Chapter 10

The letters BD, with an arrow pointing from B towards D, represent a ray that starts at B and extends infinitely in the direction of point D.

$\stackrel{\to <!--\; \to \; -->}{BD}$ is the ray that starts at point $B$ and extends infinitely in the direction of point $D$.

The letters AB, with a straight line above, represent a line segment that starts at point A and ends at point B.

$\stackrel{¯<!--\; ¯\; -->}{AB}$ represents the line segment that starts at point $A$ and ends at point $B$.

The letters BA, with an arrow pointing from A towards B, represent a ray that starts at A and extends infinitely in the direction of point B.

$\stackrel{\leftarrow <!--\; \leftarrow \; -->}{BA}$ represents the ray that starts at point $A$ and extends infinitely in the direction of point $B$.

The letters AD, with a double arrow above, represent a line that contains the points A and D. The line extends infinitely in both directions.

$\stackrel{\leftrightarrow <!--\; \leftrightarrow \; -->}{AD}$ represents a line that contains the points $A$ and $D$. Notice the arrowheads on both ends of the line above $\mathit{A}\mathit{D}$, which means that the line continues infinitely in both directions.

You are looking for the intersection of line segment $\stackrel{¯<!--\; ¯\; -->}{AB}$ and line segment $\stackrel{¯<!--\; ¯\; -->}{BC}$.

Sketch the two line segments separately above the line.

Intersection includes only the elements that are common to both drawings.

For this intersection, only point B is in common.

$\stackrel{¯<!--\; ¯\; -->}{AB}\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{BC}=B$

$\stackrel{¯<!--\; ¯\; -->}{AB}\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{BC}=B$

You are looking for the union of ray $\stackrel{\to <!--\; \to \; -->}{BC}$ and the ray $\stackrel{\leftarrow <!--\; \leftarrow \; -->}{CA}$.

Be careful. They listed the points of the second ray looking backwards!

Sketch the two rays separately above the line.

Union includes all the elements that are in either drawing.

For this union, the rays cover the entire line.

$\stackrel{\to <!--\; \to \; -->}{BC}\cup <!--\; \cup \; -->\stackrel{\leftarrow <!--\; \leftarrow \; -->}{CA}=\stackrel{\leftrightarrow <!--\; \leftrightarrow \; -->}{AC}$

$\stackrel{\to <!--\; \to \; -->}{BC}\cup <!--\; \cup \; -->\stackrel{\leftarrow <!--\; \leftarrow \; -->}{CA}=\stackrel{\leftrightarrow <!--\; \leftrightarrow \; -->}{AC}$

You are looking for the intersection of line segment $\stackrel{¯<!--\; ¯\; -->}{BC}$ and line segment $\stackrel{¯<!--\; ¯\; -->}{AC}$.

Sketch the two line segments separately above the line.

Intersection includes only the elements that are common to both drawings.

For this intersection, they have the line segment $\stackrel{¯<!--\; ¯\; -->}{BC}$ in common.

$\stackrel{¯<!--\; ¯\; -->}{BC}\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{AC}=\stackrel{¯<!--\; ¯\; -->}{BC}$

$\stackrel{¯<!--\; ¯\; -->}{BC}\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{AC}=\stackrel{¯<!--\; ¯\; -->}{BC}$

Acute angles have a measure between 0° and 90°, not including 90°.

Acute angles: ∠AOB, ∠AOC, ∠BOC, ∠BOD, ∠COD, ∠DOE, ∠FOE

Obtuse angles have a measure between 90° and 180°, not inclusively.

Obtuse angles: ∠AOE, ∠BOF, ∠COF

Right angles measure exactly 90°.

Right angles: ∠AOD, ∠BOE, ∠DOF

Straight angles measure exactly 180°.

Straight angles: ∠AOF

Straight angles measure exactly 180°.

