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PSAT Math : How to find the area of a polygon

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

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.

Possible Answers:

28.26 cm²

7.74 cm²

3.69 cm²

15.48 cm²

12.14 cm²

Correct answer:

7.74 cm²

Explanation:

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.

A_square = 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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Example Question #11 : Other Polygons

Gre10

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

Possible Answers:

144 – 6π

144 – 36π

36 – 6π

144 – 9π

36 – 9π

Correct answer:

144 – 36π

Explanation:

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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Example Question #1 : Other Polygons

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?

Possible Answers:

5 times

25 times

625 times

2 times

4 times

Correct answer:

25 times

Explanation:

First find the area of both squares using the formula A=s^2 .

For square A, s = 5.

A_A=5^2=25

For square B, s = 25.

A_B=25^2=625

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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Example Question #2 : Other Polygons

Sat_math_picture3

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

Possible Answers:

Correct answer:

Explanation:

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.

Sat_math_picture3

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.

12-6-3=3

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.

A=48+6+3=57

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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Example Question #2 : Other Polygons

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.

Possible Answers:

$\small 18\pi-36$

$\small 9\pi-6$

$9\pi-3\sqrt{2}$

$\small 18\pi-6$

$\small 9\pi-36$

Correct answer:

$\small 18\pi-36$

Explanation:

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

Thingy

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

Possible Answers:

$7x^{2}$

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

$5x^{2}$

$\left ( 1+ \sqrt{5}\right )x^{2}$

$6x^{2}$

Correct answer:

$6x^{2}$

Explanation:

$\overline{BE }$ is both one side of Square BCDE 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 ABCDE is the composite of $\Delta ABE$ and Square BCDE , 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 BCDE 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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Example Question #7 : Other Polygons

Stuff_1

Note: Figure NOT drawn to scale.

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

What percent of Rectangle RCAN is shaded?

Possible Answers:

$72 \%$

$80 \%$

$60 \%$

$66 \frac{2}{3} \%$

$75 \%$

Correct answer:

$66 \frac{2}{3} \%$

Explanation:

The area of ETAN , 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

RN = CA = 6

and, subsequently,

RC = NA = 12

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

CT = 3 .

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

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

RE + 2 (RE)= 12

3 (RE)= 12

RE = 4 and EC = 8

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 72-12-12 = 48 , which is

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

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PSAT Math : How to find the area of a polygon

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

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

Example Question #11 : Other Polygons

Example Question #1 : Other Polygons

Example Question #2 : Other Polygons

Example Question #2 : Other Polygons

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

Example Question #7 : Other Polygons

All PSAT Math Resources

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