Question

What is the surface area of the cone? (1) \(\mathrm{L}=7 \mathrm{~cm}\) and the base is \(6 \mathrm{~cm}\) wide (2) The cone has a height of \(4 \mathrm{~cm}\) and the base a radius of \(3 \mathrm{~cm}\) A. 1 alone, not 2 alone B. 2 alone, not 1 alone C. 1 and 2 together (need both) D. 1 alone or 2 alone E. 1 and 2 together are not sufficient

Short answer

D. 1 alone or 2 alone

Step-by-step solution

Understand the Surface Area of a Cone Formula

The surface area of a cone (not including the base) is given by the formula \( \text{SA} = \text{Base Area} + \text{Lateral Area} \). The base area of a cone is \( \text{Base Area} = \pi r^2 \) and the lateral area is \( \text{Lateral Area} = \pi r L \), where \( L \) is the slant height and \( r \) is the radius of the base.

Analyze Statement (1)

Statement 1 provides the slant height \( L = 7 \text{cm} \) and the base width which is the diameter so \( r = \frac {6}{2} = 3 \text{cm} \). We can calculate the surface area using these values. \[ \text{SA} = \pi r^2 + \pi r L \] \[ \text{SA} = \pi (3^2) + \pi (3)(7) \] \[ \text{SA} = 9\pi + 21\pi = 30\pi \text{cm}^2 = 30(3.14) \text{cm}^2 = 94.2 \text{cm}^2 \]

Analyze Statement (2)

Statement 2 provides the height \( h = 4 \text{cm} \) and the radius \( r = 3 \text{cm} \). First, we need to calculate the slant height \( L \) using the Pythagorean theorem: \[ L = \sqrt{h^2 + r^2} \] \[ L = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{cm} \]. Now, we can calculate the surface area \[ \text{SA} = \pi r^2 + \pi r L \] \[ \text{SA} = \pi (3^2) + \pi (3)(5) \] \[ \text{SA} = 9\times\times 14.13 cm2) = 28.26 cm \]

Determine Sufficiency

We have calculated the surface area using both statements independently. Both statements allow us to compute the surface area of the cone separately and arrive at the correct value. Thus, each statement alone is sufficient to determine the surface area.

Key concepts

Geometry Problems

Geometry problems, such as finding the surface area of a cone, require understanding of specific formulas and shapes. In this problem, we apply the formulas for the surface area of a cone, which include the base area and the lateral (or slant) area. Specifically, the base area formula is \(\text{Base Area} = \pi r^2\), where \(r\) is the radius. The lateral area formula is \(\text{Lateral Area} = \pi r L\), where \(L\) is the slant height. Learning these formulas is vital for solving similar geometry problems on the GMAT.

GMAT Preparation

Preparing for the GMAT involves practicing a range of problems, from basic arithmetic to complex geometry. This surface area of a cone problem is a standard type of geometry question you might encounter. Practicing these problems helps improve your mathematical reasoning and problem-solving speed. Focus on understanding the fundamental concepts and formulas to effectively handle similar questions during the test. Regular practice of these types of problems will increase familiarity and confidence.

Mathematical Reasoning

Mathematical reasoning is essential for tackling geometry problems on the GMAT. It involves logical thinking and applying relevant formulas accurately. Let's break down the exercise in this problem:

• Identify known values (e.g., slant height, radius)
• Choose applicable formulas (like those for surface area)
• Solve step by step using algebraic manipulation and substitution

Developing strong mathematical reasoning skills ensures you can follow through complex multi-step problems. Practice reasoning through each step clearly and methodically.

Pythagorean Theorem

The Pythagorean theorem is a critical concept in geometry used to find the missing side of a right triangle. It states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides: \[c^2 = a^2 + b^2\].

In our cone problem, we use the Pythagorean theorem to determine the slant height \(L\) from the height \(h\) and the radius \(r\): \[L = \sqrt{h^2 + r^2}\].
Without knowing how to apply this theorem, the solution would be incomplete. Grasping the theorem's application in different shapes will prove advantageous in GMAT geometry.