$5x+2x+5=180$

$7x=175$

$x={25}^{\circ <!--\; \circ \; -->}$

$2x+5$ = $2\left(25\right)+5=50+5={55}^{\circ <!--\; \circ \; -->}$

$5x=5\left(25\right)={125}^{\circ <!--\; \circ \; -->}$

The angles measure 125° and 55°.

Right angles measure exactly 90°.

$6x+9x+7x+2=90$

$22x=88$

$x=4$

$6x=6\left(4\right)={24}^{\circ <!--\; \circ \; -->}$

$9x=\left(4\right)={36}^{\circ <!--\; \circ \; -->}$

$7x+2=7\left(4\right)+2=28+2={30}^{\circ <!--\; \circ \; -->}$

The angles measure 24°, 36°, and 30°.

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}(6x)={24}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}(9x)={36}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}(7x+2)={30}^{\circ <!--\; \circ \; -->}$

Vertical angles have equal measures. The measure of angle 4 is 67°.

Angles 3 and 4 are supplementary, so sum of the measures is180°.

The measure of angle 3 = 180° – 67° = 113°

Since vertical angles have equal measures, angles 3 and 1 have the same measure.

m∡3 = m∡1 = 113°

m∡4 = 67°

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}4={67}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}1={113}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}3={113}^{\circ <!--\; \circ \; -->}$

The two angles form a straight angle. To find the supplementary angle, subtract it from 180.

$180-<!--\; -\; -->50=130$

The angle measures 130°.

${130}^{\circ <!--\; \circ \; -->}$

Angle 1 and angle 5 are corresponding. Corresponding angles have equal measures.

m∡5 = 120°

Angle 1 and angle 4 are vertical angles. Vertical angles have equal measures.

m∡4 = 120°

Angle 5 and angle 8 are vertical angles. Vertical angles have equal measures.

m∡8 = 120°

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}5={120}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}4={120}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}8={120}^{\circ <!--\; \circ \; -->}$

Angle 1 and angle 2 are supplementary. The sum of the measures of supplementary angles is 180°

180° – 48° = 132°

m∡1 = 132°

Angle 1 and angle 5 are corresponding. Corresponding angles have equal measures.

m∡5 = 132°

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}1={132}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}5={132}^{\circ <!--\; \circ \; -->}$

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}x={66}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}(x+1)={67}^{\circ <!--\; \circ \; -->}$, $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}(x-<!--\; -\; -->19)={47}^{\circ <!--\; \circ \; -->}$

The sum of the angle measures of a triangle must be 180°.

The angles measure 47°, 67°, and 66°.

Angle 1 and the 145° angle are supplementary. The sum of the measures of supplementary angles is 180°.

180° – 145° = 35°

m∡1 = 35°

Angle 3 and the 60° are vertical angles. Vertical angles have equal measures.

m∡3 = 60°

The sum of the angle measures of a triangle must be 180°. The sum of measures of angles 1, 2, and 3 must be 180°. Subtract the sum of 35 and 60 from 180 to get the measure of angle 2.

180° – (35° + 60°) = 85°

m∡2 = 85°

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}1={35}^{\circ <!--\; \circ \; -->}$; $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}2=2={85}^{\circ <!--\; \circ \; -->}$; $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}3={60}^{\circ <!--\; \circ \; -->}$

The sum of the angle measures of a triangle must be 180°. Since two of the angles in triangle ABC have equal measures to two of the angles in triangle DEF, you know the third pair of angles are also of equal measure. The sides are also equal. These are congruent triangles.

Triangle $ABC$ is congruent to triangle $DEF$.

Two sides are marked as congruent. The angle between the two sides is marked as congruent. This is the Side-Angle-Side (SAS) congruence theorem. If two sides of a triangle and the angle between them are equal to the corresponding two sides and included angle of the second triangle, then the triangles are congruent.

SAS

Two angles are marked as congruent as well as the side between the two angles. This is the Angle-Side-Angle (ASA) congruence theorem. If two angles and the side between them in one triangle are congruent to the two corresponding angles and the side between them in a second triangle, then the two triangles are congruent.

