Right Triangles - SSAT Upper Level Quantitative

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Question

The base and height of a right triangle are each 1 inch. What is the hypotenuse?

Answer

You need to use the Pythagorean Theorem, which is $a^{2}+b^{^{2}} = c^{^{2}}$ .

Add the first two values and you get . Take the square root of both sides and you get $\sqrt{2}$ .

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Question

Give the perimeter of the above parallelogram if .

Answer

By the $30^{\circ}-60^{\circ}-90^{\circ}$ Theorem:

$AB = CD= BD\cdot \frac{\sqrt{3}}{3}= 24 \cdot \frac{\sqrt{3}}{3}= 8\sqrt{3}$ , and

$BC = AD = 2\cdot AB = 2 \cdot 8\sqrt{3} = 16 \sqrt{3}$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8\sqrt{3}+ 8\sqrt{3}+ 16\sqrt{3}+16\sqrt{3} = 48\sqrt{3}$

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Question

A right triangle has legs with lengths of units and units. What is the length of the hypotenuse?

Answer

$\text{Hypotenuse}=\sqrt{\text{(leg 1)}^2+\text{(leg 2)}^2}$

Using the numbers given to us by the question,

$\text{Hypotenuse}=\sqrt{2^2+5^2}=\sqrt{29}=5.38$ units

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Question

A right triangle has legs with the lengths and . Find the length of the hypotenuse.

Answer

Use the Pythagorean Theorem to find the length of the hypotenuse.

$(\text{Hypotenuse})^2=(\text{leg 1})^2+(\text{leg 2})^2$

$\text{Hypotenuse}=\sqrt{4^2+5^2}=\sqrt{41}:cm$

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Question

Find the length of the hypotenuse in the right triangle below.

Answer

Use the Pythagorean Theorem to find the hypotenuse.

$16^2+20^2=\text{Hypotenuse}^2$

$656=\text{Hypotenuse}^2$

$\sqrt{656}=25.61=\text{Hypotenuse}$

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Question

If James traveled north $30\textup { miles}$ and John traveled $40\textup { miles}$ west from the same town, how many miles away will they be from each other when they reach their destinations?

Answer

The distances when put together create a right triangle.

The distance between them will be the hypotenuse or the diagonal side.

You use Pythagorean Theorem or $a^{2}+b^{2}=c^{2}$ to find the length.

So you plug and for and which gives you,

$900+1600=c^{2}$ or $2500=c^{2}$ .

Then you find the square root of each side and that gives you your answer of .

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Question

One angle of a right triangle has measure $68^{\circ }$ . Give the measures of the other two angles.

Answer

One of the angles of a right triangle is by definition a right, or $90^{\circ }$ , angle, so this is the measure of one of the missing angles. Since the measures of the angles of a triangle total $180^{\circ }$ , if we let the measure of the third angle be , then:

The other two angles measure $22 ^{\circ }, 90^{\circ }$ .

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Question

One angle of a right triangle has measure $120^{\circ }$ . Give the measures of the other two angles.

Answer

A right triangle must have one right angle and two acute angles; this means that no angle of a right triangle can be obtuse. But since $120^{\circ } > 90^{\circ }$ , it is obtuse. This makes it impossible for a right triangle to have a $120^{\circ }$ angle.

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Question

Find the degree measure of in the right triangle below.

Answer

The total number of degrees in a triangle is .

While $47^{\circ}$ is provided as the measure of one of the angles in the diagram, you are also told that the triangle is a right triangle, meaning that it must contain a $90^{\circ}$ angle as well. To find the value of , subtract the other two degree measures from .

$x=180-90-47=43^{\circ}$

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Question

Find the angle value of .

Answer

All the angles in a triangle must add up to 180 degrees.

$90^{\circ}+47^{\circ}+v=180^{\circ}$

$137^{\circ}+v=180^{\circ}$

$137^{\circ}+v-137^{\circ}=180^{\circ}-137^{\circ}$

$v=43^{\circ}$

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Question

Find the angle value of .

Answer

All the angles in a triangle adds up to $180^{\circ}$ .

$90^{\circ}+32^{\circ}+w=180^{\circ}$

$122^{\circ}+w=180^{\circ}$

$122^{\circ}+w-122^{\circ}=180^{\circ}-122^{\circ}$

$w=58^{\circ}$

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Question

Find the angle value of .

Answer

All the angles in a triangle add up to degrees.

$90^{\circ}+44^{\circ}+y=180^{\circ}$

$134^{\circ}+y=180^{\circ}$

$134^{\circ}+y-134^{\circ}=180^{\circ}-134^{\circ}$

$y=46^{\circ}$

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Question

Find the angle measure of .

Answer

All the angles in a triangle add up to $180^{\circ}$ .

$90^{\circ}+59^{\circ}+z=180^{\circ}$

$149^{\circ}+z=180^{\circ}$

$149^{\circ}+z-149^{\circ}=180^{\circ}-149^{\circ}$

$z=31^{\circ}$

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Question

If a right triangle is similar to a right triangle, which of the other triangles must also be a similar triangle?

Answer

For the triangles to be similar, the dimensions of all sides must have the same ratio by dividing the 3-4-5 triangle.

The 6-8-10 triangle will have a scale factor of 2 since all dimensions are doubled the original 3-4-5 triangle.

The only correct answer that will yield similar ratios is the triangle with a scale factor of 4 from the 3-4-5 triangle.

The other answers will yield different ratios.

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Question

What is the main difference between a right triangle and an isosceles triangle?

Answer

By definition, a right triangle has to have one right angle, or a $90^{\circ}$ angle, and an isosceles triangle has equal base angles and two equal side lengths.

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Question

$\bigtriangleup ABC \cong \bigtriangleup DEF$ , where $\angle B$ is a right angle, , and .

