Finding Sides - SAT Subject Test in Math I

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Question

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

Answer

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

$AC = \sqrt{97}: m$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

Answer

The altitude perpendicular to the hypotenuse of a right triangle divides that triangle into two smaller triangles similar to each other and the large triangle. Therefore, the sides are in proportion. The hypotenuse of the triangle is equal to

$\sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169} =13$

Therefore, we can set up, and solve for in, a proportion statement involving the shorter side and hypotenuse of the large triangle and the larger of the two smaller triangles:

$\frac{N}{12} = \frac{5}{13}$

$\frac{N}{12} \times 12 = \frac{5}{13} \times 12$

$N = \frac{60}{13} = 4\frac{8}{13}$

This is not one of the choices.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate the length of the hypotenuse of the blue triangle.

Answer

The inscribed rectangle is a 20 by 20 square. Since opposite sides of the square are parallel, the corresponding angles of the two smaller right triangles are congruent; therefore, the two triangles are similar and, by definition, their sides are in proportion.

The small top triangle has legs 10 and 20. Therefore, the length of its hypotenuse can be determined using the Pythagorean Theorem:

$c = \sqrt{10^{2}+ 20^{2}} = \sqrt{100 + 400} = \sqrt{500} = \sqrt{100} \cdot \sqrt{5} = 10 \sqrt{5}$

The small top triangle has short leg 10 and hypotenuse $10 \sqrt{5}$ . The blue triangle has short leg 20 and unknown hypotenuse , where can be calculated with the proportion statement

$\frac{N}{10\sqrt{5}} = \frac{20}{10}$

$\frac{N}{10\sqrt{5}} \cdot 10\sqrt{5} = \frac{20}{10} \cdot 10\sqrt{5}$

$N = 20 \sqrt{5}$

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Question

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

Answer

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.

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Question

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

.

Evaluate .

Answer

By the Law of Sines,

$\frac{BC}{\sin a^{\circ } } = \frac{AC}{\sin b^{\circ }}$

Substitute and solve for :

$\frac{BC}{\sin 60^{\circ } } = \frac{25}{\sin 45^{\circ }}$

$\frac{BC}{\sin 60^{\circ } } \cdot \sin 60^{\circ } = \frac{25}{\sin 45^{\circ }} \cdot \sin 60^{\circ }$

$BC = \frac{25\sin 60^{\circ }}{\sin 45^{\circ }}$

$BC= \frac{25 \cdot \frac{\sqrt{3}}{2} }{ \frac{\sqrt{2}}{2}}$

$BC= \frac{ \frac{25 \sqrt{3}}{2} }{ \frac{\sqrt{2}}{2}}$

$BC= \frac{25 \sqrt{3}}{2} \cdot \frac{2}{\sqrt{2}}$

$BC= \frac{25\sqrt{3}}{ \sqrt{2} }$

$BC = \frac{25\sqrt{3}}{ \sqrt{2} } \cdot \frac{\sqrt{2}}{\sqrt{2}}$

$BC = \frac{25\sqrt{6}}{ 2 }$

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Question

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

Evaluate . Round to the nearest tenth, if applicable.

Answer

By the Law of Cosines,

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

Substitute :

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

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

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

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

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

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Question

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

Answer

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 :

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

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

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

$x^{2} \approx 41.88$

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

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Question

In triangle $\Delta GHJ$ , $m\angle G = 50^{\circ }$ and $m \angle H = 65^{\circ }$ .

Which of the following statements is true about the lengths of the sides of $\Delta GHJ$ ?

Answer

In a triangle, the shortest side is opposite the angle of least measure; the longest side is opposite the angle of greatest measure. Therefore, if we order the angles, we can order their opposite sides similarly.

Since the measures of the three interior angles of a triangle must total $180^{\circ }$ ,

$m \angle J + m \angle G + m \angle H = 180^{\circ }$

$m \angle J + 50 ^{\circ }+ 65^{\circ } = 180^{\circ }$

$m \angle J + 115^{\circ } = 180^{\circ }$

$m \angle J =65^{\circ }$

$\angle G$ has the least degree measure, so its opposite side, $\overline{HJ}$ , is the shortest. $\angle H \cong \angle J$ , so by the Isosceles Triangle Theorem, their opposite sides $\overline{GJ}$ and $\overline{GH}$ are congruent. Therefore, the correct choice is

.

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Question

Assume quadrilateral $\small ABCD$ is a rhombus. The perimeter of $\small ABCD$ is , and the length of one of its diagonals is $\small 12$ . What is the area of $\small ABCD$ ?

Answer

To solve for the area of the rhombus $\small ABCD$ , we must use the equation $\small A=\frac{pq}{2}$ , where $\small p$ and $\small q$ are the diagonals of the rhombus. Since the perimeter of the rhombus is $\small 40$ , and by definition all 4 sides of a rhombus have the same length, we know that the length of each side is $\small 10$ . We can find the length of the other diagonal if we recognize that the two diagonals combined with a side edge form a right triangle. The length of the hypotenuse is $\small 10$ , and each leg of the triangle is equal to one-half of each diagonal. We can therefore set up an equation involving Pythagorean's Theorem as follows:

$\small 10^2=x^2+6^2$ , where $\small x$ is equal to one-half the length of the unknown diagonal.