ASA

Step 1: You know the corresponding angles have equal measures from the drawings. That is enough to determine similarity.

Step 2: Set up proportions between the triangles to see if they have the same scaling factor. If they are the same, the triangles are similar.

$\frac{3.7}{4.9}\approx <!--\; \approx \; -->0.76$

$\frac{5}{6.6}\approx <!--\; \approx \; -->0.76$

$\frac{3.4}{4.5}\approx <!--\; \approx \; -->0.76$

The scale factor is approximately the same. The triangles are similar.

The triangles are similar.

Step 1: You know the corresponding angles have equal measures from the drawings. That is enough to determine similarity.

Step 2: Set up proportions to find the missing values.

$x=6$, $y=24$

Set up a proportion to find the missing value. It helps some students to draw two separate triangles.

The tree is 86 feet high.

Set up a proportion to find the missing value. It helps some students to draw two separate triangles. Let x be the height of the tree in feet.

60 ft

A rectangle is a parallelogram with four right angles and two sets of parallel sides.

rectangle

A pentagon is a polygon with five sides.

pentagon

A heptagon is a polygon with seven sides.

heptagon

A parallelogram is a quadrilateral with two sets of parallel sides.

parallelogram

Shapes 1, 2, 4, and 6 are triangles since they have three sides.

Shape 3 is a pentagon since it has five sides.

Shape 5 is a parallelogram. A parallelogram is a quadrilateral with two sets of parallel sides.

Shape 7 is a rectangle. A rectangle is a parallelogram with four right angles and two sets of parallel sides.

Shapes 1, 2, 4, and 6 are triangles; shape 3 is a pentagon; shape 5 is a parallelogram; and shape 7 is a rectangle.

The formula for the perimeter P for a rectangle with length L and width W is $P=2L+2W$.

A square is a rectangle whose length and width have equal measures.

$P=2L+2W=2\left(30\mathrm{i}\mathrm{n}\right)+2\left(30\mathrm{i}\mathrm{n}\right)=60\mathrm{i}\mathrm{n}+60\mathrm{i}\mathrm{n}=120\mathrm{i}\mathrm{n}$

The perimeter is 120 inches.

120 in

A regular heptagon has seven sides of equal length. Therefore, the perimeter of a regular heptagon with a side length of 3.2 inches is:

$P=7(3.2\mathrm{i}\mathrm{n})=22.4\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}$

22.4 in

A pentagon has five sides, so $n=5$.

The measure of each interior angle of a regular polygon with n sides is given by $a=\frac{\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{n}$.

$a=\frac{\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{n}=\frac{\left(5-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{5}=\frac{\left(3\right){180}^{\circ <!--\; \circ \; -->}}{5}=\frac{540}{5}={108}^{\circ <!--\; \circ \; -->}$

The sum of the interior angles of a polygon with n sides is $S=\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}$.

$S=\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}=\left(5-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}=\left(3\right){180}^{\circ <!--\; \circ \; -->}={540}^{\circ <!--\; \circ \; -->}$

Each interior angle of a pentagon is 108°. The sum of the interior angles of a pentagon is 540°.

$a={108}^{\circ <!--\; \circ \; -->}$
The sum of the interior angles is ${540}^{\circ <!--\; \circ \; -->}$.

Step 1: Find the sum of the interior angles. This is a quadrilateral, so $n=4$.

The sum of the interior angles of a polygon with n sides is $S=\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}$.

$S=\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}=\left(4-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}=\left(2\right){180}^{\circ <!--\; \circ \; -->}={360}^{\circ <!--\; \circ \; -->}$

Step 2: Add up the angles and set it equal to the sum of the angles. Then solve for the unknowns.

