Calculating the area of a polygon - GMAT Quantitative Reasoning

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Question

What is the maximum possible area of a quadrilateral with a perimeter of 48?

Answer

A quadrilateral with the maximum area, given a specific perimeter, is a square. Since $Area=side^{2}$ and a square has four equal sides, the max area is

$(\frac{48}{4})^{2} =12^{2}=144$

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Question

The perimeter of a regular hexagon is 72 centimeters. To the nearest square centimeter, what is its area?

Answer

This regular hexagon can be seen as being made up of six equilateral triangles, each formed by a side and two radii; each has sidelength $\small 72\div6=12$ centimeters. The area of one triangle is

$\small \small A = \frac{s^{2} \sqrt{3}}{4} = \frac{12^{2} \sqrt{3}}{4} \approx 62.4$

There are six such triangles, so multiply this by 6:

$\small 62.4 \cdot 6 \approx 374$

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Question

A man wants to design a room such that, looking from above, it appears as a trapezoid with a square attached (shown below). The area of the entire room is to be 100 square meters. The red line shown bisects the dotted line and has a length of 15. How many of the following answers are possible values for the length of one side of the square?

a) 5

b) 6

c) 7

d) 8

Figure is not to scale, but the trapezoidal figure will be similar in dimensions to the one shown.

Answer

Let denote the length of one side of a square. This is also the top of the trapezoid. Let denote the bottom of the trapezoid. Finally, let be the height of the trapezoid. The area of the trapezoid is then $\frac{x+y}{2}\cdot h$ while the area of the square is $x^{2}$ .

We then have the total area as 100, so: $\frac{x+y}{2}\cdot h + x^{2} = 100$

Now we know that the red line has length 15. is the region of this line that is in the trapezoid. What we notice, however, is that the remainder is precisely the length of one side of a square. So or

Rewriting the previous equation: $\frac{x+y}{2}\cdot (15-x)+ x^{2} = 100$

This is now an equation of 2 variables and we can easily cross out answers by plugging in possible values. What we find is that for $x=5\ and\ 6$ , $y=10\ and\ 8.\overline{2}$ respectively. For we get , which is too small ( must be greater than ). For we get $y=2.\overline{285714}\ \ or\ \frac{16}{7}$ .

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Question

What is the area of a regular hexagon with sidelength 10?

Answer

A regular hexagon can be seen as a composite of six equilateral triangles, each of whose sidelength is the sidelength of the hexagon:

Each of the triangles has area

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

Substitute to get $A =\frac{ 10^{2}\sqrt{3}}{4}= \frac{ 100 \sqrt{3}}{4} = 25 \sqrt{3}$

Multiply this by 6: $25 \sqrt{3} \times 6 = 150 \sqrt{3}$ , the area of the hexagon.

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Question

What is the area of a regular octagon with sidelength 10?

Answer

The area of a regular polygon is equal to one-half the product of its apothem - the perpendicular distance from the center to a side - and its perimeter.

The perimeter of the octagon is $10 \times 8 = 80$

From the diagram below, the apothem of the octagon is .

is one half of the sidelength, or 5. can be seen to be the length of a leg of a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with hypotenuse 10, or

$\frac{10 }{ \sqrt{ 2}}= \frac{10 \cdot \sqrt{ 2}}{ \sqrt{ 2} \cdot \sqrt{ 2}} = \frac{10 \cdot \sqrt{ 2}}{ 2} = 5\sqrt{ 2}$

This makes the apothem $5 + 5\sqrt{2}$ .

The area is therefore $\frac{1}{2} \cdot 80 \cdot \left ( 5 + 5\sqrt{2} \right ) = 200 + 200 \sqrt{2}$

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Question

What is the area of the figure with vertices ?

Answer

This figure can be seen as a composite of two simple shapes: the rectangle with vertices , and the triangle with vertices .

The rectangle has length and height , so its area is the product of these dimensions, or $9 \times 10 = 90$ .

The triangle has as its base the length of the horizontal segment connecting and , which is ; its height is the vertical distance from the other vertex to this segment, which is . The area of this triangle is half the product of the base and the height, which is $\frac{1}{2} \times 10 \times 2 = 10$ .

Add the areas of the rectangle and the triangle to get the total area:

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Question

Note: Figure NOT drawn to scale

What is the area of the above figure?

Assume all angles shown in the figure are right angles.

Answer

This figure can be seen as a smaller rectangle cut out of a larger one; refer to the diagram below.

We can fill in the missing sidelengths using the fact that a rectangle has congruent opposite sides. Once this is done, we can multiply length times height of both rectangles to get the area of each, and subtract areas:

$A = 25 \cdot 10 - 15 \cdot 3 = 250 - 45 = 205$ square feet

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Question

The following picture represents a garden with a wall built around it. The garden is represented by , the gray area,; and the wall is represented by the white area.

and are both squares and the area of the garden is equal to the area of the wall.

The length of is .

Find the area of the wall.

Answer

AB's length is 7 so the area of ABCD is:

$ABCD = 7\cdot 7=49$ .

The garden area (EFGH) is equal to the wall area .

So

,

therefore

$EFGH = 0.5 \cdot ABCD$

$EFGH = 0.5 \cdot 49 = 24.5$ .

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