Finding Sides

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SAT Subject Test in Math I › Finding Sides

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Which of the following describes a triangle with sides one kilometer, 100 meters, and 100 meters?

The triangle cannot exist.

CORRECT

The triangle is acute and equilateral.

The triangle is obtuse and isosceles, but not equilateral.

The triangle is acute and isosceles, but not equilateral.

The triangle is obtuse and scalene.

Explanation

One kilometer is equal to 1,000 meters, so the triangle has sides of length 100, 100, and 1,000. However,

That is, the sum of the least two sidelengths is not greater than the third. This violates the Triangle Inequality, and this triangle cannot exist.

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

Evaluate . Round to the nearest tenth, if applicable.

$AC \approx 14.2$

CORRECT

$AC \approx 10.3$

$AC \approx 18.0$

$AC \approx 32.4$

$AC \approx 30.4$

Explanation

By the Law of Cosines,

$\left (AC \right )^{2} =\left (AB \right )^{2}+ \left (BC \right )^{2} - 2 (AB)(BC) \cos b^{\circ }$

Substitute :

$\left (AC \right )^{2} =15^{2}+20^{2} - 2 \cdot15 \cdot 20 \cos 45^{\circ }$

$\left (AC \right )^{2} =225+400 - 600 \cdot \frac{\sqrt{2}}{2}$

$\left (AC \right )^{2} \approx 625 -424.26$

$\left (AC \right )^{2} \approx 200.74$

$AC \approx \sqrt{200.74} \approx 14.2$

Pentagon

The above figure is a regular pentagon. Evaluate to the nearest tenth.

$x \approx 6.5$

CORRECT

$x \approx 6.1$

$x \approx 6.9$

$x \approx 5.7$

$x \approx 5.4$

Explanation

Two sides of the triangle formed measure 4 each; the included angle is one angle of the regular pentagon, which measures

$m = \left [\frac{180(5-2)}{5} \right ] ^{\circ }= 108^{\circ }$

The length of the third side can be found by applying the Law of Cosines:

$x^{2} = a^{2} + b^{2} - 2ab \cos C^{\circ }$

where :

$x^{2} = 4^{2} + 4^{2} - 2 \cdot 4 \cdot 4 \cos 108^{\circ }$

$x^{2} = 16 +16 -32 \cos 108^{\circ }$

$x^{2} \approx 32 -32 \left ( -0.3090\right )$

$x^{2} \approx 41.88$

$x \approx \sqrt{41.88} \approx 6.5$

is rhombus with side lengths in meters. $\overline{AB} = 13$ and $\overline{BD} = 10$ . What is the length, in meters, of $\overline{AC}$ ?

Rhombus_1

CORRECT

Explanation

A rhombus is a quadrilateral with four sides of equal length. Rhombuses have diagonals that bisect each other at right angles.

Rhombus_2

Thus, we can consider the right triangle to find the length of diagonal $\overline{AC}$ . From the given information, each of the sides of the rhombus measures meters and $\overline{BD} = 10$ .

Because the diagonals bisect each other, we know:

$\overline{AD} = 13$

$\overline{ED} = 5$

Using the Pythagorean theorem,

$\overline{AE}^2 + 5^2 = 13^2$

$\overline{AE}^2 = 144$

$\overline{AE} = 12$

$\overline{AC} = 24$

Rhombus_1

is a rhombus with side length . Diagonal $\overline{BD}$ has a length of . Find the length of diagonal $\overline{AC}$ .

CORRECT

$10\sqrt{2}:cm$

$2\sqrt{61}:cm$

$2\sqrt{34}:cm$

Explanation

A rhombus is a quadrilateral with four sides of equal length. Rhombuses have diagonals that bisect each other at right angles.

Rhombus_2

Thus, we can consider the right triangle to find the length of diagonal $\overline{AC}$ . From the problem, we are given that the sides are and $\overline{BD} = 12:cm$ . Because the diagonals bisect each other, we know:

$\overline{AD} = 10:cm$

$\overline{ED} = 6:cm$

Using the Pythagorean Theorem,

$(\overline{AE})^2 + (6:cm)^2 = (10:cm)^2$

$(\overline{AE})^2 + 36:cm^2 = 100:cm^2$

$(\overline{AE})^2 = 64:cm^2$

$\overline{AE} = 8:cm$

$\overline{AC} = 16:cm$

Regular Hexagon has perimeter 120. $\overline{AB }$ has as its midpoint; segment $\overline{MC}$ is drawn. To the nearest tenth, give the length of $\overline{MC}$ .

CORRECT

Explanation

The perimeter of the regular hexagon is 80, so each side measures one sixth of this, or 20. Also, since is the midpoint of $\overline{AB}$ , $AM = BM = \frac{1}{2}AB = \frac{1}{2} \cdot 20 = 10$ .