$-<!--\; -\; -->\left(5x+1\right)=-<!--\; -\; -->\left(5\left(-<!--\; -\; -->9\right)+1\right)=-<!--\; -\; -->\left(-<!--\; -\; -->45+1\right)=-<!--\; -\; -->\left(-<!--\; -\; -->44\right)={44}^{\circ <!--\; \circ \; -->}$

$-<!--\; -\; -->6x=-<!--\; -\; -->6\left(-<!--\; -\; -->9\right)={54}^{\circ <!--\; \circ \; -->}$

$120+x=120+\left(-<!--\; -\; -->9\right)={111}^{\circ <!--\; \circ \; -->}$

The interior angles measure 44°, 54°, and 111°.

We have the sum of interior angles is ${360}^{\circ <!--\; \circ \; -->}$. Then, $x=-<!--\; -\; -->9.$ The other angles measure ${54}^{\circ <!--\; \circ \; -->}$, ${111}^{\circ <!--\; \circ \; -->}$, ${44}^{\circ <!--\; \circ \; -->}$.

The sum of the measures of the exterior angles of a regular polygon with n sides equals 360°.

You could stop there. The measure of each of a regular polygon with n sides is $b=\frac{{360}^{\circ <!--\; \circ \; -->}}{n}$.

A heptagon has seven sides, so n = 7.

$b=\frac{{360}^{\circ <!--\; \circ \; -->}}{n}=\frac{{360}^{\circ <!--\; \circ \; -->}}{7}\approx <!--\; \approx \; -->{51.43}^{\circ <!--\; \circ \; -->}$

To get back to the sum of all the exterior angles, multiply by 7.

7(51.43) is approximately 360°.

$7(51.43)={360}^{\circ <!--\; \circ \; -->}$

The circumference of a circle is $C=2\mathrm{\pi <!--\; \pi \; -->}r$ where r is the radius.

$C=2\mathrm{\pi <!--\; \pi \; -->}r=2\mathrm{\pi <!--\; \pi \; -->}\left(2.25\mathrm{c}\mathrm{m}\right)=4.5\mathrm{c}\mathrm{m}\approx <!--\; \approx \; -->14.14\mathrm{c}\mathrm{m}$

$C=2(2.25)\mathrm{\pi <!--\; \pi \; -->}=14.14\text{ cm}$

The circumference of a circle is $C=2\mathrm{\pi <!--\; \pi \; -->}r$ where r is the radius.

$r=2.5\text{cm}$

The trim will cover 2 feet along the bottom as well as the 5.3-foot sides, plus the semicircle at the top. The semicircle has a diameter of 2 feet.

The circumference of a circle with diameter d is $C=\mathrm{\pi <!--\; \pi \; -->}d$.

The distance around the semicircle would be half of that.

$\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{r}\mathrm{c}\mathrm{l}\mathrm{e}=\frac{1}{2}\mathrm{\pi <!--\; \pi \; -->}\left(2\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}\right)\approx <!--\; \approx \; -->3.14\mathrm{f}\mathrm{t}$

The total trim needed: $2\mathrm{f}\mathrm{t}+(5.3\mathrm{f}\mathrm{t})(2)+3.14\mathrm{f}\mathrm{t}=15.74\mathrm{f}\mathrm{t}$

You need 15.74 feet of trim.

You need to buy 15.74 feet of trim.

For regular polygons to be able to tesselate, the shapes must meet at a vertex so that the interior angles have a sum of 360°.

A heptagon has seven sides, so $n=7$.

The measure of each interior angle of a regular polygon with n sides is given by $a=\frac{\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{n}$.

$a=\frac{\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{n}=\frac{\left(7-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{7}=\frac{\left(5\right){180}^{\circ <!--\; \circ \; -->}}{7}=\frac{900}{7}\approx <!--\; \approx \; -->{127.57}^{\circ <!--\; \circ \; -->}$

Since the interior angles of a heptagon are not integer factors of 360°, it is impossible to tesselate with heptagons by themselves.

No

For regular polygons to be able to tesselate, the shapes must meet at a vertex so that the interior angles have a sum of 360°.

A dodecagons has 12 sides, so $n=12$.

The measure of each interior angle of a regular polygon with n sides is given by $a=\frac{\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{n}$.

