This GRE quant practice question is a More Than One Answer correct question in geometry. Concept tested: Area of circle, right triangle, square, rectangle, parallelogram and rhombus.

#### Directions: Select one or more answer choices according to the specific question directions.

If the question does not specify how many answer choices to select, select all that apply.

- The correct answer may be just one of the choices or as many as all of the choices, depending on the question.
- No credit is given unless you select all of the correct choices and no others.

If the question specifies how many answer choices to select, select exactly that number of choices.

#### Question: The area for which of the following will necessarily be more than 50 square units

Indicate __all__ such events.

- Circle whose circumference is 22 units.
- Parallelogram whose adjacent sides measure 20 units and 10 units.
- Rhombus whose perimeter is 52 units.
- Rectangle whose perimeter is 50 units.
- Square whose perimeter is 32 units.
- Right triangle whose hypotenuse measures 17 units.

#### Explanatory Answer

Video Explanation will be added soon#### Choice A

##### Circle whose circumference is 22 units.

Circumference of a circle = 2 π r, where r is the radius of the circle.

2 π r = 22

Expressing π as \\frac{22}{7}) and solving for r we get r = 22 * \\frac{7}{22})*\\frac{1}{2}) = 3.5 units.

Area of a circle whose radius is 3.5 units = π (3.5)^{2}.

Expressing π as \\frac{22}{7}) we get area = \\frac{22}{7}) x (3.5)^{2}

i.e., the area of the circle = 11 * 3.5 = 38.5 square units.

Choice A is __NOT__ one of the answers.

#### Choice B

##### Parallelogram whose adjacent sides measure 20 units and 10 units.

Area of the parallelogram = a * b * Sin θ, where 'a' and 'b' are the measures of the adjacent sides of the parallelogram and θ is the angle between these two adjacent sides.

Area of this parallelogram = 20 * 10 * Sin θ where 0^{o} ≤ θ ≤ 180^{o}

As 0^{o} ≤ θ ≤ 180^{o}, 0 ≤ Sin θ ≤ 1

For instance, for some θ between 0^{o} and 180^{o}, Sin θ will be 0.1.

The area for that parallelogram = 20 * 10 * 0.1 = 20 square units.

Therefore, if the adjacent sides of parallelogram measure 20 and 10, its area __need not__ be more than 50 square units.

Choice B is __NOT__ one of the answers.

#### Choice C

##### Rhombus whose perimeter is 52 units.

Perimeter of a rhombus = 52 units.

All sides of a rhombus are equal in length.

So, each side of the rhombus measures 13 units.

Area of a rhombus = a^{2} Sin θ, where 'a' is the side of the rhombus and θ is the angle between any two adjacent sides of the rhombus.

So, area of the rhombus = 13² * Sin θ

0^{o} ≤ θ ≤ 180^{o}

Therefore, 0 ≤ Sin θ ≤ 1

For instance, for some θ between 0^{o} and 180^{o}, Sin θ will be 0.1.

For such a rhombus, the area = 13 * 13 * 0.1 = 16.9.

Therefore, if the perimeter of the rhombus is 52, its area need not be more than 50 square units.

Choice C is __NOT__ one of the answers.

#### Choice D

##### Rectangle whose perimeter is 50 units

Perimeter of a rectangle = 2(l + b).

2(l + b) = 50 or (l + b) = 25.

If we can find one instance in which the area is less than 50 square units, it is enough to rule out this choice.

Let l = 24 and b = 1.

Area of this rectangle = l * b = 24 * 1 = 24 square units.

Choice D is __NOT__ one of the answers.

#### Choice E

##### Square whose perimeter is 32 units

Perimeter of a square = 4a, where a is the side of the square.

4a = 32 or a = 8 units.

Area of a square = a^{2} = 8^{2} = 64 square units.

Choice E is one of the choices where the area exceeds 50 square units.

#### Choice F

##### Right triangle whose hypotenuse measures 17 units

Watch out for the temptation to jump to the conclusion that the triangle is a 15, 8, 17 right triangle - it need not be.

Let us see whether a right triangle with a hypotenuse of 17 units could have an area less than 50 square units.

Let us say the other two sides measure 'a' and 'b' units.

a + b > 17 and a^{2} + b^{2} = 289

The greater the difference between 'a' and 'b', smaller the area.

Let a^{2} = 285 and b^{2} = 4.

So, a = 16.88 and b = 2. (Do not compute the value of 'a' accurately. We know a^{2} lies between 256 and 289. So, a should lie between 16 and 17).

Even if one rounded off 'a' to be 16, we get a + b > 17. The basic condition for 3 line segments to form sides of a triangle is satisfied.

The area of the triangle = \\frac{1}{2} * 16.88 * 2 = 16.88 sq units.

Choice F is __NOT__ one of the answers.