Calculating the area of a polygon - GMAT Quantitative

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 centimeters. The area of one triangle is

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

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 while the area of the square is .

We then have the total area as 100, so:

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:

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 , respectively. For we get , which is too small ( must be greater than ). For we get .

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

Substitute to get

Multiply this by 6: , the area of the hexagon.

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

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 triangle with hypotenuse 10, or

This makes the apothem .

The area is therefore

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 .

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 .

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

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:

square feet

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

.

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

So

,

therefore

.

A quadrilateral with the maximum area, given a specific perimeter, is a square. Since and a square has four equal sides, the max area is

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 centimeters. The area of one triangle is

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

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 while the area of the square is .

We then have the total area as 100, so:

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:

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 , respectively. For we get , which is too small ( must be greater than ). For we get .

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

Substitute to get

Multiply this by 6: , the area of the hexagon.

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

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 triangle with hypotenuse 10, or

This makes the apothem .

The area is therefore

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 .

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 .

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

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:

square feet

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

.

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

So

,

therefore

.

A quadrilateral with the maximum area, given a specific perimeter, is a square. Since and a square has four equal sides, the max area is

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 centimeters. The area of one triangle is

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

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 while the area of the square is .

We then have the total area as 100, so:

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:

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 , respectively. For we get , which is too small ( must be greater than ). For we get .

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

Substitute to get

Multiply this by 6: , the area of the hexagon.