$a=\frac{\left(n-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{n}=\frac{\left(12-<!--\; -\; -->2\right){180}^{\circ <!--\; \circ \; -->}}{12}=\frac{\left(10\right){180}^{\circ <!--\; \circ \; -->}}{12}=\frac{1,{080}^{\circ <!--\; \circ \; -->}}{12}={90}^{\circ <!--\; \circ \; -->}$

Since the interior angles of a dodecagon are integer factors of 360°, it makes tessellations within reach. However, dodecagons by themselves cannot tesselate. If you add equilateral triangles between them, they can fill the gaps.

Not by themselves, but by adding an equilateral triangle, the two regular polygons do tessellate the plane without gaps.

$8{\text{cm}}^2$

The area of a triangle is $A=\frac{1}{2}bh$, where b represents the base and h represents the height.

$A=\frac{1}{2}bh=\frac{1}{2}\left(4\mathrm{c}\mathrm{m}\right)\left(4\mathrm{c}\mathrm{m}\right)=8\mathrm{c}{\mathrm{m}}^2$

The area is 8 square centimeters.

The area of a rectangle is $A=LW$where L represents the length and W represents the width.

The length is 18 feet.

Break the width expression into its components.

width = $\frac{1}{3}\left(18\mathrm{f}\mathrm{t}\right)=6\mathrm{f}\mathrm{t}$

$A=LW=\left(18\mathrm{f}\mathrm{t}\right)\left(6\mathrm{f}\mathrm{t}\right)=108\mathrm{f}{\mathrm{t}}^2$

The area is 108 square feet.

108 ft²

45 boxes at a cost of $2,025.00

The area of a rectangle is $A=LW$where L represents the length and W represents the width.

$A=LW=\left(30\mathrm{f}\mathrm{t}\right)\left(15\mathrm{f}\mathrm{t}\right)=450\mathrm{f}{\mathrm{t}}^2$

Make conversion factors out of the equivalences of one box costing $45 and covering 10 square feet to answer the questions.

$450\cancel{\mathrm{f}{\mathrm{t}}^2}\left(\frac{1\mathrm{b}\mathrm{o}\mathrm{x}}{10\cancel{\mathrm{f}{\mathrm{t}}^2}}\right)=45\mathrm{b}\mathrm{o}\mathrm{x}\mathrm{e}\mathrm{s}$

$45\cancel{\mathrm{b}\mathrm{o}\mathrm{x}\mathrm{e}\mathrm{s}}\left(\frac{\mathrm{\$<!--\; \$\; -->}45}{10\cancel{\mathrm{b}\mathrm{o}\mathrm{x}\mathrm{e}\mathrm{s}}}\right)=\mathrm{\$<!--\; \$\; -->}2,025$

You need 45 boxes of tiles. It will cost you $2,025.

The area of a parallelogram with base b and height h is $A=bh$.

$A=bh=\left(15\mathrm{i}\mathrm{n}\right)\left(18\mathrm{i}\mathrm{n}\right)=270\mathrm{i}{\mathrm{n}}^2$

The area of the parallelogram is 270 square inches.

$270\text{i}{\text{n}}^2$

13,671 square feet; cost is $19,872.95.

First, convert the dimensions to feet. There are 3 feet in one yard.

$31\cancel{\mathrm{y}\mathrm{a}\mathrm{r}\mathrm{d}}\left(\frac{3\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}}{1\cancel{\mathrm{y}\mathrm{a}\mathrm{r}\mathrm{d}}}\right)=93\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}$

$49\cancel{\mathrm{y}\mathrm{a}\mathrm{r}\mathrm{d}}\left(\frac{3\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}}{1\cancel{\mathrm{y}\mathrm{a}\mathrm{r}\mathrm{d}}}\right)=147\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}$

The area of a parallelogram with base b and height h is $A=bh$.

$A=bh=\left(93\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}\right)\left(147\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}\right)=13,671\mathrm{f}{\mathrm{t}}^2$

There are two charges per square foot: $0.45 and $1.00, for a total of $1.45 per square foot.

