14 * 32 = 28.28 cm2

The resulting difference is 7.74 cm2

← Didn't Know|Knew It →

Question

In the square above, the radius of each half-circle is 6 inches. What is the area of the shaded region?

Tap to reveal answer

Answer

We can find the area of the shaded region by subtracting the area of the semicircles, which is much easier to find. Two semi-circles are equivalent to one full circle. Thus we can just use the area formula, where r = 6:

π(62) → 36π

Now we must subtract the area of the semi-circles from the total area of the square. Since we know that the radius also covers half of a side, 6(2) = 12 is the full length of a side of the square. Squaring this, 122 = 144. Subtracting the area of the circles, we get our final terms,

= 144 – 36π

← Didn't Know|Knew It →

Question

If Bailey paints the wall shaped like above and uses one bucket per 5 square units, how many buckets does Bailey need?

Tap to reveal answer

Answer

To solve, we will need to find the area of the wall. We can do this by finding the areas of each section and adding them together. Break the area into a rectange and two triangles.

The area of the rectangle will be equal to the base times the height. The area of each triangle will be one half its given base times its height.

$A_{rec}=bh$

$A_{tri}=\frac{1}{2}bh$

For the rectangle, the base is 12 and the height is 4 (both given in the figure).

$A_{rec}=(12)(4)=48$

The triangle to the right has a given base of 6, but we need to solve for its height. The height will be equal to the difference between the total height (6) and the height of the rectangle (4).

$h_{tri}=6-4=2$

We now have the base and height of the triangle to the right, allowing us to calculate its area.

$A_{tri_r}=\frac{1}{2}(6)(2)=6$

Now we need to solve the triangle to the left. We solved for its height (2), but we still need to solve for its base. The total base of the rectangle is 12. Subtract the base of the right-side triangle (6) and the small segment at the top of the rectangle (3) from this total length to solve for the base of the left triangle.

The left-side triangle has a base of 3 and a height of 2, allowing us to calculate its area.

$A_{tri_l}=\frac{1}{2}(3)(2)=3$

Add together the two triangles and the rectangle to find the total area.

We know that each bucket of paint will cover 5 square units, and we have 57 square units total. Divide to find how many buckets are required.

$\frac{57}{5}=11.4$

We will need 11 full buckets and part of a twelfth bucket to cover the wall, meaning that we will need 12 buckets total.

← Didn't Know|Knew It →

Question

A square is inscribed within a circle with a radius $\small 3\sqrt{2}$ . Find the area of the circle that is not covered by the square.

Tap to reveal answer

Answer

First, find the area of the circle.

$A=\pi(3\sqrt{2})^2$

$\small A=18\pi$

Next, find the length of 1 side of the square using the Pythagorean Theorem. Two radii from the center of the circle to adjacent corners of the square will create a right angle at the center of the circle. The radii will be the legs of the triangle and the side of the square will be the hypotenuse.

$\small (3\sqrt{2})^2 +(3\sqrt{2})^2 =c^2$

$\small 18+18=36=c^2$

$\small c=6$

Find the area of the square.

$\small A=c^2=6^2=36$

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

$\small 18\pi-36$

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Question

Refer to the above figure. Quadrilateral is a square. Give the area of Polygon in terms of .

Tap to reveal answer

Answer

$\overline{BE }$ is both one side of Square and the hypotenuse of $\Delta ABE$ ; its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

$BE = \sqrt{(AB)^{2}+ \left ( AE\right )^{2}}$

$BE = \sqrt{x^{2}+ \left ( 2x\right )^{2}} = \sqrt{x^{2}+ 4x ^{2}} = \sqrt{5x ^{2}} = x\sqrt{5}$

Polygon is the composite of $\Delta ABE$ and Square , so its area is the sum of those of the two figures.

The area of $\Delta ABE$ is half the product of its legs:

$A= \frac{1}{2} \cdot x \cdot 2x = x^{2}$

The area of Square is the square of the length of a side:

$A = \left ( x\sqrt{5} \right )^{2} = 5x^{2}$

Add the areas:

$x^{2}+ 5x^{2} = 6x^{2}$

This is the area of the polygon.

← Didn't Know|Knew It →

Question

Note: Figure NOT drawn to scale.

The above figure shows Rectangle ; is the midpoint of $\overline{CA}$ ; $EC = 2 \cdot RE$ ; $RC = 2 \cdot RN$ .

What percent of Rectangle is shaded?

Tap to reveal answer

Answer

The area of , the shaded region in question, is that of the rectangle minus those of $\Delta ERN$ and $\Delta ECT$ . We look at both.

The answer is independent of the sidelengths of the rectangle, so to ease calculations - this will become more apparent later - we will arbitrarily assign to the rectangle the dimensions

and, subsequently,

and, since is the midpoint of $\overline{CA}$ ,

.

The area of Rectangle is equal to $RN \cdot RC = 6 \cdot 12 = 72$ .

Since $EC = 2 \cdot RE$ , and ,

and

The area of $\Delta ERN$ is equal to $\frac{1}{2}(RE)(RN)= \frac{1}{2} \cdot 4 \cdot 6 = 12$ .

The area of $\Delta ECT$ is equal to $\frac{1}{2}(EC)(CT)= \frac{1}{2} \cdot 8 \cdot 3 = 12$

The area of the shaded region is therefore , which is

$\frac{48}{72} \times 100 % = 66 \frac{2}{3} %$ of the rectangle.

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