Plane Geometry

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PSAT Math › Plane Geometry

Questions 1 - 10
1

In order to get to work, Jeff leaves home and drives 4 miles due north, then 3 miles due east, followed by 6 miles due north and, finally, 7 miles due east. What is the straight line distance from Jeff’s work to his home?

2√5

11

10√2

15

6√2

Explanation

Jeff drives a total of 10 miles north and 10 miles east. Using the Pythagorean theorem (a2+b2=c2), the direct route from Jeff’s home to his work can be calculated. 102+102=c2. 200=c2. √200=c. √100Ÿ√2=c. 10√2=c

2

Thingy_3

Refer to the above diagram. .

Give the area of Quadrilateral .

Explanation

, since both are right; by the Corresponding Angles Theorem, , and Quadrilateral is a trapezoid.

By the Angle-Angle Similarity Postulate, since

and

(by reflexivity),

,

and since corresponding sides of similar triangles are in proportion,

, the larger base of the trapozoid;

The smaller base is .

, the height of the trapezoid.

The area of the trapezoid is

3

All of the following could be the possible side lengths of a triangle EXCEPT:

Explanation

The length of the third side of a triangle must always be between (but not equal to) the sum and the difference of the other two sides.

For instance, take the example of 2, 6, and 7.

and . Therefore, the third side length must be greater than 4 and less than 8. Because 7 is greater than 4 and less than 8, it is possible for these to be the side lengths of a triangle.

The 5, 7, 12 answer choice is the only option for which this is not the case.

and . Therefore, the third side length must be between 2 and 12. Because it is equal to the sum, not less than the sum, it is not possible that these could be the side lengths of a triangle.

4

Stuff_1

Note: Figure NOT drawn to scale.

The above figure shows Rectangle ; is the midpoint of ; ; .

What percent of Rectangle is shaded?

Explanation

The area of , the shaded region in question, is that of the rectangle minus those of and . We look at both.

The answer is independent of the sidelengths of the rectangle, so to ease calculations - this will become more apparent later - we will arbitrarily assign to the rectangle the dimensions

and, subsequently,

and, since is the midpoint of ,

.

The area of Rectangle is equal to .

Since , and ,

and

The area of is equal to .

The area of is equal to

The area of the shaded region is therefore , which is

of the rectangle.

5

What is the area of a square that has a diagonal whose endpoints in the coordinate plane are located at (-8, 6) and (2, -4)?

100

100√2

50√2

50

200√2

Explanation

Square_part1

Square_part2

Square_part3

6

The base angle of an isosceles triangle is 15 less than three times the vertex angle. What is the vertex angle?

Explanation

Every triangle contains 180 degrees. An isosceles triangle has one vertex angle and two congruent base angles.

Let = vertex angle and = base angle

So the equation to solve becomes .

7

A square with a side length of 4 inches is inscribed in a circle, as shown below. What is the area of the unshaded region inside of the circle, in square inches?

Act_math_01

8π - 16

4π-4

8π-4

2π-4

8π-8

Explanation

Using the Pythagorean Theorem, the diameter of the circle (also the diagonal of the square) can be found to be 4√2. Thus, the radius of the circle is half of the diameter, or 2√2. The area of the circle is then π(2√2)2, which equals 8π. Next, the area of the square must be subtracted from the entire circle, yielding an area of 8π-16 square inches.

8

Circle_graph_area3

100_π_

50_π_

25_π_

10_π_

20_π_

Explanation

Circle_graph_area2

9

Circle_graph_area3

100_π_

50_π_

25_π_

10_π_

20_π_

Explanation

Circle_graph_area2

10

A square with a side length of 4 inches is inscribed in a circle, as shown below. What is the area of the unshaded region inside of the circle, in square inches?

Act_math_01

8π - 16

4π-4

8π-4

2π-4

8π-8

Explanation

Using the Pythagorean Theorem, the diameter of the circle (also the diagonal of the square) can be found to be 4√2. Thus, the radius of the circle is half of the diameter, or 2√2. The area of the circle is then π(2√2)2, which equals 8π. Next, the area of the square must be subtracted from the entire circle, yielding an area of 8π-16 square inches.

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