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SAT II Math II : Area

Study concepts, example questions & explanations for SAT II Math II

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Example Question #11 : Geometry

Find the area of a triangle with a base length of x^2 and a height of $\frac{1}{2}x$ .

Possible Answers:

$\frac{1}{4}x^3$

$\frac{1}{2}x^3$

x^3

4x^2

x^2

Correct answer:

$\frac{1}{4}x^3$

Explanation:

Write the formula for the area of a triangle.

$A=\frac{BH}{2}$

Substitute the dimensions.

$A = \frac{x^2\cdot \frac{1}{2}x}{2} = \frac{1}{2}(x^2\cdot \frac{1}{2}x)$

The answer is: $\frac{1}{4}x^3$

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Example Question #12 : Geometry

Find the area of a circle with a radius of $\frac{3}{2}$ .

Possible Answers:

$9\pi$

$3\pi$

$\frac{9}{2}\pi$

$\frac{9}{4}\pi$

Correct answer:

$\frac{9}{4}\pi$

Explanation:

Write the formula for the area of a circle.

$A=\pi r^2$

Substitute the radius into the equation.

$A=\pi (\frac{3}{2})^2 = \frac{9}{4} \pi$

The area is: $\frac{9}{4}\pi$

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Example Question #11 : Area

Determine the area of a rectangle if the length is x+3 and the height is $\frac{1}{4}(x+3)$ .

Possible Answers:

$\frac{1}{4}x^2 + 6x + \frac{9}{4}$

$\frac{1}{2}x^2 +\frac{1}{2}x+ \frac{3}{2}$

$\frac{1}{4}x^2 + \frac{3}{2}x + \frac{9}{4}$

$\frac{1}{2}x^2 + \frac{3}{2}$

$\frac{1}{4}x^2 + 3x + \frac{9}{4}$

Correct answer:

$\frac{1}{4}x^2 + \frac{3}{2}x + \frac{9}{4}$

Explanation:

The area of a rectangle is: A =LH

Substitute the length and height into the formula.

$A = (x+3)\cdot \frac{1}{4} (x+3)$

We will move the constant to the front and apply the FOIL method to simplify the binomials.

$A = \frac{1}{4} (x+3)(x+3) = \frac{1}{4} (x^2+6x+9)$

Distribute the fraction through all the terms of the trinomial.

The answer is: $\frac{1}{4}x^2 + \frac{3}{2}x + \frac{9}{4}$

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Example Question #11 : Geometry

What's the area of triangle with a side of $\frac{1}{2}x+\frac{1}{4}$ and a height of 2x+2 ?

Possible Answers:

x^2+3x+1

$x+\frac{1}{2}$

$\frac{1}{2}x^2+\frac{3}{4}x+\frac{1}{4}$

$\frac{1}{2}x+\frac{1}{4}$

$x^2+\frac{3}{2}x+\frac{1}{2}$

Correct answer:

$\frac{1}{2}x^2+\frac{3}{4}x+\frac{1}{4}$

Explanation:

Write the formula for the area of a triangle.

$A= \frac{BH}{2}$

Substitute the dimensions.

$A= \frac{(\frac{1}{2}x+\frac{1}{4})(2x+2)}{2}=\frac{1}{2}(\frac{1}{2}x+\frac{1}{4})(2x+2)$

$A= (\frac{1}{2}x+\frac{1}{4})(x+1)$

Use the FOIL method to expand this.

$A = (\frac{1}{2}x)(x)+ (\frac{1}{2}x)(1)+ (\frac{1}{4})(x)+ (\frac{1}{4})(1)$

Simplify the terms.

$A = \frac{1}{2}x^2+\frac{1}{2}x+\frac{1}{4}x+\frac{1}{4}$

Combine like-terms.

The answer is: $\frac{1}{2}x^2+\frac{3}{4}x+\frac{1}{4}$

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Example Question #12 : Area

Determine the side of a square with an area of 100x .

Possible Answers:

$\sqrt{10x}$

10x^2

10000x^2

10x

$10\sqrt{x}$

Correct answer:

$10\sqrt{x}$

Explanation:

Write the formula for the area of a square.

A = s^2

Substitute the area into the equation.

100x= s^2

Square root both sides.

$\sqrt{100x}= \sqrt{s^2}$

$s = 10\sqrt{x}$

The answer is: $10\sqrt{x}$

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Example Question #11 : Area

Find the area of a circle with a radius of 3x^3 .

Possible Answers:

$9\pi x^6$

$6\pi x^6$

$9\pi^2 x^6$

$9\pi x^9$

$6\pi^2 x^6$

Correct answer:

$9\pi x^6$

Explanation:

The area of a circle is $A=\pi r^2$ .

Substitute the radius and solve for the area.

$A=\pi (3x^3)^2 = \pi (3x^3)(3x^3) = 9\pi x^6$

The answer is: $9\pi x^6$

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Example Question #11 : Geometry

Determine the area of a triangle with a base of 6, and a height of $\frac{2}{3}$ .

Possible Answers:

$\frac{2}{9}$

$\frac{1}{9}$

Correct answer:

Explanation:

Write the formula for the area of a triangle.

$A =\frac{1}{2}bh$

Substitute the base and height into the equation.

