Home / Digital SAT Math Practice Questions -Advanced : Area and volume

Digital SAT Math Practice Questions -Advanced : Area and volume

SAT MAth Practice questions – all topics

  • Geometry and Trigonometry Weightage: 15% Questions: 5-7
    • Area and volume
    • Lines, angles, and triangles
    • Right triangles and trigonometry
    • Circles

SAT MAth and English  – full syllabus practice tests

  Question  Hard

The rectangular garden shown has a width of 50 feet and a length of 45 feet and is surrounded by a paved path with a uniform width of \(x\) feet. If the combined area of the garden and the paved path is 2646 square feet, what is the value of \(x\) ?

▶️Answer/Explanation

Ans:2

Total Area of rectangle will be length \(\times\) breadth \(=546\)

As \(\mathrm{x}\) denotes denotes distance which cant be negative.

  Question  Hard

A sphere and a right circular cylinder both have radius \(r\). The height of the cylinder is 18. For what value of \(r\) will the volume of the sphere be twice the volume of the cylinder?
A) 6.75
B) 13.5
C) 27
D) 54

▶️Answer/Explanation

Ans: C

A sphere and a right circular cylinder both have radius \(r\). The height of the cylinder is 18. We need to find the value of \(r\) for which the volume of the sphere is twice the volume of the cylinder.

The volume \(V_{\text{sphere}}\) of a sphere is given by:
\[
V_{\text{sphere}} = \frac{4}{3}\pi r^3
\]

The volume \(V_{\text{cylinder}}\) of a right circular cylinder is given by:
\[
V_{\text{cylinder}} = \pi r^2 h
\]
where \(h = 18\). So,
\[
V_{\text{cylinder}} = \pi r^2 \cdot 18 = 18\pi r^2
\]

We want the volume of the sphere to be twice the volume of the cylinder:
\[
\frac{4}{3}\pi r^3 = 2 \cdot 18\pi r^2
\]

Simplify and solve for \(r\):
\[
\frac{4}{3}\pi r^3 = 36\pi r^2
\]

Divide both sides by \(\pi\):
\[
\frac{4}{3} r^3 = 36 r^2
\]

Divide both sides by \(r^2\) (assuming \(r \neq 0\)):
\[
\frac{4}{3} r = 36
\]

Multiply both sides by \(\frac{3}{4}\):
\[
r = 36 \cdot \frac{3}{4}
\]
\[
r = 27
\]

So, the correct answer is:
\[
\boxed{27}
\]

  Question Hard

The function \(f(x)=2 x+2(2 x+1)\) gives the perimeter, in meters, of a rectangle that has a width of \(x\) meters. Which of the following is the best interpretation of \(2 x+1\) in this context?
A) The length, in meters, of the rectangle
B) The width, in meters, of the rectangle
C) Half the perimeter, in meters, of the rectangle
D) Twice the length, in meters, of the rectangle

▶️Answer/Explanation

Ans: A

The function given is:
\[
f(x) = 2x + 2(2x + 1)
\]

This represents the perimeter of a rectangle. The perimeter \(P\) of a rectangle is given by:
\[
P = 2(\text{length} + \text{width})
\]

In the given function, \(x\) is the width of the rectangle. The expression inside the function must be equal to twice the sum of the length and width.

The function can be simplified as:
\[
f(x) = 2x + 2(2x + 1) = 2x + 4x + 2 = 6x + 2
\]

So, the expression \(2x + 1\) in this context represents the length of the rectangle. Therefore, the best interpretation of \(2x + 1\) is:

\[
\boxed{A}
\]

  Question  Hard

A cylinder and a sphere both have the same radius \(r\), where \(r>0\). The cylinder has a height of 16. The volume of the sphere is half the volume of the cylinder. What is the value of \(r\) ?

▶️Answer/Explanation

Ans:6

Let’s denote the radius of both the cylinder and the sphere as \(r\). The volume of a cylinder is given by \(\pi r^2 h\) and the volume of a sphere is \(\frac{4}{3} \pi r^3\).

Given that the volume of the sphere is half the volume of the cylinder:

\[\frac{1}{2} \pi r^2 \cdot 16 = \frac{1}{2} \cdot \frac{4}{3} \pi r^3\]

Cancelling out the common terms and simplifying:

\[8r = \frac{4}{3} r^2\]

Multiplying both sides by \(3\) to eliminate the fraction:

\[24r = 4r^2\]

Rearranging the equation:

\[4r^2 – 24r = 0\]

Factoring out \(4r\):

\[4r(r – 6) = 0\]

So, either \(4r = 0\) or \(r – 6 = 0\).

Since \(r\) must be greater than \(0\), the only valid solution is \(r = 6\).

 Question Hard

Trapezoid ABCD is similar to trapezoid PQRS. The length of each side of trapezoid PQRS is 3 times the length of its corresponding side of trapezoid ABCD. The area of trapezoid ABCD is 6 square centimeters. What is the area, in square centimeters, of trapezoid PQRS?
A. 9
B. 18
C. 54
D. 216

▶️Answer/Explanation

Ans: C

Since the trapezoids are similar, the ratio of their areas is the square of the ratio of corresponding side lengths. If each side of PQRS is 3 times the corresponding side of ABCD, then the ratio of areas is \(3^2 = 9\).

Given that the area of trapezoid ABCD is 6 square centimeters, the area of trapezoid PQRS is \(6 \times 9 = 54\) square centimeters. So, the answer is C. \(54\).

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