Also, each interior angle of a regular hexagon measures $120^{\circ }$ .

Below is the hexagon in question, with indicated and $\overline{MC}$ constructed; all relevant measures are marked.

Hexagon

A triangle $\bigtriangleup MBC$ is formed with , , and included angle measure $\angle B= 120 ^{\circ }$ . The length of the remaining side can be calculated using the Law of Cosines:

$c^{2} = a^{2} + b^{2} -2 ab \cos \gamma$

where and are the lengths of two sides, is the measure of their included angle, and is the length of the third side.

Setting $a = M B= 10, b= BC= 20 , \gamma= m \angle B = 120 ^{\circ }$ , and , substitute and evaluate :

$(CM)^{2} = 10^{2} + 20 ^{2} -2 (10) (20) \cos 120^{\circ }$

$(CM)^{2} = 700$ ;

Taking the square root of both sides:

$CM \approx\sqrt{ 700} \approx 26.5$ ,

the correct choice.

Regular Pentagon has perimeter 80. $\overline{AB }$ and $\overline{CD}$ have and as midpoints, respectively; segment $\overline{MN}$ is drawn. Give the length of $\overline{MN}$ to the nearest tenth.

CORRECT

Explanation

The perimeter of the regular pentagon is 80, so each side measures one fifth of this, or 16. Since is the midpoint of $\overline{AB}$ , $AM = BM = \frac{1}{2}AB = \frac{1}{2} \cdot 16 = 8$ .

Similarly, .

Also, each interior angle of a regular pentagon measures $108^{\circ }$ .

Below is the pentagon with the midpoints and , and with $\overline{MN}$ constructed. Note that perpendiculars have been drawn to $\overline{MN}$ from and , with feet at points and , respectively.

Pentagon 2

is a rectangle, so .

$\sin \angle MBX = \frac{MX}{MB}$ , or $MX = MB \cdot \sin \angle MBX$

$m \angle MBX = m \angle MBC- m \angle XBC = 108 ^{\circ } - 90 ^{\circ } = 18^{\circ }$ . Substituting:

$MX = 8 \cdot \sin 18^{\circ }$

$\approx 8 \cdot 0.3090$

$\approx 2.47$

For the same reason,

$NY \approx 2.47$ .

Adding the segment lengths:

$MN = MX + XY + NY \approx 2.47+ 16 + 2.47 \approx 20.94$ ,

Rond answer to the nearest tenth.

$MN \approx 20.9$

What is the length of the diagonals of trapezoid ? Assume the figure is an isoceles trapezoid.

Trapezoid

$\sqrt{97}m$

CORRECT

$\sqrt{98}m$

$\sqrt{99}m$

$\sqrt{96}m$

$\sqrt{95}m$

Explanation

To find the length of the diagonal, we need to use the Pythagorean Theorem. Therefore, we need to sketch the following triangle within trapezoid :

Trapezoid

We know that the base of the triangle has length . By subtracting the top of the trapezoid from the bottom of the trapezoid, we get:

Dividing by two, we have the length of each additional side on the bottom of the trapezoid:

$\frac{6: m}{2} = 3: m$

Adding these two values together, we get .

The formula for the length of diagonal uses the Pythagoreon Theorem:

, where is the point between and representing the base of the triangle.

Plugging in our values, we get:

$AC = \sqrt{97}: m$

If the two legs of a right triangle are and , find the third side.

CORRECT

Explanation

Step 1: Recall the formula used to find the missing side(s) of a right triangle...

Step 2: Identify the legs and the hypotenuse in the formula...

are the legs, and is the hypotenuse.

Step 3: Plug in the values of a and b given in the question...

$\Rightarrow 25+144=c^2$

$\Rightarrow 169=c^2$

$\Rightarrow c=\pm 13$

Step 3: A special rule about all triangles...

For any triangle, the measurements of any of the sides CANNOT BE zero.

So, the missing side is , or

Regular Pentagon has perimeter 80. $\overline{AB }$ has as its midpoint; segment $\overline{MC}$ is drawn. To the nearest tenth, give the length of $\overline{MC}$ .

CORRECT

Explanation

The perimeter of the regular pentagon is 80, so each side measures one fifth of this, or 16. Also, since is the midpoint of $\overline{AB}$ , $AM = BM = \frac{1}{2}AB = \frac{1}{2} \cdot 16 = 8$ .

Also, each interior angle of a regular pentagon measures $108^{\circ }$ .

Below is the pentagon in question, with indicated and $\overline{MC}$ constructed; all relevant measures are marked.

Pentagon 2

A triangle $\bigtriangleup MBC$ is formed with , , and included angle measure $\angle B= 108^{\circ }$ . The length of the remaining side can be calculated using the law of cosines: