Measurement and Geometry - HiSET

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ such that $\frac{AB}{DE} = \frac{BC}{EF}$ .

Does sufficient information exist to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , and if so, by what postulate or theorem?

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Answer

We are given that, between the triangles, two pairs of corresponding sides are proportional. Without knowing anything else, the proportionality of two pairs of sides is insufficient to prove that the triangles are similar.

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ such that $\frac{AB}{DE} = \frac{BC}{EF}$ and $\angle B \cong \angle E$ .

Does sufficient information exist to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , and if so, by what postulate or theorem?

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Answer

We are given that, between the triangles, two pairs of corresponding sides are proportional, and that a pair of corresponding angles are congruent. The angles that are congruent are the included angles of their respective sides. By the SAS Similarity Postulate, this is enough to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ such that $\angle C \cong \angle F$ and $\angle A \cong \angle D$ .

Does sufficient information exist to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , and if so, by what postulate or theorem?

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Answer

We are given that, between the triangles, two pairs of corresponding angles are congruent. By the AA Similarity Postulate, this is enough to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ such that $\frac{AB}{DE} = \frac{BC}{EF}$ and $\angle C \cong \angle F$ .

Does sufficient information exist to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , and if so, by what postulate or theorem?

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Answer

We are given that, between the triangles, two pairs of corresponding sides are proportional, and that a pair of corresponding angles are congruent. If the angles were the included angles of the triangles, then the SAS Similarity Theorem could be applied to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ ; however, the two congruent angles are nonincluded, and there is no "SSA" statement that can be applied to prove similarity. Without further information, it cannot be proved that the triangles are similar.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ with scale factor 4:5, with $\bigtriangleup ABC$ the smaller triangle.

Complete the sentence: the perimeter of $\bigtriangleup ABC$ is % less than that of $\bigtriangleup DEF$ .

(Select the closest whole percent)

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Answer

The ratio of the perimeters of two similar triangles is equal to their scale factor. This is factor is $\frac{4}{5}$ .

This makes the perimeter of the smaller triangle $\bigtriangleup ABC$ equal to $\frac{4}{5} \times 100 % = 80 %$ of that of larger triangle $\bigtriangleup DEF$ —or, equivalently, less.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ with scale factor 5:4, with $\bigtriangleup ABC$ the larger triangle.

Complete the sentence: the perimeter of $\bigtriangleup ABC$ is % greater than that of $\bigtriangleup DEF$ .

(Select the closest whole percent).

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Answer

The ratio of the perimeters of two similar triangles is equal to their scale factor. This is factor is $\frac{5}{4}$ .

This makes the perimeter of larger triangle $\bigtriangleup ABC$ equal to $\frac{5}{4} \times 100 % = 125 %$ of that of smaller triangle $\bigtriangleup DEF$ —or, equivalently, 25% greater.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ with scale factor 4:5, with $\bigtriangleup ABC$ the smaller triangle.

Complete the sentence: the area of $\bigtriangleup ABC$ is % less than that of $\bigtriangleup DEF$ .

(Select the closest whole percent).

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Answer

The ratio of the areas of two similar triangles is equal to the square of their scale factor. The scale factor is equal to $\frac{4} {5}$ , so the ratio of the areas is the square of this, or $\left ( \frac{4}{5} \right ) ^{2} = \frac{16}{25}$ .

This makes the area of smaller triangle $\bigtriangleup ABC$ equal to $\frac{16}{25} \times 100 % = 64 %$ of that of larger triangle $\bigtriangleup DEF$ —or, equivalently, less.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ with scale factor 5:4, with $\bigtriangleup ABC$ the larger triangle.

Complete the sentence: the area of $\bigtriangleup ABC$ is % greater than that of $\bigtriangleup DEF$ .

(Select the closest whole percent)

Tap to reveal answer

Answer

The ratio of the areas of two similar triangles is equal to the square of their scale factor. The scale factor is equal to $\frac{5}{4}$ , so the ratio of the areas is the square of this, or $\left ( \frac{5}{4} \right ) ^{2} = \frac{25}{16}$ .

This makes the area of larger triangle $\bigtriangleup ABC$ equal to $\frac{25}{16} \times 100 % = 156 \frac{1}{4} %$ of that of smaller triangle $\bigtriangleup DEF$ —or, equivalently, $56 \frac{1}{4} %$ greater.

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Question

A triangle has sides of length 8 and 12; the triangle is scalene and obtuse. Which of the following could be the length of its third side?

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Answer

A scalene triangle has three sides of different lengths. The triangle is known to have sides of length 8 and 12, so this eliminates 8 and 12 as the correct choices for the length of the third side.

The sum of the lengths of the two smallest sides must exceed the length of the third side. 4 can be eliminated as a the correct choice, since , violating this condition.

This leaves 6 and 10 as possible answers. For a triangle to be obtuse, it must hold that if are its sidelengths, the greatest of the three,

$a^{2}+b^{2} < c^{2}$ .

If the length of the third side is 10, setting , we see that

$10^{2}+ 8^{2}= 100 + 64 = 164 > 144 = 12^{2}$ ,

violating this condition.

If the length of the third side is 6, setting , we see that

$6^{2}+ 8^{2}= 36 + 64 = 100 < 144 = 12^{2}$ ,

satisfying this condition.

This makes 6 the correct choice.

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Question

Two of the angles of a triangle are congruent; the third has measure ten degrees greater than either one of the first two. What is the measure of the third angle?

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Answer

Let be the measure of the third angle. Since its measure is ten degrees greater than either of the others, then the common measure of the other two is . The sum of the measures of the angles of a triangle is 180 degrees, so

To solve for , first ungroup and collect like terms:

Isolate ; first add 20:

Divide by 3:

$3x \div 3 = 200 \div 3$

Since $200 \div 3 = 66 \textup{ R }2$ ,

$x = 66 \frac{2}{3}$ .

The third angle measures $66 \frac{2}{3} ^{\circ }$ .

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Question

A quadrilateral is shown, and the angle measures of 3 interior angles are given. Find x, the missing angle measure.

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Answer

The sum of the measures of the interior angles of a quadrilateral is 360 degrees. The sum of the measures of the interior angles of any polygon can be determined using the following formula:

$Sum \ of\ Interior \ Angles = 180^{\circ}(n-2)$ , where is the number of sides.

For example, with a quadrilateral, which has 4 sides, you obtain the following calculation:

$Sum \ of\ Interior \ Angles = 180^{\circ}(4-2)=180^{\circ}\cdot2=360^{\circ}$

Solving for requires setting up an algebraic equation, adding all 4 angles to equal 360 degrees:

$50^{\circ}+67^{\circ}+115^{\circ}+x^{\circ}=360^{\circ}$

Solving for is straightforward: subtract the values of the 3 known angles from both sides:

$x^{\circ}= 360^{\circ}-50^{\circ}-67^{\circ}-115^{\circ}$

$x^{\circ}=128^{\circ}$

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Question

is the center of the above circle. Calculate $m \angle AOB$ .

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Answer

$\angle AOB$ is the central angle that intercepts $\overarc{AB}$ , so

$m \angle AOB = m \overarc{AB}$ .

Therefore, we need to find $m \overarc{AB}$ to obtain our answer.

If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{(360- x )- x }{2} =45$

$\frac{360- 2 x }{2} =45$

Multiply both sides by 2:

$\frac{360- 2 x }{2} \cdot 2 =45 \cdot 2$

Subtract 360 from both sides:

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

,

the measure of $\overarc{AB}$ and, consequently, that of $\angle AOB$ .

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Question

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle DBO$ .

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Answer

Consider the figure below, which adds some radii of the heptagon (and circle):

$\overline{OB}$ , as a radius of a regular polygon, bisects $\angle EBD$ . The measure of this angle can be calculated using the formula

$m = \frac{180 (n-2)}{n}$ ,

where :

$m \angle EBD = \frac{180^{\circ } (7-2)}{7} = \frac{180^{\circ } (5)}{7} = \frac{900}{7}^{\circ } = 1 28 \frac{4}{7}^{\circ }$

Consequently,

$\angle DBO = \frac{1}{2} \cdot m \angle EBD = \frac{1}{2} \cdot 1 28 \frac{4}{7}^{\circ } = 64 \frac{2}{7}^{\circ }$ ,

the correct response.

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Question

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOB$ .

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Answer

Examine the diagram below, which divides $\angle AOB$ into three congruent angles, one of which is $\angle 1$ :

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$ . $\angle AOC$ has measure three times this; that is,

$m \angle AOC = 3 \times51 \frac{3}{7}^{\circ } = 154 \frac{2}{7}^{\circ }$

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Question

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOC$ .

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Answer

Examine the diagram below, which divides $\angle AOC$ into two congruent angles, one of which is $\angle 1$ :

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$ . $\angle AOC$ has measure twice this; that is,

$m \angle AOC = 2 \times51 \frac{3}{7}^{\circ } = 102 \frac{6}{7}^{\circ }$

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Question

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOB$ .

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Answer

Examine the diagram below, which divides $\angle AOB$ into two congruent angles, one of which is $\angle 1$ :

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$ . $\angle AOB$ has measure twice this; that is,

$m \angle AOB = 2 \times 36 ^{\circ } = 72 ^{\circ }$ .

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Question

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle OBC$ .

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Answer

Consider the triangle $\bigtriangleup OBC$ . Since $\overline{OB}$ and $\overline{OC}$ are radii, they are congruent, and by the Isosceles Triangle Theorem, $m \angle OBC = m \angle OCB$ .

Now, examine the figure below, which divides $\angle BOC$ into three congruent angles, one of which is $\angle 1$ :

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$ . $\angle BOC$ has measure three times this; that is,

$m \angle BOC = 3 \times 36 ^{\circ } = 108 ^{\circ }$ .

The measures of the interior angles of a triangle total $108^{\circ }$ , so

$m \angle OCB + m \angle OBC+m \angle BOC = 180$

Substituting 108 for $\angle BOC$ and $m \angle OBC$ for $m \angle OCB$ :

$m \angle OBC+ m \angle OBC+108 = 180$

$2 m \angle OBC+108 = 180$

$2 m \angle OBC+108- 108 = 180- 108$

$2 m \angle OBC=72$

$2 m \angle OBC \div 2 =72 \div 2$

$m \angle OBC= 36$

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Question

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle CBE$ .

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Answer

Through symmetry, it can be seen that Quadrilateral is a trapezoid, such that $\overline{BC} || \overline{DE }$ . By the Same-Side Interior Angle Theorem, $\angle CBE$ and $\angle DEB$ are supplementary - that is,

$m \angle CBE +\angle DEB = 180$ .

The measure of $\angle DEB$ can be calculated using the formula

$m = \frac{180 (n-2)}{n}$ ,

where :

$m \angle DEB= \frac{180^{\circ } (10-2)}{10} = \frac{180^{\circ } (8)}{10} = \frac{1,440}{10}^{\circ } =144 ^{\circ }$

Substituting:

$m \angle CBE +144= 180$

$m \angle CBE +144 - 144 = 180 - 144$

$m \angle CBE = 36$

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Question

If two angles are supplementary and one angle measures $95^{\circ}$ , what is the measurement of the second angle?

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Answer

Step 1: Define supplementary angles. Supplementary angles are two angles whose sum is $180^{\circ}$ .

Step 2: Find the other angle by subtracting the given angle from the maximum sum of the two angles.

So, $180^{\circ}-95^{\circ}=85^\circ$

The missing angle (or second angle) is $85^{\circ}$

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Question

$\angle 1$ and $\angle 2$ are complementary angles.

$\angle 1$ and $\angle 3$ are supplementary angles.

$\angle 2 \cong \angle 4$

$m \angle 3= 112^{\circ }$

Evaluate $m \angle 4$ .

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Answer

$\angle 1$ and $\angle 3$ are supplementary angles, so, by definition,

$m \angle 1 + m \angle 3 = 180^{\circ }$

$m \angle 3= 112^{\circ }$ , so substitute and solve for $m \angle 1$ :

$m \angle 1 + 112^{\circ } = 180^{\circ }$

$m \angle 1 + 112^{\circ } -112^{\circ } = 180^{\circ }-112^{\circ }$

$m \angle 1 = 68 ^{\circ }$

$\angle 1$ and $\angle 2$ are complementary angles, so, by definition,

$m \angle 1 + m \angle 2 = 90^{\circ }$

Substitute and solve for $m \angle 2$ :

$68 ^{\circ } + m \angle 2 = 90^{\circ }$

$68 ^{\circ } + m \angle 2 - 68 ^{\circ } = 90^{\circ }-68 ^{\circ }$

$m \angle 2 =22^{\circ }$

$\angle 2 \cong \angle 4$ - that is, the angles have the same measure. Therefore,

$m \angle 4 =22^{\circ }$ .

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