Which of the following is true?

Answer

$\bigtriangleup ABC \cong \bigtriangleup DEF$ , and corresponding parts of congruent triangles are congruent.

Since $\angle B$ is a right angle, so is $\angle E$ . and ; since , it follows that . $\bigtriangleup DEF$ is an isosceles right triangle; consequently, $m \angle A = m \angle C = 45^{\circ }$ .

$\bigtriangleup DEF$ is a 45-45-90 triangle with hypotenuse of length . By the 45-45-90 Triangle Theorem, the length of each leg is equal to that of the hypotenuse divided by $\sqrt{2}$ ; therefore,

$DE=EF = \frac{DF}{\sqrt{2}} = \frac{20}{\sqrt{2}} =\frac{20 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{20 \sqrt{2}}{2} = 10 \sqrt{2}$

$DE = 20\sqrt{2}$ is eliminated as the correct choice.

Also, the perimeter of $\bigtriangleup DEF$ is

$P = DE+EF+DF = 10 \sqrt{2} + 10 \sqrt{2} + 20 = 20+ 20 \sqrt{2} \ne 40$ .

This eliminates the perimeter of $\bigtriangleup DEF$ being 40 as the correct choice.

Also, $m \angle F = 60^{\circ }$ is eliminated as the correct choice, since the triangle is 45-45-90.

The area of $\bigtriangleup DEF$ is half the product of the lengths of its legs:

$A = \frac{1}{2} \cdot DE \cdot EF$

$= \frac{1}{2} \cdot 10 \sqrt{2}\cdot 10 \sqrt{2}$

$= \frac{1}{2} \cdot 10\cdot 10 \cdot \sqrt{2} \cdot \sqrt{2}$

$= \frac{1}{2} \cdot 10\cdot 10 \cdot 2$

The correct choice is the statement that $\bigtriangleup DEF$ has area 100.

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ with right angles $\angle B$ and $\angle E$ ; $\angle A \cong \angle D$ .

Which of the following statements alone, along with this given information, would prove that $\bigtriangleup ABC \cong \bigtriangleup DEF$ ?

I) $\overline{AB } \cong \overline{DE}$

II) $\overline{BC }\cong \overline{EF}$

III) $\overline{AC } \cong \overline{DF}$

Answer

$\angle A \cong \angle D$ ; $\angle B \cong \angle E$ since both are right angles.

Given that two pairs of corresponding angles are congruent and any one side of corresponding sides is congruent, it follows that the triangles are congruent. In the case of Statement I, the included sides are congruent, so by the Angle-Side-Angle Congruence Postulate, $\bigtriangleup ABC \cong \bigtriangleup DEF$ . In the case of the other two statements, a pair of nonincluded sides are congruent, so by the Angle-Angle-Side Congruence Theorem, $\bigtriangleup ABC \cong \bigtriangleup DEF$ . Therefore, the correct choice is I, II, or III.

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Question

Given:

$\bigtriangleup ABC$ , where $\angle B$ is a right angle; ;

$\bigtriangleup DEF$ , where $\angle E$ is a right angle and ;

$\bigtriangleup GHI$ , where $\angle H$ is a right angle and $\bigtriangleup GHI$ has perimeter 60;

$\bigtriangleup JKL$ , where $\angle K$ is a right angle and $\bigtriangleup JKL$ has area 120;

$\bigtriangleup MNO$ , where $\angle N$ is a right triangle and $\angle M \cong \angle A$

Which of the following must be a false statement?

Answer

$\bigtriangleup ABC$ has as its leg lengths 10 and 24, so the length of its hypotenuse, $\overline{AC}$ , is

$AC = \sqrt{(AB)^{2}+(BC)^{2}} = \sqrt{10^{2}+24^{2}} = \sqrt{100+576} = \sqrt{676 } = 26$

Its perimeter is the sum of its sidelengths:

Its area is half the product of the lengths of its legs:

$A = \frac{1}{2} \cdot 10 \cdot 24 = 120$

$\bigtriangleup GHI$ and $\bigtriangleup JKL$ have the same perimeter and area, respectively, as $\bigtriangleup ABC$ ; also, between $\bigtriangleup MNO$ and $\bigtriangleup ABC$ , corresponding angles are congruent. In the absence of other information, none of these three triangles can be eliminated as being congruent to $\bigtriangleup ABC$ .

However, and . Therefore, $\overline{AC} \ncong \overline{DF}$ . Since a pair of corresponding sides is noncongruent, it follows that $\bigtriangleup DEF \ncong \bigtriangleup ABC$ .

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Question

A right triangle has a hypotenuse of 39 and one leg is 36. What is the length of the other leg?

Answer

You may recognize these numbers as multiples of 13 and 12 (each by a factor of 3) and remember that sides of length 5, 12 and 13 make a special right triangle. So the other leg would be 15 $(5\times 3)$ .

If you don't remember this, you can use Pythagorean theorem:

$39^{2}-36^{2}= 1521-1296=225$

$\sqrt{225}=15$

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Question

A right triangle has a hypotenuse of and one leg has a length of . What is the length of the other leg?

Answer

When calculating the lengths of sides of a right triangle, we can use the Pythagorean Theorem as follows:

$a^{2}+b^{2}=c^{2}$ , where and are legs of the triangle and is the hypotenuse.

Plugging in our given values:

$9^{2}+b^{2}=12^{2}$

Subtracting from each side of the equation:

$b^{2}=12^{2}-9^{2}$

$b^{2}=144-81$

$b^{2}=63$

Taking the square root of each side of the equation:

$b=\sqrt{63}$

Simplifying the square root:

$b=\sqrt{9\times 7}$

$b=3\sqrt{7}$

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