# GRE® MTOA Practice - Geometry

Concept:Area of circles, squares, rectangles, parallelogram, right triangle, rhombus.

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.

1. Circle whose circumference is 22 units.
2. Parallelogram whose adjacent sides measure 20 units and 10 units.
3. Rhombus whose perimeter is 52 units.
4. Rectangle whose perimeter is 50 units.
5. Square whose perimeter is 32 units.
6. Right triangle whose hypotenuse measures 17 units.

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 0o ≤ θ ≤ 180o
As 0o ≤ θ ≤ 180o, 0 ≤ Sin θ ≤ 1
For instance, for some θ between 0o and 180o, 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 = a2 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
0o ≤ θ ≤ 180o
Therefore, 0 ≤ Sin θ ≤ 1
For instance, for some θ between 0o and 180o, 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 = a2 = 82 = 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 a2 + b2 = 289
The greater the difference between 'a' and 'b', smaller the area.
Let a2 = 285 and b2 = 4.
So, a = 16.88 and b = 2. (Do not compute the value of 'a' accurately. We know a2 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.