SAT PRACTICE QUESTIONS ON SHAPE

Problem 1 :

What is the radius of a circle that has a circumference of π?

A) 1/4      B) 1/2     C) 1     D) 2     E) 4

Solution:

Given that the circumference of circle is π.

We know that the circumference of circle is 2πr, where r is radius of circle.

 Circumference = π = 2πr

r = 1/2

So, option (B) is correct.

Problem 2 :

SAT-ques-geometry-q24.png

In the figure above, arc SBT is one quarter of a circle with center R and radius 6. If the length plus the width of rectangle ABCR is 8, then the perimeter of the shaded region is

A) 8 + 3π        B) 10 + 3π      C) 14 + 3π     D) 1 + 6π

E) 12 + 6π

Solution:

Given, radius = 6

Perimeter of quarter circle=2𝜋r4=2𝜋(6)4=3𝜋

AS = 6 - a , CT = 6 - b , AC = RB = 6 ( radius )

Perimeter of the shaded region = AS + AC + CT + C(SBT)

Perimeter = ( 6 - a ) + 6 + ( 6 - b ) + 3π

Perimeter = 18 - ( a + b ) + 3π

length + width = 8

Perimeter = 18 - 8 + 3π

 Perimeter = 10 + 3π

So, option (B) is correct.

Problem 3 :

In the xy-coordinate plane, what is the area of the square with opposite vertices at (-2, -2) and (2, 2) ?

A) 4    B) 8    C) 16     D) 32   E) 64

Solution:

By finding the distance between the above two points, we get the side length of the square.

So, option (D) is correct.

Problem 4 :

In rectangle ABCD, point E is the midpoint of BC. If the area of quadrilateral ABED is 2/3, what is the area of rectangle ABCD?

A)  1/2      B)  3/4      C)  8/9     D)  1       E)  8/3

Solution :

SAT-practice-question4.png

Area of rectangle = AB x AD

Area of quadrilateral = 1/2 x height x sum of parallel sides

1/2 x AB x (AD + BE) = 2/3

1/2 x AB x (AD + 1/2 x AD) = 2/3

1/2 x AB x (3/2 x AD) = 2/3

AB x AD = (2/3) x (4/3)

AB x AD = (8/9)

So, option C is correct.

Problem 5 :

SAT-ques-geometry-q27.png

In right circular cylinder above has diameter d and height h. Of the following expressions, which represents the volume of the smallest rectangular box that completely contains the cylinder?

A) dh      B) d2h       C) dh2      D) d2h2       E) (d + h)2

Solution:

Diameter of the cylinder will be the width of the rectangular box, length of the rectangular box is h.

Volume of rectangular box = length x width x height

= h x d x d

= h d2

So, option B is correct.

Problem 6 :

If the volume of a cube is 8, what is the shortest distance from the center of the cube to the base of the cube?

A) 1     B) 2    C) 4    D) √2    E) 2√2   

Solution:

Given, volume of a cube = 8

Volume of cube = a3

a3 = 8

a = 2

Shortest distance from the center of the cube to the base is half of its side length a

= a/2

= 2/2

= 1 unit

So, option (A) is correct.

Problem 7 :

SAT-ques-geometry-q29.png

In the figure above, point A is the center of the circle and segments BD and CE are diameters. Which of the following statements is true?

A) CA > 6    B) ED > 4    C) BA  < 4   D) CA = 4   E) ED = 4

Solution:

In triangles AED and BCA.

AD = AE = AC = AB (radii)

∠EAD = ∠BAC (A)

AE = AC (S)

AD = AB (S)

Using SAS, the triangles are congruent.

So, ED = 4, option E.

Problem 8 :

SAT-ques-geometry-q30.png

In the figure above, the two circles are tangent at point B and AC = 6. If the circumference of the circle with center A is twice the circumference of the circle with center C, what is the length of BC?

A) 1    B) 2    C) 3     D) 4    E) 6

Solution:

Let BC (r) be the radius of smaller circle and AB (R) is the radius of the larger circle.

R + r = 6

Circumference of the larger circle = 2(circumference of the smaller circle)

2πR = 2(2πr)

R = 2r

R = 2(6-R)

R = 12 - 2R

3R = 12

R = 4

BC = r 

AB + BC = 6

R + r = 6

BC = 6 - 4

BC = 2

Problem 9 :

SAT-ques-geometry-q31.png

The figure above shows part of a circle whose circumference is 45. If arcs of length 2 and length b continue to alternate around the entire circle so that there are 18 arcs of each length, what is the degree measure of each of the arcs of length b ?

A) 4°    B) 6°    C) 10°    D) 16°    E) 20°

Solution:

You know that 18(2 + b) = 45, so you can calculate b:

18(2 + b) = 45

36 + 18b = 45

18b = 9

b = 0.5

0.5/45 = x/360

x = 4

Problem 10 :

SAT-ques-geometry-q32.png

In the figure above, the circle with center O is inscribed in square ABCD. What is the area of the shaded portion of the circle?

A) 𝜋4B) 𝜋2C) 𝜋D) 3𝜋2E) 2𝜋

Solution:

Area of sector = θ/360  πr2

Side length of the square = diameter of the circle

radius = 1

= (90/360) π1(1)2

= π/4

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