Right Triangles - GMAT Quantitative Reasoning

Let one of the angles measure ; then the other angle measures . The sum of the measures of the acute angles of a triangle is , so we can set up and solve this equation:

The acute angles measure ; since we want the smaller of the two, is the correct choice.

By definition, one of the angles of a right triangle must be 90 degrees. We are given the measure of another angle in the problem, so we now know the measure of two angles in the triangle. Because the sum of the angles of any triangle is 180 degrees, we can then solve for the measure of the only remaining unknown angle:

Triangle ABC is an isosceles right triangle. Therefore, its angles at the basis BC will always be .

This stems from the fact that the sums of the angles of a triangle are and in our case with ABC a right and isosceles triangle, , therefore for the two remaining angles are equal.

There are 90 degrees left, therefore to find the measure of each angle we do the following,

.

Here, we can tell the size of the angles by recognizing the length of the sides indicative of a right triangle with angles .

Indeed, the length of the sides are and . Any triangle with sides , where is a constant, will have angles .

In our case . Therefore, angle will be the smallest of the three possible angles, since it is between the two longest sides ( the hypotenuse and AB, which is longer than AC). Therefore the larger angle will be thus arriving at our final answer.

The measures of the acute angles of a right triangle have sum , so

Along with , a system of linear equations can be formed and solved as follows:

Let be the measure of . The sum of the measures of the acute angles of a right triangle, and , is , so

Assume Statement 1 alone. This can be rewritten:

Assume Statement 2 alone. This can be rewritten:

An acute angle must have measure less than .

Assume Statement 1 alone. We are looking for a whole number that is a multiple of both 2 and 7, and is less than 90. There are several such numbers - 14 and 28, for example. There is no way of eliminating any of them, so the question is left unanswered.

Statement 2 alone provides insufficient information for similar reasons, since there are several whole numbers less than 90 that are multiples of 3 and 4 - 12 and 24, for example - with no way of eliminating any of them.

Now assume both statements to be true. 3, 4, and 7 are relatively prime - the greatest common factor of the four is 1 - so in order to find the least common multiple of the four, we need to multiply them. This product is

,

which is also a multiple of 2. Any other multiple of all four numbers must be a multiple of 84, but any other positive multiple of 84 is greater than 90. Therefore, from the two statements together, it can be deduced that has measure .

Since are right triangles, we know that is an isosceles right triangle. So we know that the lengths of and are 2 cm, so we can get the length of by using the Pythagorean Theorem:

is the midpoint of , so the length of is .

Now we can use the Pythagorean Theorem again to solve for : .

Finally, we have all the elements needed to solve for the area of :

Becaue this is a right triangle and since the hypotenuse is 10 and one side is 6, the other side will be 8. This can be found using the Pythagorean Theorem, or multiplying a 3-4-5 triangle by 2 to get a 6-8-10 triangle. Since the area of a right triangle is half of the product of the two sides, we have

This is a right triangle with legs along the - and -axes, so the area of each can be calculated by taking one-half the product of the two legs.

The vertical leg has length ; the horizontal leg has length .

Now calculate the area:

Let be the sidelength of the square. Then its area is .

If the hypotenuse of a triangle is , its shorter leg is half that, or ; its longer leg is times the shorter leg, or . The area of the triangle is half the product of the legs, or

The ratio of the area of the square to that of the triangle is

or

or

The equation for the area of a right triangle is:

In this case, our values are:

Plugging this into the equation leaves us with:

which can be rewritten as

In order to calculate the area of a right triangle, we need to know the height and the base. We are only given the length of the base, so we first need to use the Pythagorean theorem with the given hypotenuse to find the length of the height:

Now that we have the height and the base, we can plug these values into the formula for the area of a right triangle:

To find the area, use the following formula:

The hypotenuse of the large right triangle is

The area of the large right triangle is half the product of its base and its height. The base can be any side of the triangle; the height would be the length of the altitude, which is the perpendicular segment from the opposite vertex to that base.

Therefore, the area of the triangle can be calculated as half the product of the legs:

Or half the product of the hypotenuse and the length of the dashed line.

To calculate , we can set these expressions equal to each other:

Using the formula for the area of a right triangle, we can plug in the given values and solve for the height of the triangle:

The height AE, divides the triangle ABC, in two triangles AEC and AEB with same proportions as the original triangle ABC, this property holds true for any right triangle.

In other words, .

Therefore, we can calculate, the length of AE:

.

As we have previously seen, the height of a right triangle divides a it into two similar triangles with sides of same proportion.

Therefore, we can set up the following equality: or .

By plugging in the numbers, we get, or .

Since here ABC is a isosceles right triangle, its height is half the size of the hypotenuse.

We just need to apply the Pythagorean Theorem to get the length of BC, and divide this length by two.

, so .

Therefore, and the final answer is .