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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

Example Question #21 : 2 Dimensional Geometry

On the XY plane, line segment AB has endpoints (0, a) and (b, 0). If a square is drawn with segment AB as a side, in terms of a and b what is the area of the square?

Possible Answers:

$\displaystyle A=2(ab)^2$

Cannot be determined

$\displaystyle A=b^2+a^2$

$\displaystyle A=(a + b) (a - b)$

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

Correct answer:

$\displaystyle A=b^2+a^2$

Explanation:

Since the question is asking for area of the square with side length AB, recall the formula for the area of a square.

$A=\text{side}^2$ $\displaystyle A=\text{side}^2$

Now, use the distance formula to calculate the length of AB.

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ $\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

let

$\$x_1,y_1)=(0,a) \\(x_2,y_2)=(b,0)$ \(\displaystyle \\(x_1,y_1)=(0,a) \\(x_2,y_2)=(b,0)$

Now substitute the values into the distance formula.

$\\d=\sqrt{(b-0)^2+(0-2)^2} \\d=\sqrt{b^2+a^2}$ $\displaystyle \\d=\sqrt{(b-0)^2+(0-2)^2} \\d=\sqrt{b^2+a^2}$

From here substitute the side length value into the area formula.

$\\A=\text{side}^2 \\A=\left(\sqrt{b^2+a^2}\right)^2 \\A=b^2+a^2$ $\displaystyle \\A=\text{side}^2 \\A=\left(\sqrt{b^2+a^2}\right)^2 \\A=b^2+a^2$

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Example Question #451 : Sat Subject Test In Math I

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

AB = 13 $\displaystyle AB = 13$

BC = 25 $\displaystyle BC = 25$

AC = 17 $\displaystyle AC = 17$

Possible Answers:

$\displaystyle 55$

$\displaystyle 29$

102 $\displaystyle 102$

Cannot be determined

$\displaystyle 74$

Correct answer:

102 $\displaystyle 102$

Explanation:

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

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

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

Setting, $\displaystyle a = AB = 13$, $\displaystyle b = AC = 17$, and $\displaystyle c= BC = 25$,

$s = \frac{13+17+25}{2} = \frac{55}{2} = 27.5$ $\displaystyle s = \frac{13+17+25}{2} = \frac{55}{2} = 27.5$

and, substituting in Heron's formula:

$A = \sqrt{27.5(27.5-13)(27.5-17)(27.5-25 )}$ $\displaystyle A = \sqrt{27.5(27.5-13)(27.5-17)(27.5-25 )}$

$= \sqrt{27.5(14.5)(10.5)(2.5 )}$ $\displaystyle = \sqrt{27.5(14.5)(10.5)(2.5 )}$

$= \sqrt{10,467.1875}$ $\displaystyle = \sqrt{10,467.1875}$

$\approx 102$ $\displaystyle \approx 102$

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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 #21 : 2 Dimensional Geometry

Example Question #451 : Sat Subject Test In Math I

All SAT II Math I Resources

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