$13,671\cancel{\text{f}{\text{t}}^2}\left(\frac{\mathrm{\$<!--\; \$\; -->}1.45}{1\cancel{\text{f}{\text{t}}^2}}\right)=\mathrm{\$<!--\; \$\; -->}19,822$

Add the $50 flat fee for equipment.

$\text{\$19,822 + \$50 = \$19,872}\text{.95}$

You will buy 13,671 square feet of sod. The total cost is $19,872.95.

The area of a trapezoid with height h and parallel bases a and b is $A=\frac{1}{2}h(a+b)$.

$A=\frac{1}{2}h(a+b)=A=\frac{1}{2}\left(12\mathrm{i}\mathrm{n}\right)(6\mathrm{i}\mathrm{n}+16\mathrm{i}\mathrm{n})=\left(6\mathrm{i}\mathrm{n}\right)\left(22\mathrm{i}\mathrm{n}\right)=132\mathrm{i}{\mathrm{n}}^2$

The area is 132 square inches.

$132{\text{in}}^2$

The area of a rhombus with diagonals $d_1$ and $d_2$ is $A=\frac{d_1{d}_2}{2}$.

The measure of $d_2$ is 10 inches.

$d_2=10$

A kite is a rhombus. The area of a rhombus with diagonals $d_1$ and $d_2$ is $A=\frac{d_1{d}_2}{2}$.

The measure of $d_2$ is 40 inches.

40 in

The apothem of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The area of a regular polygon with apothem a and perimeter p is $A=\frac{1}{2}ap$.

A regular pentagon has five sides of equal length. Therefore, the perimeter of a regular pentagon with a side length of 7 cm is:

$p=5(7\mathrm{c}\mathrm{m})=35\mathrm{c}\mathrm{m}$

$A=\frac{1}{2}ap=\frac{1}{2}\left(5.5\mathrm{c}\mathrm{m}\right)\left(35\mathrm{c}\mathrm{m}\right)=96.25\mathrm{c}{\mathrm{m}}^2$

The area is 96.25 square centimeters.

$A=96.25\text{c}{\text{m}}^2$

First, convert the dimensions to yards. There are 3 feet in one yard.

$15\cancel{\text{feet}}\left(\frac{\text{1 yard}}{\text{3 }\cancel{\text{feet}}}\right)=5\text{ yards}$

$18\cancel{\text{feet}}\left(\frac{\text{1 yard}}{\text{3 }\cancel{\text{feet}}}\right)=6\text{ yards}$

The area of a rectangle is $A=LW$where L represents the length and W represents the width.

$A=LW=\left(5\text{ yd}\right)\left(\text{6 yd}\right)=30\text{ y}{\text{d}}^2$

You need 30 square yards of carpet.

$30\text{y}{\text{d}}^2$

The area of a circle with radius r is $A=\mathrm{\pi <!--\; \pi \; -->}{r}^2$.

$A=\mathrm{\pi <!--\; \pi \; -->}{r}^2=\mathrm{\pi <!--\; \pi \; -->}{\left(3\text{ cm}\right)}^2=9\mathrm{\pi <!--\; \pi \; -->}\text{ c}{\text{m}}^2\approx <!--\; \approx \; -->28.3\text{ c}{\text{m}}^2$

The area of the circle is approximately 28.3 square centimeters.

$28.3\text{c}{\text{m}}^2$

Find the cost of one square inch of pizza for both pizzas. To do that, divide the cost of each pizza by the area of each pizza. The area of a circle with radius r is $A=\mathrm{\pi <!--\; \pi \; -->}{r}^2$.

If the diameter is 9 inches, the radius is 4.5 inches.

If the diameter is 15 inches, the radius is 7.5 inches.

The area of a circle with radius r is $A=\mathrm{\pi <!--\; \pi \; -->}{r}^2$.

