Triangles - GRE Quantitative Reasoning

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.

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.

A triangle with a fixed perimeter does not have to have a fixed area. For example, a triangle with sides 3, 4, and 5 has a perimeter of 12 and an area of 6. A triangle with sides 4, 4, and 4 also has a perimeter of 12 but not an area of 6. Thus the answer cannot be determined.

To find the perimeter, add up the sides, here 5 + 12 + 13 = 30. To find the area, multiply the two legs together and divide by 2, here (5 * 12)/2 = 30.

If B is a midpoint of AC, then we know AB is 12. Moreover, triangles ACE and ABD share angle DAB and have right angles which makes them similar triangles. Thus, their sides will all be proportional, and BD is 4. 1/2bh gives us 1/2 * 12 * 4, or 24.

Area = 1/2(base)(height). You could use Pythagorean theorem to find the height or, if you know the special right triangles, recognize the 5-12-13. The area = 1/2(12)(5) = 30.

Quantity A: area = base * height / 2 = 10 * 24 / 2 = 120

Quantity B: 2 * area = 2 * base * height / 2 = base * height = 5 * 12 = 60

Therefore Quantity A is greater.

Quantity A: Area = 1/2 * b * h = 1/2 * 6 * 7 = 42/2 = 21

Quantity B: Area = 1/2 * (_b_1 + _b_2) * h = 1/2 * (6 + 10) * 6 = 48

Half of the area = 48/2 = 24

Quantity B is greater.

This is easier to see when the triangle is divided into six parts (blue). Each one contains an angle which is half of 120 degrees and contains a 90 degree angle. This means each triangle is a 30/60/90 triangle with its long side equal to the radius of the circle. Knowing that means that the height of each triangle is and the base is .

Applying and multiplying by 6 gives ).

The question doesn't tell us if this is a right triangle, so we can't assume that it is. But there is a formula to find the area when we don't know the height: area = \[p(p – a)(p – b)(p – c)\]1/2, where a, b, and c are the side lengths and p is half of the perimeter. The perimeter is 18 + 24 + 30 = 72, so p = 72/2 = 36.

Then area = \[36(36 – 18)(36 – 24)(36 – 30)\]1/2 = \[36 * 12 * 6 * 18\]1/2 = 216.

Since we are told nothing about the angles we cannot assume that these are isosceles triangles and are open to possibilites such as that shown below in which the left side would be larger. If both were isosceles triangles then the right side would be larger.

All sides of an equilateral triangle are equal, so all sides of this triangle equal 4.

Area = 1/2 base * height, so we need to calculate the height: this is easy for an equilateral triangle, since you can bisect any such triangle into two identical 30:60:90 triangles.

The ratio of lengths of a 30:60:90 triangle is 1:√3:2. The side of the equilateral triangle is 4, and we divided the base in half when we bisected the triangle, so that give us a length of 2, so our triangle must have sides of 2, 4, and 2√3; thus we have our height.

One of our 30:60:90 triangles will have a base of 2 and a height of 2√3. Half the base is 1, so 1 * 2√3 = 2√3.

We have two of these triangles, since we divided the original triangle, so the total area is 2 * 2√3 = 4√3.

You can also solve for the area of any equilateral triangle by applying the formula (s2√3)/4, where s = the length of any side.

An equilateral triangle can be considered to be 2 identical 30-60-90 triangles, giving the triangle a height of . From there, use the formula for the area of a triangle:

To solve this equation, first note that a line drawn from the origin to a vertex of the equilateral triangle will bisect the angle of the vertex. Furthermore, the length of this line is equal to the radius:

That this creates in turn is a 30-60-90 right triangle. Recall that the ratio of the sides of a 30-60-90 triangle is given as:

Therefore, the length of the side can be found to be

This is also one half of the base of the triangle, so the base of the triangle can be found to be:

Furthermore, the length of the side is:

The vertical section rising from the origin is the length of the radius, which when combined with the shorter section above gives the height of the triangle:

The area of a triangle is given by one half the base times the height, so we can find the answer as follows: