GRE® Select One or More Answers Question

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

GRE Select One or More Answers (SOMA) | Directions ▼

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.

1. The correct answer may be just one of the choices or as many as all of the choices, depending on the question.
2. 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 4 : The area for which of the following will necessarily be more than 50 square units.
Indicate all such expressions

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

Play Video: GRE Geometry Select One or More Answer Video Solution

---

---

Explanatory Answer | GRE Geometry Practice Question 4

Option 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; Take π = \\frac{22}{7})
→ r = 22 × \\frac{7}{22}) × \\frac{1}{2}) = 3.5 units
Area of a circle = π × r² (Take π = \\frac{22}{7}) and radius = 3.5)
Area = \\frac{22}{7}) × (3.5)² = \\frac{22}{7}) × 3.5 × 3.5 = 11 × 3.5 = 38.5 square units
Choice A is NOT one of the answers.

---

Option 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° < θ < 180°
If we find one instance in which the area is less than 50 sq. units, it is enough to rule out this choice.
As 0° < θ < 180°, 0 < Sin θ ≤ 1
For instance, for some θ between 0° and 180°, Sin θ will be 0.1. (When θ = 5.73°, sin θ = 0.1)
The area for that parallelogram = 20 × 10 × 0.1 = 20 square units.
Choice B is NOT one of the answers.

---

Option C: Rhombus whose perimeter is 52 units

All sides of a rhombus are equal in length.
Perimeter of rhombus = 4a = 52 → Side of a rhombus, a = 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.
Area of the rhombus = 13² × Sin θ , where 0° < θ < 180°
As 0° < θ < 180°, 0 < Sin θ ≤ 1
For instance, for some θ between 0° and 180°, Sin θ will be 0.1.
For such a rhombus, the area = 13 × 13 × 0.1 = 16.9.
Choice C is NOT one of the answers.

---

Option D: Rectangle whose perimeter is 50 units

Perimeter of a rectangle = 2(l + b)
2(l + b) = 50
(l + b) = 25
If we find one instance in which the area is less than 50 sq. 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.

---

Option 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
Area of a square = 64 square units
The value of area exceeds 50 square units.
Choice E is one of the answers.

---

Option F: Right triangle whose hypotenuse measures 17 units

As the hypotenuse is 17, the triangle could be 15, 8, 17 right triangle. (It is a Pythagorean Triplet)
The area of 15, 8, 17 right triangle is 60 sq units.
Let’s see whether we can find a different right triangle whose hypotenuse is 17 units and area is less than 50 sq units.

Conditions to be satisfied
1. Hypotenuse 17; let the other two sides of triangle be a and b.
2. Hypotenuse 17; let the other two sides of triangle be a and b.
3. Satisfy Pythagoras Theorem: a² + b² = 172 → 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.
For 16.88, 2, 17 right triangle, a + b > 17.
So, the basic condition for 3-line segments to form sides of a triangle is satisfied.
The area of the triangle = \\frac{1}{2}) × b × h = \\frac{1}{2}) × 16.88 × 2 = 16.88 sq units.
Choice F is NOT one of the answers.