Right Triangles - SSAT Upper Level Quantitative (Math)

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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

$\bigtriangleup ABC \cong \bigtriangleup DEF$ , 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

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

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

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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

What is the area of a right triangle whose hypotenuse is 13 inches and whose legs each measure a number of inches equal to an integer?

Answer

We are looking for a Pythagorean triple - that is, three integers that satisfy the relationship $a^{2}+b^{2} = c^{2}$ . We know that , and the only Pythagorean triple with is . The legs of the triangle are therefore 5 and 12, and the area of the right triangle is

$A = \frac{1}{2} bh= \frac{1}{2} \cdot 5\cdot 12 = 30$

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Question

Right Triangle A has legs of lengths 10 inches and 14 inches; Right Triangle B has legs of length 20 inches and 13 inches; Rectangle C has length 30 inches. The area of Rectangle C is the sum of the areas of the two right triangles. What is the height of Rectangle C?

Answer

The area of a right triangle is half the product of its legs. The area of Right Triangle A is equal to $\frac{1}{2} \times 10 \times 14 = 70$ square inches; that of Right Triangle B is equal to $\frac{1}{2} \times 20 \times 13 = 130$ square inches. The sum of the areas is square inches, which is the area of Rectangle C.

The area of a rectangle is the product of its length and its height. Therefore, the height is the quotient of the area and the length, which, for Rectangle C, is $200 \div 30 = 6 \frac{2}{3}$ inches.

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Question

Right Triangle A has hypotenuse 25 inches and one leg of length 24 inches; Right Triangle B has hypotenuse 15 inches and one leg of length 9 inches; Rectangle C has length 16 inches. The area of Rectangle C is the sum of the areas of the two right triangles. What is the width of Rectangle C?

Answer

The area of a right triangle is half the product of its legs. In each case, we know the length of one leg and the hypotenuse, so we need to apply the Pythagorean Theorem to find the second leg, then take half the product of the legs:

Right Triangle A:

The length of the second leg is

$\sqrt{25^{2} -24 ^{2}} = \sqrt{625 - 576} = \sqrt{49} = 7$ inches.

The area is

$\frac{1}{2} \times 7 \times 24 = 84$ square inches.

Right Triangle B:

The length of the second leg is

$\sqrt{15^{2} -9 ^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12$ inches.

The area is

$\frac{1}{2} \times 9 \times 12 = 54$ square inches.

The sum of the areas is square inches.

The area of a rectangle is the product of its length and its height. Therefore, the height is the quotient of the area and the length, which, for Rectangle C, is $138 \div 16 = 8 \frac{5}{8}$ inches.

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Question

A right triangle has leg lengths of $10\text{ and }12$ . What is the area of this triangle?

Answer

Since the legs of a right triangle form a right angle, you can use these as the base and the height of the triangle.

$\text{Area}=\frac{1}{2}(\text{base})(\text{height})$

$\text{Area}=\frac{1}{2}(10)(12)=\frac{1}{2}(120)=60$

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Question

A right triangle has leg lengths of and . Find the area of the right triangle.

Answer

The legs of a right triangle also make up its base and its height.

$\text{Area}=\frac{1}{2}\text{base}\times\text{height}$

$\text{Area}=\frac{1}{2}\times6:cm\times9:cm=27:cm^2$

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Question

A right triangle has leg lengths of and . Find the area of this triangle.

Answer

The legs of a right triangle are also its height and its base.

$\text{Area}=\frac{1}{2}\text{base}\times\text{height}$

$\text{Area}=\frac{1}{2}(9:cm)(12:cm)=54:cm^2$

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Question

A given right triangle has two legs of lengths and , respectively. What is the area of the triangle?

Answer

The area of a right triangle with a base and a height can be found with the formula $A=\frac{1}{2}bh$ . Since the two legs of a right triangle are perpendicular to each other, we can use these as the base and height in the formula. Therefore:

$A=\frac{1}{2}bh$

$A=\frac{1}{2}(6:cm)(8:cm)$

$A=\frac{1}{2}(48:cm^2)$

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Question

A right triangle has two legs of lengths and , respectively. What is the area of the right triangle?

Answer

The area of a right triangle with a base and a height can be found with the formula $A=\frac{1}{2}bh$ . Since the two legs of a right triangle are perpendicular to each other, we can use these as the base and height in the formula. Therefore:

$A=\frac{1}{2}(9:cm)(12:cm)$

$A=\frac{1}{2}(108:cm^{2})$

$A=54:cm^{2}$

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