Area

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SAT Subject Test in Math II › Area

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Determine the area of a circle with a diameter of .

$9x^2 \pi$

CORRECT

$9x\pi$

$36x^2\pi$

$18x^2\pi$

$6x^2\pi$

Explanation

Write the formula for the area of a circle.

$A= \pi r^2$

The radius is half the diameter, .

Substitute the radius into the equation.

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

The answer is: $9x^2 \pi$

Find the area of a circle with a radius of .

$9\pi x^6$

CORRECT

$6\pi x^6$

$9\pi^2 x^6$

$6\pi^2 x^6$

$9\pi x^9$

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: $9\pi x^6$

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

CORRECT

$\frac{1}{9}$

$\frac{2}{9}$

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:

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

CORRECT

Explanation

The area of a triangle, given its three sidelengths, can be calculated using Heron's formula:

$A = \sqrt{s(s-a)(s-b)(s-c)}$ ,

where , , and are the lengths of the sides, and $s = \frac{a+b+ c}{2}$ .

Setting , , and , evaluate :

$s = \frac{20+26+32}{2} = \frac{78}{2} = 39$

and, substituting in Heron's formula:

$A = \sqrt{39(39-20)(39-26)(39-32)}$

$= \sqrt{39(19)(13)(7)}$

$= \sqrt{67,431 }$

$\approx 259.7$

To the nearest whole, this is 260.

Determine the side of a square with an area of .

$10\sqrt{x}$

CORRECT

$\sqrt{10x}$

Explanation

Write the formula for the area of a square.

Substitute the area into the equation.

Square root both sides.

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

$s = 10\sqrt{x}$

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

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

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

CORRECT

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

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

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

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

Explanation

The area of a rectangle is:

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

Determine the area of a circle with a radius of $\frac{3}{5}\pi$ .

$\frac{9}{25}\pi ^3$

CORRECT

$\frac{9}{25}\pi ^2$

$\frac{6}{25}\pi ^2$

$\frac{9}{5}\pi ^3$

$\frac{9}{5}\pi ^2$

Explanation

Write the formula of the area of a circle.

$A= \pi r^2$

Substitute the radius.

$A= \pi (\frac{3}{5}\pi)^2=\pi (\frac{3}{5}\pi)(\frac{3}{5}\pi)$

The answer is: $\frac{9}{25}\pi ^3$

Hexagon

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

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

CORRECT

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

$125 \sqrt{2}$

Explanation

Quadrilateral 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 , and . Quadrilateral is a trapezoid with bases of length 10 and 20. Since $\overline{MN}$ has its endpoints at the midpoints of the legs of Trapezoid , 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 $h =PQ = \frac{5}{2} \sqrt{3} , B = MN = 15, b = BC = 10$ :

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

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

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

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

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

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

CORRECT

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

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

Right_triangle_3

Note: Figure NOT drawn to scale.

Refer to the above diagram. , , and $\angle ABC$ and $\angle ADB$ are right angles. What percent of $\Delta ABC$ is colored red?

CORRECT

$16 \frac{2}{3} %$

$33 \frac{1}{3}%$

Explanation

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$ .

The area of $\Delta ADB$ , the shaded region, is half the products of its legs:

$A= \frac{1}{2} (AD) \times (DB ) =\frac{1}{2} \times 6 \times 12 = 36$

The area of $\Delta ABC$ is half the product of its hypoteuse, which we can see as the base, and the length of corresponding altitude $\overline{BD}$ :