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Advanced Geometry : Advanced Geometry

Study concepts, example questions & explanations for Advanced Geometry

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Example Questions

Example Question #1 : Graphing

Solve and graph the following inequality:

$2x - 7 \geq 6$

Possible Answers:

$x\geq13$

$x\geq6.5$

$x\leq 13$

x=6.5

$x\leq6.5$

Correct answer:

$x\geq6.5$

Explanation:

To solve the inequality, the first step is to add to both sides:

$2x - 7 +7 \geq 6+7$

$2x \geq 13$

The second step is to divide both sides by :

$\frac{2x}{2} \geq \frac{13}{2}$

$x \geq 6.5$

To graph the inequality, you draw a straight number line. Fill in the numbers from 6.5 to infinity. Infinity can be designated by a ray. Be sure to fill in the number 6.5 , since the equation indicated greater than OR equal to.

The graph should look like:

Number_line

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Example Question #201 : Coordinate Geometry

Points (3,0) and (0,3) lie on a circle. Which of the following could be the equation of that circle?

Possible Answers:

$(x-3)^{2}+(y+3)^{2}=3$

$(x+3)^{2}+(y+3)^{2}=9$

$(x-3)^{2}+(y+3)^{2}=9$

$(x-3)^{2}+(y-3)^{2}=3$

$(x-3)^{2}+(y-3)^{2}=9$

Correct answer:

$(x-3)^{2}+(y-3)^{2}=9$

Explanation:

If we plug the points (3,0) and (0,3) into each equation, we find that these points work only in the equation $(x-3)^{2}+(y-3)^{2}=9$ . This circle has a radius of and is centered at (3,3) .

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Example Question #1 : Graphing

Which of the following lines is perpendicular to the line y=3x+4 ?

Possible Answers:

$y=\frac{x}{3}+4$

$y=\frac{x}{3}+3$

$y=\frac{-x}{3}+2$

y=3x+6

y=3x+2

Correct answer:

$y=\frac{-x}{3}+2$

Explanation:

The key here is to look for the line whose slope is the negative reciprocal of the original slope.

In this case, $\frac{-1}{3}$ is the negative reciprocal of .

Therefore, the equation of the line which is perpendicular to the original equation is:

$y=\frac{-x}{3}+2$

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Example Question #1 : How To Graph A Two Step Inequality

Which inequality does this graph represent?

Inequality a

Possible Answers:

$0\geq y \geq -\frac{3}{5}x+3$ ;

$0 \geq y \geq \frac{1}{2}+1$

$0<y\leq-\frac{1}{2}x+1$ ;

$0<y\leq \frac{3}{5}x+3$

$-\frac{3}{5}x+3 \geq y \geq \frac{1}{2}x+1$

$0<y\leq \frac{1}{2}x+1$ ;

$0<y\leq-\frac{3}{5}x+3$

Correct answer:

$0<y\leq \frac{1}{2}x+1$ ;

$0<y\leq-\frac{3}{5}x+3$

Explanation:

The two lines represented are $y=\frac{1}{2}x+1$ and $y=-\frac{3}5{}x+3$ . The shaded region is below both lines but above y=0.

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Example Question #1 : How To Graph A Two Step Inequality

What is the area of the shaded region for the following inequality:

$y\leq -\frac{3}{5}x+3$ ;

$0\leq y \leq \frac{1}{2}x+1$

Possible Answers:

7.5

5.2

0.5

Correct answer:

Explanation:

This inequality will produce the following graph:

Inequality a

The shaded area is a triangle with base 7 and height 2.

To find the area, plug these values into the area formula for a triangle, $\frac{1}{2}bh$ .

In this case, we are evaluating $\frac{1}{2}\cdot 7\cdot 2$ , which equals 7.

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Example Question #291 : Advanced Geometry

What is the area of the shaded region for this system of inequalities:

$0\leq y \leq \frac{1}{2}x+3$ ; $\frac{1}{2}x \leq y \leq 5$

Possible Answers:

$\infty$

Correct answer:

Explanation:

Once graphed, the inequality will look like this:

Inequality b

To find the area, it is easiest to consider it as 2 congruent triangles with base 6 and height 3.

The total area will then be

$2\cdot \frac{1}{2}(6)(3)$ , or just $6\cdot 3 = 18$ .

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Example Question #1 : How To Graph A Two Step Inequality

Find the -intercept for the following:

3x-2y=11

Possible Answers:

$\left (0,\frac{-11}{2} \right )$

$\left (0,-2 \right )$

$\left (0,\frac{11}{2} \right )$

$\left (0,2 \right )$

Correct answer:

$\left (0,\frac{-11}{2} \right )$

Explanation:

$When\ finding\ the\ y-intercept,\ you\ must\ plug\ in\ x=0\ and\ solve\ for\ y.$

$When\ x=0\ the\ equation\ becomes: 3(0)-2y=11$ .

$The\ next\ step\ becomes: -2y=11$ .

$Dividing\ both\ sides\ by\ -2\ gives\ us\ the\ y-int\ which\ is\ y=-\frac{11}{2}$ .

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Example Question #201 : Graphing

Let D be the region on the (x,y) coordinate plane that contains the solutions to the following inequalities:

$x\leq k$ , where is a positive constant

$0\leq y\leq 12x$

Which of the following expressions, in terms of , is equivalent to the area of D?

Possible Answers:

12k^2

3k^2

8k^2

6k^2

4k^2

Correct answer:

6k^2

Explanation:

Inequality_region1

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Example Question #1 : Kites

Screen_shot_2015-03-06_at_6.20.29_pm

The area of the rectangle is , what is the area of the kite?

Possible Answers:

Correct answer:

Explanation:

The area of a kite is half the product of the diagonals.

$A=\frac{p \cdot q}{2}$

The diagonals of the kite are the height and width of the rectangle it is superimposed in, and we know that because the area of a rectangle is base times height.

Therefore our equation becomes:

$A=\frac{l \cdot w}{2}$ .

We also know the area of the rectangle is $A=l \cdot w=80$ . Substituting this value in we get the following:

$A=\frac{l \cdot w}{2}=\frac{80}{2}=40$

Thus,, the area of the kite is .

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Example Question #1 : Kites

Given: Quadrilateral KITE such that $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , $\angle T$ is a right angle, and diagonal $\overline{I E}$ has length 24.

Give the length of diagonal $\overline{KT}$ .

Possible Answers:

None of the other responses is correct.

$12+ 12\sqrt{3}$

$12+ 12\sqrt{2}$

$12\sqrt{2}+ 12\sqrt{3}$

Correct answer:

$12+ 12\sqrt{3}$

Explanation:

The Quadrilateral KITE is shown below with its diagonals $\overline{I E}$ and $\overline{KT}$ .

. We call the point of intersection :

Kite

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, $\overline{KT}$ bisects the $60 ^{\circ }$ and $90^{\circ }$ angles of the kite. Consequently, $\bigtriangleup KXI$ is a 30-60-90 triangle and $\bigtriangleup TXI$ is a 45-45-90 triangle. Also, the diagonal that connects the common vertices of the pairs of adjacent sides bisects the other diagonal, making the midpoint of $\overline{IE}$ . Therefore,

$IX = \frac{1}{2} \cdot IE = \frac{1}{2} \cdot 24 = 12$ .

By the 30-60-90 Theorem, since $\overline{IX}$ and $\overline{KX}$ are the short and long legs of $\bigtriangleup KXI$ ,

$KX = IX \cdot \sqrt{3}= 12 \sqrt{3}$

By the 45-45-90 Theorem, since $\overline{IX}$ and $\overline{XT}$ are the legs of a 45-45-90 Theorem,

XT = IX = 12 .

The diagonal $\overline{KT}$ has length

$KT =XT + KX= 12+ 12\sqrt{3}$ .

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Advanced Geometry : Advanced Geometry

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #1 : Graphing

Example Question #201 : Coordinate Geometry

Example Question #1 : Graphing

Example Question #1 : How To Graph A Two Step Inequality

Example Question #1 : How To Graph A Two Step Inequality

Example Question #291 : Advanced Geometry

Example Question #1 : How To Graph A Two Step Inequality

Example Question #201 : Graphing

Example Question #1 : Kites

Example Question #1 : Kites

All Advanced Geometry Resources

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