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SAT II Math I : Area

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

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Example Questions

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Example Question #1 : How To Find The Area Of A Kite

What is the area of the following kite?

Kites

Possible Answers:

30m^2

20m^2

10m^2

8m^2

15m^2

Correct answer:

10m^2

Explanation:

The formula for the area of a kite:

$A = \frac{1}{2} (a)(b)$ ,

where represents the length of one diagonal and represents the length of the other diagonal.

Plugging in our values, we get:

$A = \frac{1}{2} (4\: m)(5\: m)$

$A = \frac{1}{2} (20\: m^2) = 10\: m^2$

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Example Question #2 : How To Find The Area Of A Kite

Find the area of a kite if the diagonal dimensions are x+3 and x-6 .

Possible Answers:

x^2+3x-18

$(x^2-3x-18)\sqrt2$

$\frac{x^2-3x+18}{2}$

$\frac{x^2-3x-18}{2}$

$\frac{x^2+3x-18}{2}$

Correct answer:

$\frac{x^2-3x-18}{2}$

Explanation:

The area of the kite is given below. The FOIL method will need to be used to simplify the binomial.

$A=\frac{d1\cdot d2}{2}=\frac{(x+3)(x-6)}{2}= \frac{x^2-6x+3x-18}{2}$

$A= \frac{x^2-3x-18}{2}$

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Example Question #1 : How To Find The Area Of A Kite

The diagonals of a kite are $\sqrt10\:cm$ and $\sqrt5\:cm$ . Find the area.

Possible Answers:

$\frac{5\sqrt2}{2}\:cm^2$

$5\:cm^2$

$5\sqrt2\:cm^2$

$\sqrt5\:cm^2$

$\frac{2\sqrt5}{2}\:cm^2$

Correct answer:

$\frac{5\sqrt2}{2}\:cm^2$

Explanation:

The formula for the area for a kite is

$A=\frac{d_1\cdot d_2}{2}$ , where d_1 and d_2 are the lengths of the kite's two diagonals. We are given the length of these diagonals in the problem, so we can substitute them into the formula and solve for the area:

$A = \frac{d1\cdot d2}{2} = \frac{\sqrt10 \cdot \sqrt5}{2} = \frac{\sqrt50}{2} = \frac{5\sqrt2}{2}\:cm^2$

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Example Question #11 : Geometry

Find the area of a kite with diagonal lengths of a+b and 2a-2b .

Possible Answers:

a^2+b^2

a^2-ab-b^2

2a^2+2b^2

a^2-b^2

2a^2-2b^2

Correct answer:

a^2-b^2

Explanation:

Write the formula for the area of a kite.

$A= \frac{d1\cdot d2}{2}$

Plug in the given diagonals.

$A= \frac{(a+b)\cdot (2a-2b)}{2}$

Pull out a common factor of two in 2a-2b and simplify.

$A= \frac{(a+b)\cdot2(a-b)}{2}$

A=(a+b)(a-b)

Use the FOIL method to simplify.

A=(a+b)(a-b)= (a)(a)+(a)(-b)+(b)(a)+(b)(-b)

A= a^2-b^2

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Example Question #15 : Rhombuses

Find the area of a rhombus if the diagonals lengths are $20\:cm$ and $40\:cm$ .

Possible Answers:

$800\:cm^2$

$400\:cm^2$

$900\:cm^2$

$600\:cm^2$

$200\:cm^2$

Correct answer:

$400\:cm^2$

Explanation:

Write the formula for the area of a rhombus:

$A=\frac{d_1 \cdot d_2}{2}$

Substitute the given lengths of the diagonals and solve:

$A=\frac{d1 \cdot d2}{2} = \frac{20\:cm \cdot 40\:cm}{2} =\frac{800\:cm^2}{2} = 400\:cm^2$

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Example Question #1 : How To Find The Area Of A Rhombus

Find the area of a rhombus if the diagonals lengths are and 5a^2 .

Possible Answers:

10a^3

2a+5a^2

7a^3

5a^2

5a^3

Correct answer:

5a^3

Explanation:

Write the formula for finding the area of a rhombus. Substitute the diagonals and evaluate.

$A=\frac{d1\cdot d2}{2}= \frac{2a \cdot 5a^2}{2}= 5a^3$

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Example Question #11 : Geometry

Give the area of $\bigtriangleup ABC$ to the nearest whole square unit, where:

AB = 19

AC = 25

$m \angle A = 57^{\circ }$

Possible Answers:

129

199

259

398

238

Correct answer:

199

Explanation:

The area of a triangle with two sides of lengths and and included angle of measure $\gamma$ can be calculated using the formula

$A = \frac{1}{2} ab \sin \gamma$ .

Setting a = AB = 19 , b = AC = 25 , and $\gamma = m \angle A = 57^{\circ }$ , then evaluating :

$A = \frac{1}{2} (19 )(25)\sin 57^{\circ }$

$A = \frac{1}{2} (19 )(25) (0.8387)$

$A \approx 199$ .

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Example Question #11 : Area

Find the area of a triangle with a height of $\frac{1}{2}$ and a base of $\frac{1}{4}$ .

Possible Answers:

$\frac{3}{4}$

$\frac{3}{8}$

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{16}$

Explanation:

Write the formula for the area of a triangle.

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

Substitute the base and height into the formula.

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

Simplify the fractions.

The answer is: $\frac{1}{16}$

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Example Question #12 : Geometry

Find the area of a circle if the circumference is 3.

Possible Answers:

$\frac{3}{2\pi}$

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

$\frac{3}{2\pi^2}$

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

$\frac{9}{4\pi^2}$

Correct answer:

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

Explanation:

Write the formula for the circumference of a circle.

$C=2\pi r$

Substitute the circumference.

$3=2\pi r$

Divide by $2\pi$ on both sides.

$\frac{3}{2\pi}= r$

We will need the formula for the area of the circle.

$A=\pi r^2$

Substitute the radius to find the area.

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

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

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Example Question #13 : Area

What is the area of a square if the length of the side is $\pi\sqrt2$ ?

Possible Answers:

$2\pi$

$2\pi ^2$

$\sqrt{2\pi}$

$4\pi ^2$

$4\pi$

Correct answer:

$2\pi ^2$

Explanation:

Write the formula for the area of a square.

A=s^2

Substitute the side into the formula.

$A=(\pi\sqrt2)^2 = \pi^2 (2) = 2\pi ^2$

The answer is: $2\pi ^2$

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SAT II Math I : Area

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

All SAT II Math I Resources

Example Questions

Example Question #1 : How To Find The Area Of A Kite

Example Question #2 : How To Find The Area Of A Kite

Example Question #1 : How To Find The Area Of A Kite

Example Question #11 : Geometry

Example Question #15 : Rhombuses

Example Question #1 : How To Find The Area Of A Rhombus

Example Question #11 : Geometry

Example Question #11 : Area

Example Question #12 : Geometry

Example Question #13 : Area

All SAT II Math I Resources

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