The 15-inch pizza is cheaper at roughly $0.08 per square inch compared to the smaller pizza at roughly $0.17 per square inch.

the 15-inch pizza

The door is a rectangle plus a semicircle.

The rectangular area is:

$A=LW=\left(\text{3}\text{.5 ft}\right)\left(\text{6}\text{.25 ft}\right)=21.875\text{ f}{\text{t}}^2$

The area of a circle with radius r is $A=\mathrm{\pi <!--\; \pi \; -->}{r}^2$, but you want half of that.

$\frac{1}{2}A=\frac{1}{2}\mathrm{\pi <!--\; \pi \; -->}{r}^2=\frac{1}{2}\mathrm{\pi <!--\; \pi \; -->}{\left(\frac{3.5}{2}\text{ ft}\right)}^2\approx <!--\; \approx \; -->4.8105\text{ f}{\text{t}}^2$

Add the two results: $\text{(21}\text{.875 + 4}\text{.8105)}$ square feet

The total area is approximately 26.7 square feet.

$A=21.875+4.8=26.7\text{f}{\text{t}}^2$

Step 1: Find the area of the square.

The area of a square with side s is $A=s^2$

$A=s^2={\left(10\text{ cm}\right)}^2=100\text{ c}{\text{m}}^2$

Step 2: Find the area of the circle.

The area of a circle with radius r is $A=\mathrm{\pi <!--\; \pi \; -->}{r}^2$.

$A=\mathrm{\pi <!--\; \pi \; -->}{r}^2=\mathrm{\pi <!--\; \pi \; -->}{\left(\text{5 cm}\right)}^2=25\mathrm{\pi <!--\; \pi \; -->}\text{ c}{\text{m}}^2$ (Keeping this exact makes your final answer more accurate.)

Step 3: Subtract the area of the circle from the area of the square.

$100\text{ c}{\text{m}}^2-<!--\; -\; -->25\mathrm{\pi <!--\; \pi \; -->}\text{ c}{\text{m}}^2=21.46\text{ c}{\text{m}}^2$

The shaded area is approximately 21.46 square centimeters.

$21.46\text{c}{\text{m}}^2$

The area of a parallelogram with base b and height h is $A=bh$.

$A=bh=\left(\text{25 feet}\right)\left(\text{12 feet}\right)=300\text{ f}{\text{t}}^2$

You are only covering half of the parallelogram, so you only need 150 square foot of sod.

Make a conversion factor that uses the fact that sod costs $50 a bag and one bag covers 25 square feet.

$150\cancel{\text{f}{\text{t}}^2}\left(\frac{\mathrm{\$<!--\; \$\; -->}50}{\text{25 }\cancel{\text{f}{\text{t}}^2}}\right)=\mathrm{\$<!--\; \$\; -->}300$

The sod will cost $300.

$300

The surface area of a right prism where B is the area of the base, p is the perimeter of the base, and h is the height is $SA=2B+ph$.

The rectangular base has an area $\text{ = LW = (6 cm) (15 cm) = 90 c}{\text{m}}^{\text{2}}$.

The perimeter of the base $\text{ = }p\text{ = 2(6 cm) + 2(15 cm) = 42 cm}$.

The surface area $=2B+ph=2\left(90\text{ c}{\text{m}}^2\right)+\left(42\text{ cm}\right)\left(6\text{ cm}\right)=180\text{ c}{\text{m}}^2+252\text{ c}{\text{m}}^2=432\text{ c}{\text{m}}^2$.

The surface area is 432 square centimeters.

The volume of a right prism is $V=Bh=\left(90\text{ c}{\text{m}}^2\right)\left(6\text{ cm}\right)=540\text{ c}{\text{m}}^3$.

The volume is 540 cubic centimeters.

The surface area of a right prism where B is the area of the base, p is the perimeter of the base, and h is the height is $SA=2B+ph$.

The area of the triangular base is $A=\frac{1}{2}bh$, where b represents the base and h represents the height.

