Triangles - GRE Quantitative Reasoning

Since this is a 30-60-90 triangle, we know that the length of the side opposite the 60 degree angle is $\sqrt{3}$ times the side opposite the 30 degree angle. Thus, 5$\sqrt{3}$, which is about 8.66. This is larger than 7.5.

The ratio of the sides of a 30-60-90 triangle is x:x{$\sqrt{3}$}:2x, with the hypotenuse being 2x. Thus, the perimeter of this triangle would be x+x$\sqrt{3}$+2x=3x$\sqrt{3}$. Since the triangle depicted in this problem has a perimeter of 3$\sqrt{3}$, x must equal 1, which would make the hypotenuse equal to 2.

Use the Pythagorean theorem, $a^2$ + $b^2$ = $c^2$ , with a=x-4 and c=x+4, and solve for b.

$(x-4)^2$ + $b^2$ = $(x+4)^2$

Rearrange to isolate $b^{2}$:

$b^2$ = $(x+4)^2$ - $(x-4)^2$

$b^2$ = (x+4)(x+4) - (x-4)(x-4)

Use FOIL to multiply out:

$b^2$ = $(x^2$ + 8x + 16) - $(x^2$ - 8x + 16)

Distribute the minus sign to rewrite without parentheses:

$b^2$ = $x^2$ + 8x + 16 - $x^2$ + 8x - 16

Combine like terms:

$b^2$ = 16x

Take the square root of both sides:

b = 4$\sqrt{x}$

At first sight, it's tempting to assume this is a right triangle and to thus use the Pythagorean Theorem to find a length of 5 for the missing side.

However, the triangle was not stated to be a right triangle in the problem statement, and no indication was given in the drawing to indicate that it was a right triangle either, such as a square demarcation in the vertex opposite the side measuring 13.

Thus there is not enough information to give the length of the missing side. When taking standardized math tests, be careful making assumptions about information that is not given.

Draw the triangles

Triangle 1 is a 6,8,10 right triangle with 10 as the hypotenuse

Triangle 2: 3 is half of 6, 5 is half of 10; x must be half of 8

Since this is a 30-60-90 triangle, we know that the length of the side opposite the 60 degree angle is $\sqrt{3}$ times the side opposite the 30 degree angle. Thus, 5$\sqrt{3}$, which is about 8.66. This is larger than 7.5.

The ratio of the sides of a 30-60-90 triangle is x:x{$\sqrt{3}$}:2x, with the hypotenuse being 2x. Thus, the perimeter of this triangle would be x+x$\sqrt{3}$+2x=3x$\sqrt{3}$. Since the triangle depicted in this problem has a perimeter of 3$\sqrt{3}$, x must equal 1, which would make the hypotenuse equal to 2.

Use the Pythagorean theorem, $a^2$ + $b^2$ = $c^2$ , with a=x-4 and c=x+4, and solve for b.

$(x-4)^2$ + $b^2$ = $(x+4)^2$

Rearrange to isolate $b^{2}$:

$b^2$ = $(x+4)^2$ - $(x-4)^2$

$b^2$ = (x+4)(x+4) - (x-4)(x-4)

Use FOIL to multiply out:

$b^2$ = $(x^2$ + 8x + 16) - $(x^2$ - 8x + 16)

Distribute the minus sign to rewrite without parentheses:

$b^2$ = $x^2$ + 8x + 16 - $x^2$ + 8x - 16

Combine like terms:

$b^2$ = 16x

Take the square root of both sides:

b = 4$\sqrt{x}$

At first sight, it's tempting to assume this is a right triangle and to thus use the Pythagorean Theorem to find a length of 5 for the missing side.

However, the triangle was not stated to be a right triangle in the problem statement, and no indication was given in the drawing to indicate that it was a right triangle either, such as a square demarcation in the vertex opposite the side measuring 13.

Thus there is not enough information to give the length of the missing side. When taking standardized math tests, be careful making assumptions about information that is not given.

If an equilateral triangle is divided in 2, it forms two 30-60-90 triangles. Therefore, the side of the equilateral triangle is the same as the hypotenuse of a 30-60-90 triangle. The side lengths of a 30-60-90 triangle adhere to the ratio x: x√3 :2x. since we know the hypothesis is 4, we also know that the base is 2 and the height is 2√3.

To find the top inner angle, recognize that a straight line contains 180o; thus we can subtract 180 – 115 = 65o. Since we are given the other interior angle of 30 degrees, we can add the two we know: 65 + 30 = 95o.

180 - 95 = 85

We know the angles in a triangle must add up to 180, so we can solve for x.

3_x_ + 4_x_ + 10 + 8_x_ + 20 = 180

15_x_ + 30 = 180

15_x_ = 150

x = 10

The given triangle is obtuse. Thus, angle is greater than 90 degrees. A triangle has a sum of 180 degrees, so angle + angle + angle = 180. Since angle C is greater than 90 then angle + angle must be less than 90 and it follows that Quantity A is greater.

An isosceles triangle always has two equal angles. As there cannot be another 110° (the triangle cannot have over 180° total), the other two angles must equal eachother. 180° - 110° = 70°. 70° represents the other two angles, so it needs to be divided in 2 to get the answer of 35°.

Since the triangle is isosceles, and <A is located at the top of the triangle that is on its base, <B and <C are equivalent. Since <B is 84 degrees, <C is also. There are 180 degrees in a triangle so 180 - 84 - 84 = 12 degrees.

Since we are given expressions for the two congruent angles of the isosceles triangle, we can set the expressions equal to see how x relates to y. We get,

x – 3 = y – 7 --> y = x + 4

Logically, y must be the greater number if it takes x an additional 4 units to reach its value (knowing they are both positive integers).

We know that an isosceles triangel has two equal sides and thus two equal angles opposite those equal sides. Because there is one obtuse angle of 112 degrees we automatically know that this angle is the vertex. If you sum any triangle's interior angles, you always get 180 degrees.

180 – 112 = 68 degrees. Thus there are 68 degrees left for the two equal angles. Each angle must therefore measure 34 degrees.

The area of an equilateral triangle is .

So let's set-up an equation to solve for .

Cross multiply.

The cancels out and we get .

Then take square root on both sides and we get as the final answer.

By having a height in an equilateral triangle, the angle is bisected therefore creating two triangles.

The height is opposite the angle . We can set-up a proportion.

Side opposite is and the side of equilateral triangle which is opposite is .

Cross multiply.

Divide both sides by

Multiply top and bottom by to get rid of the radical.