We can therefore solve for $\small x$ as follows:

$\small x^2=10^2-6^2=100-36=64$

$\small x$ is therefore equal to 8, and our other diagonal is 16. We can now use both diagonals to solve for the area of the rhombus:

$\small A=\frac{(16)(12)}{2}=\frac{192}{2}=96$

The area of rhombus $\small ABCD$ is therefore equal to $\small 96$

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Question

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

Answer

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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Question

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

Answer

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

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,

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

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

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

$\overline{AE} = 8:cm$

$\overline{AC} = 16:cm$

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Question

is a rhombus. $\overline{AB} = 17:cm$ $\overline{AB} = 17$ and $\overline{AC} = 30:cm$ . Find $\overline{BD}$ .

Answer

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

Thus, we can consider the right triangle to find the length of diagonal $\overline{BD}$ . From the problem, we are given that the sides are and $\overline{AC} = 30:cm$ . Because the diagonals bisect each other, we know:

$\overline{AD} = 17:cm$

$\overline{AE} = 15:cm$

Using the Pythagorean Theorem,

$(\overline{ED})^2 + (15:cm)^2 = (17:cm)^2$

$(\overline{ED})^2 + 225:cm^2 = 289:cm^2$

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

$\overline{ED} = 8:cm$

$\overline{BD} = 16:cm$

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Question

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

Answer

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

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,

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

$\overline{AE}^2 = 144$

$\overline{AE} = 12$

$\overline{AC} = 24$

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Question

is a rhombus. $\overline{BD} = 18$ and $\overline{AC} = 80$ . Find the length of the sides.

Answer

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

Thus, we can consider the right triangle to find the length of side $\overline{AD}$ . From the problem, we are given $\overline{BD} = 18$ and $\overline{AC} = 80$ . Because the diagonals bisect each other, we know:

$\overline{AE} = 40$

$\overline{ED} = 9$

Using the Pythagorean Theorem,

$40^2 + 9^2 = \overline{AD}^2$

$1681 = \overline{AD}^2$

$41 = \overline{AD}$

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Question

A kite has two perpendicular interior diagonals. One diagonal has a measurement of and the area of the kite is $60\textup{ units}^{2}$ . Find the length of the other interior diagonal.

Answer

This problem can be solved by applying the area formula:

Since this question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal.

Thus the solution is:

$60=\frac{8\times diagonal B}{2}$

$60\times2=8\times diagonalB$

$diagonal B=\frac{120}{8}=15$

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Question

Using the kite shown above, find the length of the red (vertical) diagonal.

Answer

In order to solve this problem, first observe that the red diagonal line divides the kite into two triangles that each have side lengths of and Notice, the hypotenuse of the interior triangle is the red diagonal. Therefore, use the Pythagorean theorem: , where the length of the red diagonal.

The solution is:

$c=\sqrt{289}=\sqrt{17\times 17}=17$

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Question

A kite has two perpendicular interior diagonals. One diagonal has a measurement of and the area of the kite is $28\textup{ units}^{2}$ . Find the sum of the two perpendicular interior diagonals.

Answer

You must find the length of the missing diagonal before you can find the sum of the two perpendicular diagonals.

To find the missing diagonal, apply the area formula:

This question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal.

$28=\frac{4\times diagonal B}{2}$

$28\times2=4\times diagonalB$

$diagonal B=\frac{56}{4}=14$

Therefore, the sum of the two diagonals is:

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Question

The area of the kite shown above is $125\textup{ units}^{2}$ and the red diagonal has a length of . Find the length of the black (horizontal) diagonal.

Answer

To find the length of the black diagonal apply the area formula:

Since this question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal.

Thus the solution is:

$125=\frac{25\times diagonal B}{2}$

$125\times2=25\times diagonalB$

$diagonal B=\frac{250}{25}=10$

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Question

Find the length of the diagonals of this isosceles trapezoid, with $\textup{Height}=8$ .

Answer

To find the length of the diagonals, split the top side into 3 sections as shown below:

The two congruent sections plus 8 adds to 14. , so the two congruent sections add to 6. They must each be 3. This means that the top of the right triangle with the diagonal as a hypotenuse must be 11, since .

We can solve for the diagonal, now pictured, using Pythagorean Theorem:

take the square root of both sides

$\sqrt{185}=C$

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Question

Find the length of both diagonals of this quadrilateral.

Answer

All of the lengths with one mark have length 5, and all of the side lengths with two marks have length 4. With this knowledge, we can add side lengths together to find that one diagonal is the hypotenuse to this right triangle:

Using Pythagorean Theorem gives:

take the square root of each side

$\sqrt{97}= C$

Similarly, the other diagonal can be found with this right triangle:

Once again using Pythagorean Theorem gives an answer of $\sqrt{65}$

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