Math - Geometry | Practice Hub

A right triangle has legs of 15m and 20m. What is the length of the hypotenuse?

30m

45m

35m

40m

25m

Explanation

The Pythagorean theorem is a2 + b2 = c2, where a and b are legs of the right triangle, and c is the hypotenuse.

(15)2 + (20)2 = c2 so c2 = 625. Take the square root to get c = 25m

An isoceles triangle has a base angle that is twice the vertex angle. What is the sum of the base and vertex angles?

Explanation

All triangles have degrees. An isoceles triangle has one vertex angle and two congruent base angles.

Let vertex angle and base angle.

So the equation to solve becomes:

Thus for the vertex angle and for the base angle.

The sum of the vertex and one base angle is .

To the nearest tenth, give the area of a $60^{\circ }$ sector of a circle with diameter 18 centimeters.

$42.4 \textrm{ cm}^{2}$

$84.8 \textrm{ cm}^{2}$

$169.6 \textrm{ cm}^{2}$

$21.2 \textrm{ cm}^{2}$

$254.5 \textrm{ cm}^{2}$

Explanation

The radius of a circle with diameter 18 centimeters is half that, or 9 centimeters. The area of a $60^{\circ }$ sector of the circle is

$A = \frac{60}{360 }\cdot \pi \cdot 9^{2} =\frac{1}{6} \cdot \pi \cdot 81 \approx 42.4 \textrm{ cm}^{2}$

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 .

Find the area of the following kite:

Kite

Explanation

The formula for the area of a kite is:

$A = \frac{1}{2}(d_{1}\cdot d_{2})$

Where $d_{1}$ is the length of one diagonal and $d_{2}$ is the length of the other diagonal

Plugging in our values, we get:

$A = \frac{1}{2}(d_{1}\cdot d_{2})$

$A = \frac{1}{2}(6m\cdot 13m) = 39m^2$

The perimeter of a square is 48. What is the length of its diagonal?

$12\sqrt{2}$

$\sqrt{12}$

$24\sqrt{2}$

$48\sqrt{2}$

$2\sqrt{12}$

Explanation

Perimeter = side * 4

48 = side * 4

Side = 12

We can break up the square into two equal right triangles. The diagonal of the sqaure is then the hypotenuse of these two triangles.

Therefore, we can use the Pythagorean Theorem to solve for the diagonal:

$\sqrt{144+144}=\sqrt{hypotenuse^2}$

$hypotenuse=\sqrt{2\times 144}=\sqrt{2\times 12\times 12}=12\sqrt{2}$

A cube has a side length of meters. What is the volume of the cube?

$8\ \textup{m}^{3}$

$6\ \textup{m}^{3}$

$4\ \textup{m}^{3}$

$2\ \textup{m}^{3}$

$16\ \textup{m}^{3}$

Explanation

The formula for the volume of a cube is:

$Volume = (side)^{3}$

Since the length of one side is meters, the volume of the cube is:

$2^{3 } = 8$ meters cubed.

Find the area of the shaded region:

Screen_shot_2014-02-27_at_6.35.30_pm

$108\pi-81\sqrt{3}m^2$

$108\pi-81\sqrt{2}m^2$

$81\pi-81\sqrt{3}m^2$

$108\pi-108\sqrt{3}m^2$

$81\pi-108\sqrt{3}m^2$

Explanation

To find the area of the shaded region, you must subtract the area of the triangle from the area of the sector.

The formula for the shaded area is:

$A=A_{sector}-A_{triangle}$

$A = \pi(r^2)(part)-\frac{1}{2}(b)(h)$ ,

where is the radius of the circle, is the fraction of the sector, is the base of the triangle, and is the height of the triangle.

In order to the find the base and height of the triangle, use the formula for a triangle:

$a-a\sqrt{3}-2a$ , where is the side opposite the $\measuredangle30$ .

Plugging in our final values, we get:

$A = \pi(18m^2)(\frac{120}{360})-\frac{1}{2}(18\sqrt{3}m)(9m)$

$A = 108\pi-81\sqrt{3}m^2$

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

200√2

Explanation

Square_part1

Square_part2

Square_part3

To the nearest tenth, give the area of a $60^{\circ }$ sector of a circle with diameter 18 centimeters.

$42.4 \textrm{ cm}^{2}$

$84.8 \textrm{ cm}^{2}$

$169.6 \textrm{ cm}^{2}$

$21.2 \textrm{ cm}^{2}$

$254.5 \textrm{ cm}^{2}$

Explanation

The radius of a circle with diameter 18 centimeters is half that, or 9 centimeters. The area of a $60^{\circ }$ sector of the circle is

$A = \frac{60}{360 }\cdot \pi \cdot 9^{2} =\frac{1}{6} \cdot \pi \cdot 81 \approx 42.4 \textrm{ cm}^{2}$

Geometry

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