$A =\frac{1}{2}(6)(\frac{2}{3}) =2$

The answer is:

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Example Question #11 : 2 Dimensional Geometry

Find the area of a circle with a diameter of .

Possible Answers:

$\frac{\pi^2 }{4}$

$\frac{\pi x^2}{4}$

$\frac{\pi x^2 }{2}$

$\frac{\pi^2 x}{4}$

$\frac{\pi^2 x^2 }{2}$

Correct answer:

$\frac{\pi x^2}{4}$

Explanation:

Divide the diameter by two. This will be the radius.

$r=\frac{x}{2}$

Write the formula for the area of a circle.

$A=\pi r^2$

$A=\pi (\frac{x}{2})^2 = \pi (\frac{x^2}{4})$

The answer is: $\frac{\pi x^2}{4}$

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Example Question #11 : Area

Hexagon

In the provided diagram, hexagon ABCDEF is regular; and are the midpoints of their respective sides. The perimeter of the hexagon is ; what is the area of Quadrilateral MBCN ?

Possible Answers:

125

$\frac{125}{4} \sqrt{3}$

175

$125 \sqrt{2}$

$\frac{175}{4} \sqrt{3}$

Correct answer:

$\frac{125}{4} \sqrt{3}$

Explanation:

Quadrilateral MBCN is a trapezoid, so we need to find the lengths of its bases and its height.

The perimeter of the hexagon is , so each side of the hexagon measures one sixth of this, or .

Construct the diameters of the hexagon, which meet at center ; construct the apothem from to $\overline{BC}$ , with point of intersection . The diagram is below:

Hexagon

The six triangles formed by the diameters are equilateral, so AO = OD = BC = 10 , and AD = AO + OD = 10+10= 20 . Quadrilateral ABCD is a trapezoid with bases of length 10 and 20. Since $\overline{MN}$ has its endpoints at the midpoints of the legs of Trapezoid ABCD , it follows that $\overline{MN}$ is a midsegment, and has as its length $MN = \frac{1}{2} (AD + BC) = \frac{1}{2} (20+ 10 ) = 15$ .

The trapezoid has bases of length and ; we now need to find its height. This is the measure of $\overline{PQ}$ , which is half the length of apothem $\overline{OP}$ . is the height of an equilateral triangle $\bigtriangleup BCO$ and, consequently, the long leg of a right triangle $\bigtriangleup BCO$ . By the 30-60-90 Theorem,

$OP = BP \cdot \sqrt{3} = \frac{1}{2} \cdot BC \cdot \sqrt{3} = \frac{1}{2} \cdot 10 \cdot \sqrt{3}=5\sqrt{3}$ .

$PQ = \frac{1}{2} \cdot OP = \frac{1}{2} \cdot 5 \sqrt{3} = \frac{5}{2} \sqrt{3}$

The area of a trapezoid of height and base lengths and is

$A = \frac{1}{2} (B+b) h$ ;

Setting $h =PQ = \frac{5}{2} \sqrt{3} , B = MN = 15, b = BC = 10$ :

$A = \frac{1}{2} (B+b) h$

$= \frac{1}{2} \cdot (15 + 10 ) \cdot \frac{5}{2} \sqrt{3}$

$= \frac{25}{2} \cdot \frac{5}{2} \sqrt{3}$

$= \frac{125}{4} \sqrt{3}$

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Example Question #11 : Geometry

To the nearest whole, give the area of a regular pentagon with a perimeter of fifty.

Possible Answers:

172

101

385

Correct answer:

172

Explanation:

In a regular pentagon, called Pentagon ABCDE , construct the five perpendicular segments from each vertex to its opposite side, as shown below:

Pentagon 2

The segments divide the pentagon into ten congruent triangles.

In particular, examine $\bigtriangleup OXD$ . $\overline{OD}$ , a radius of the pentagon, bisects $\angle EDC$ , which, as the interior angle of a regular pentagon, has measure $108 ^{\circ }$ ; therefore, $m \angle ODX = 54 ^{\circ }$ . $\overline{OX}$ is an apothem and therefore bisects $\overline{CD}$ ; since the pentagon has perimeter 50, $\overline{CD}$ has length one fifth of this, or 10, and XD= 5 .

Using trigonometry,

$\tan \angle ODX = \frac{OX}{XD }$ ,

or, substituting,

$\frac{OX}{5} = \tan 54 ^{\circ }$

Solving for :

$5 \cdot \frac{OX}{5} =5 \cdot \tan 54 ^{\circ }$

$OX \approx 5 \cdot (1.3764 )$

$OX \approx 6.8819$

The area of this triangle is half the product of the lengths of legs $\overline{XD}$ and $\overline{OX}$ :

$A \approx \frac{1}{2} \cdot 5 \cdot 6.8819 \approx 17.2048$

Since the pentagon comprises ten triangles of this area, multiply:

$17.2048 \cdot 10 = 172 .048$

To the nearest whole, this is 172.

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SAT II Math II : Area

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #11 : Geometry

Example Question #12 : Geometry

Example Question #11 : Area

Example Question #11 : Geometry

Example Question #12 : Area

Example Question #11 : Area

Example Question #11 : Geometry

Example Question #11 : 2 Dimensional Geometry

Example Question #11 : Area

Example Question #11 : Geometry

All SAT II Math II Resources

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