Calculating the area of an acute / obtuse triangle - GMAT Quantitative

In this problem, the base is 12 and the height is 6. Therefore:

The figure is a composite of a rectangle and a triangle, as shown:

The rectangle has area

The triangle has area

The total area of the figure is

This can be illustrated by showing this triangle inscribed inside a rectangle whose vertices are :

The area of the white triangle is the one whose area we calculate. To do this, we need the area of the square:

The area of the red triangle:

The area of the green triangle:

And the area of the beige triangle:

The area of the white triangle will be as follows:

The only restriction on the measure of the vertex angle of an isosceles triangle is the restriction on any angle of a triangle - that it fall between and , noninclusive. If is any number in that range, each base angle, the two being congruent, will measure , which will fall in the acceptable range.

Since all of these measures fall in that range, the correct response is that all are allowed.

The easiest way to solve this is to graph the three lines and to observe the dimensions of the resulting triangle. It helps to know the coordinates of the three points of intersection, which we can do as follows:

The intersection of and the -axis - that is, the line can be found with some substitution:

The lines intersect at

The intersection of and the -axis can be found the same way:

These lines intersect at

The intersection of and can be found via the substitution method:

The lines intersect at

The triangle therefore has these three vertices. It is shown below.

As can be seen, it is a triangle with base 9 and height 12, so its area is

The vertical segment connecting and can be seen as the base of this triangle; this base has length . The height is the perpendicular (horizontal) distance from to this segment, which is 6, the same as the -coordinate of this point. The area of the triangle is therefore

.

We can view the horizontal segment connecting , and as the base; its length wiill be . The height will be the perpendicular (vertical) distance to this segment from the opposite point , which is , the -coordinate; therefore, the area of the triangle will be half the product of these two numbers, or

.

Use Heron's formula:

where , and

In this problem, the base is 12 and the height is 6. Therefore:

The figure is a composite of a rectangle and a triangle, as shown:

The rectangle has area

The triangle has area

The total area of the figure is

This can be illustrated by showing this triangle inscribed inside a rectangle whose vertices are :

The area of the white triangle is the one whose area we calculate. To do this, we need the area of the square:

The area of the red triangle:

The area of the green triangle:

And the area of the beige triangle:

The area of the white triangle will be as follows:

The only restriction on the measure of the vertex angle of an isosceles triangle is the restriction on any angle of a triangle - that it fall between and , noninclusive. If is any number in that range, each base angle, the two being congruent, will measure , which will fall in the acceptable range.

Since all of these measures fall in that range, the correct response is that all are allowed.

The easiest way to solve this is to graph the three lines and to observe the dimensions of the resulting triangle. It helps to know the coordinates of the three points of intersection, which we can do as follows:

The intersection of and the -axis - that is, the line can be found with some substitution:

The lines intersect at

The intersection of and the -axis can be found the same way:

These lines intersect at

The intersection of and can be found via the substitution method:

The lines intersect at

The triangle therefore has these three vertices. It is shown below.

As can be seen, it is a triangle with base 9 and height 12, so its area is

The vertical segment connecting and can be seen as the base of this triangle; this base has length . The height is the perpendicular (horizontal) distance from to this segment, which is 6, the same as the -coordinate of this point. The area of the triangle is therefore

.

We can view the horizontal segment connecting , and as the base; its length wiill be . The height will be the perpendicular (vertical) distance to this segment from the opposite point , which is , the -coordinate; therefore, the area of the triangle will be half the product of these two numbers, or

.

Use Heron's formula:

where , and

In this problem, the base is 12 and the height is 6. Therefore:

The figure is a composite of a rectangle and a triangle, as shown:

The rectangle has area

The triangle has area

The total area of the figure is

This can be illustrated by showing this triangle inscribed inside a rectangle whose vertices are :

The area of the white triangle is the one whose area we calculate. To do this, we need the area of the square:

The area of the red triangle:

The area of the green triangle:

And the area of the beige triangle:

The area of the white triangle will be as follows: