Acute / Obtuse Triangles - SAT Math

Card 0 of 216

Question

Two sides of a triangle have lengths 4 and 7. Which of the following represents the set of all possible lengths of the third side, x?

Answer

The set of possible lengths is: 7-4 < x < 7+4, or 3 < X < 11.

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Question

If two sides of a triangle have lengths 8 and 10, what could the length of the third side NOT be?

Answer

According to the Triangle Inequality Theorem, the sums of the lengths of any two sides of a triangle must be greater than the length of the third side. Since 10 + 8 is 18, the only length out of the answer choices that is not possible is 19.

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Question

A triangle has sides of length 8, 13, and L. Which of the following cannot equal L?

Answer

The sum of the lengths of two sides of a triangle cannot be less than the length of the third side. 8 + 4 = 12, which is less than 13.

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Question

Two sides of a triangle are 20 and 32. Which of the following CANNOT be the third side of this triangle.

Answer

Please remember the Triangle Inequality Theorem, which states that the sum of any two sides of a triangle must be greater than the third side. Therefore, the correct answer is 10 because the sum of 10 and 20 would not be greater than the third side 32.

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Question

A triangle has sides of length 5, 7, and x. Which of the following can NOT be a value of x?

Answer

The sum of the lengths of any two sides of a triangle must exceed the length of the third side; therefore, 5+7 > x, which cannot happen if x = 13.

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Question

The lengths of two sides of a triangle are 9 and 7. Which of the following could be the length of the third side?

Answer

Let us call the third side x. According to the Triangle Inequality Theorem, the sum of any two sides of a triangle must be larger than the other two sides. Thus, all of the following must be true:

x + 7 > 9

x + 9 > 7

7 + 9 > x

We can solve these three inequalities to determine the possible values of x.

x + 7 > 9

Subtract 7 from both sides.

x > 2

Now, we can look at x + 9 > 7. Subtracting 9 from both sides, we obtain

x > –2

Finally, 7 + 9 > x, which means that 16 > x.

Therefore, x must be greater than 2, greater than –2, but also less than 16. The only number that satisfies all of these requirements is 12.

The answer is 12.

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Question

The lengths of a triangle are 8, 12, and x. Which of the following inequalities shows all of the possible values of x?

Answer

According to the Triangle Inequality Theorem, the sum of any two sides of a triangle must be greater (not greater than or equal) than the remaining side. Thus, the following inequalities must all be true:

x + 8 > 12

x + 12 > 8

8 + 12 > x

Let's solve each inequality.

x + 8 > 12

Subtract 8 from both sides.

x > 4

Next, let's look at the inequality x + 12 > 8

x + 12 > 8

Subtract 12 from both sides.

x > –4

Lastly, 8 + 12 > x, which means that x < 20.

This means that x must be less than twenty, but greater than 4 and greater than –4. Since any number greater than 4 is also greater than –4, we can exclude the inequality x > –4.

To summarize, x must be greater than 4 and less than 20. We can write this as 4 < x < 20.

The answer is 4 < x < 20.

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Question

If 2 sides of the triangle are have lengths equal to 8 and 14, what is one possible length of the third side?

Answer

The sum of the lengths of 2 sides of a triangle must be greater than—but not equal to—the length of the third side. Further, the third side must be longer than the difference between the greater and the lesser of the other two sides; therefore, 20 is the only possible answer.

$4+8< 14$

$6+8= 14$

$8+14> 20$

$8+14=22$

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Question

In $\Delta ABC$ the length of AB is 15 and the length of side AC is 5. What is the least possible integer length of side BC?

Answer

Rule - the length of one side of a triangle must be greater than the differnce and less than the sum of the lengths of the other two sides.

Given lengths of two of the sides of the $\Delta ABC$ are 15 and 5. The length of the third side must be greater than 15-5 or 10 and less than 15+5 or 20.

The question asks what is the least possible integer length of BC, which would be 11

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Question

Given $\bigtriangleup ABC$ with $m \angle A > m \angle B$ and $m \angle C = 60^{\circ }$ .

Which of the following could be the correct ordering of the lengths of the sides of the triangle?

I)

II)

III)

Answer

Given two angles of unequal measure in a triangle, the side opposite the greater angle is longer than the side opposite the other angle. Therefore, we seek to find the relationship among the measures of the angles.

$m \angle C = 60^{\circ }$ , and the measures of the interior angles of a triangle total $180^{\circ }$ , so

$m \angle A + m \angle B + m \angle C = 180 ^{\circ }$

$m \angle A + m \angle B + 60 ^{\circ } = 180 ^{\circ }$

$m \angle A + m \angle B = 120 ^{\circ }$

Since $m \angle A > m \angle B$ ,

$m \angle A + m \angle A > m \angle B+ m \angle A = 120^{\circ }$

$2 m \angle A > 120^{\circ }$

$m \angle A > 60^{\circ }$ ,

and, similarly,

$m \angle B < 60^{\circ }$

Therefore,

$m \angle A > m \angle C > m \angle B$ ,

and the lengths of their opposite sides rank similarly:

.

The correct response is that only (II) can be true.

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Question

Given $\bigtriangleup ABC$ with $m \angle A > m \angle B$ and $m \angle C = 45^{\circ }$ .

Which of the following could be the correct ordering of the lengths of the sides of the triangle?

I)

II)

III)

Answer

Given two angles of unequal measure in a triangle, the side opposite the greater angle is longer than the side opposite the other angle.

If $\angle C$ were the angle of greatest measure, then $m \angle A , m \angle B < 45^{\circ }$ , and

$m \angle A + m \angle B + m \angle C < 45 ^{\circ}+ 45 ^{\circ}+ 45 ^{\circ} = 135 ^{\circ} < 180 ^{\circ}$ .

Since the measures of the angles must total $180 ^{\circ}$ , $\angle C$ cannot have the greatest measure.

$m \angle A > m \angle B$ , so we can explore two other possibilities:

$m \angle A > m \angle B > m \angle C$ , which here is possible if, for example,

$m \angle A = 85 ^{\circ }$ and $m \angle B = 50 ^{\circ }$ - since ; and,

$m \angle A > m \angle C > m \angle B$ , which here is possible if, for example,

$m \angle A = 95 ^{\circ }$ and $m \angle B = 40 ^{\circ }$ - since .

If $m \angle A > m \angle B > m \angle C$ , then ; if $m \angle A > m \angle C > m \angle B$ , then .

This makes the correct response II or III only.

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Question

Which of the given answers choices could NOT represent the sides of a triangle?

Answer

In order for three lengths to represent the sides of a triangle, they must pass the Triangle Inequality.

This means that the sum of any two sides of the triangle must exceed the length of the third side.

With the given answers, the one set of lengths that fail this test is .

$4+5 \ngtr 10$

Therefore, the lengths could not represent the sides of a triangle.

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Question

Find the height of a triangle if the area of the triangle = 18 and the base = 4.

Answer

The area of a triangle = (1/2)bh where b is base and h is height. 18 = (1/2)4h which gives us 36 = 4h so h =9.

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Question

$\text{Triangle A}$ and $\text{Triangle B}$ are similar triangles. The perimeter of Triangle A is 45” and the length of two of its sides are 15” and 10”. If the perimeter of Triangle B is 135” and what are lengths of two of its sides?

Answer

The perimeter is equal to the sum of the three sides. In similar triangles, each side is in proportion to its correlating side. The perimeters are also in equal proportion.

Perimeter A = 45” and perimeter B = 135”

The proportion of Perimeter A to Perimeter B is $\frac{45}{135}=\frac{1}{3}$ .

This applies to the sides of the triangle. Therefore to get the any side of Triangle B, just multiply the correlating side by 3.

15” x 3 = 45”

10” x 3 = 30“

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Question

If triangle ABC has vertices (0, 0), (6, 0), and (2, 3) in the xy-plane, what is the area of ABC?

Answer

Sketching ABC in the xy-plane, as pictured here, we see that it has base 6 and height 3. Since the formula for the area of a triangle is 1/2 * base * height, the area of ABC is 1/2 * 6 * 3 = 9.

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Question

The height, , of triangle in the figure is one-fourth the length of . In terms of h, what is the area of triangle ?

Answer

If $\dpi{100} \small h=\frac{1}{4}$ * $\dpi{100} \small \overline{PQ}$ , then the length of $\dpi{100} \small \overline{PQ}$ must be $\dpi{100} \small 4h$ .

Using the formula for the area of a triangle ( $\frac{1}{2}bh$ ), with $\dpi{100} \small b=4h$ , the area of the triangle must be $2h^{2}$ .

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Question

$\text{Triangle A}$ and $\text{Triangle B}$ are similar triangles. The perimeter of Triangle A is 45” and the length of two of its sides are 15” and 10”. If the perimeter of Triangle B is 135” and what are lengths of two of its sides?

Answer

The perimeter is equal to the sum of the three sides. In similar triangles, each side is in proportion to its correlating side. The perimeters are also in equal proportion.

Perimeter A = 45” and perimeter B = 135”

The proportion of Perimeter A to Perimeter B is $\frac{45}{135}=\frac{1}{3}$ .

This applies to the sides of the triangle. Therefore to get the any side of Triangle B, just multiply the correlating side by 3.

15” x 3 = 45”

10” x 3 = 30“

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Question

If triangle ABC has vertices (0, 0), (6, 0), and (2, 3) in the xy-plane, what is the area of ABC?

Answer

Sketching ABC in the xy-plane, as pictured here, we see that it has base 6 and height 3. Since the formula for the area of a triangle is 1/2 * base * height, the area of ABC is 1/2 * 6 * 3 = 9.

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Question

The height, , of triangle in the figure is one-fourth the length of . In terms of h, what is the area of triangle ?

Answer

If $\dpi{100} \small h=\frac{1}{4}$ * $\dpi{100} \small \overline{PQ}$ , then the length of $\dpi{100} \small \overline{PQ}$ must be $\dpi{100} \small 4h$ .

Using the formula for the area of a triangle ( $\frac{1}{2}bh$ ), with $\dpi{100} \small b=4h$ , the area of the triangle must be $2h^{2}$ .

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Question

If the average of the measures of two angles in a triangle is 75o, what is the measure of the third angle in this triangle?

Answer

The sum of the angles in a triangle is 180o: a + b + c = 180

In this case, the average of a and b is 75:

(a + b)/2 = 75, then multiply both sides by 2

(a + b) = 150, then substitute into first equation

150 + c = 180

c = 30

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