$A=\frac{1}{2}bh=\frac{1}{2}\left(\text{7 cm}\right)\left(\text{4 cm}\right)=\text{ 14 c}{\text{m}}^2$

The perimeter of the base $=p=6\text{ cm }+7\text{ cm }+5\text{ cm }=18\text{ cm}$.

The surface area $=2B+ph=2\left(14\text{ c}{\text{m}}^2\right)+\left(18\text{ cm}\right)\left(4\text{ cm}\right)=28\text{ c}{\text{m}}^2+72\text{ c}{\text{m}}^2=100\text{ c}{\text{m}}^2$.

The surface area is 100 square centimeters.

$100\text{cm}{}^2$

The surface area of a right prism where B is the area of the base, p is the perimeter of the base, and h is the height is $SA=2B+ph$.

The apothem of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The area of a regular polygon with apothem a and perimeter p is $A=\frac{1}{2}ap$.

A regular octagon has eight sides of equal length. Therefore, the perimeter is:

$p=8(3\text{ cm) = 24 cm}$.

$A=\frac{1}{2}ap=\frac{1}{2}\left(4\text{ cm}\right)\left(\text{24 cm}\right)=48\text{ c}{\text{m}}^2$, which in the surface area formula is B.

The surface area $=2B+ph=2\left(48\text{ c}{\text{m}}^2\right)+\left(24\text{ cm}\right)\left(10\text{ cm}\right)=96\text{ c}{\text{m}}^2+240\text{ c}{\text{m}}^2=336\text{ c}{\text{m}}^2$.

The surface area is 336 square centimeters.

The volume of a right prism is $V=Bh$ $=\left(48\text{ c}{\text{m}}^2\right)\left(10\text{ cm}\right)=480\text{ c}{\text{m}}^3$.

The volume is 480 cubic centimeters.

The surface area of a right prism where B is the area of the base, p is the perimeter of the base, and h is the height is $SA=2B+ph$.

The rectangular base has an area $\text{ = LW = (12 ft) (25 ft) = 300 f}{\text{t}}^{\text{2}}$.

The perimeter of the base $\text{ = }p\text{ = 2(12 ft) + 2(25 ft) = 74 ft}$.

The surface area $=2B+ph=2\left(300\text{ f}{\text{t}}^2\right)+\left(74\text{ ft}\right)\left(8\text{ ft}\right)=600\text{ f}{\text{t}}^2+592\text{ f}{\text{t}}^2=1,192\text{ f}{\text{t}}^2$.

The surface area is 1,192 square feet.

$\mathrm{1,192}\text{ ft}{}^2$

The surface area of a right cylinder with radius r and height h is $SA=2\mathrm{\pi <!--\; \pi \; -->}{r}^2+2\mathrm{\pi <!--\; \pi \; -->}rh$.

$SA=2\mathrm{\pi <!--\; \pi \; -->}{r}^2+2\mathrm{\pi <!--\; \pi \; -->}rh=2\mathrm{\pi <!--\; \pi \; -->}{\left(7\text{ cm}\right)}^2+2\mathrm{\pi <!--\; \pi \; -->}\left(7\text{ cm}\right)\left(\text{5 cm}\right)\approx <!--\; \approx \; -->527.78\text{ c}{\text{m}}^2$

The surface area is 527.78 square centimeters.

The volume of a right cylinder with radius r and height h is $V=\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h$.

$V=\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h=\mathrm{\pi <!--\; \pi \; -->}{\left(7\text{ cm}\right)}^2\left(\text{5 cm}\right)\approx <!--\; \approx \; -->769.69\text{ c}{\text{m}}^3$

The volume is 769.69 cubic centimeters.

The volume of a right cylinder with radius r and height h is $V=\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h$.

Step 1: Find the volume of the casserole dish available.

The depth of the dish is 4 inches, but one inch is filled with pasta. Use 3 inches for the height of the cylinder. The diameter is 10 inches, so the radius is 5 inches.

$V=\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h=\mathrm{\pi <!--\; \pi \; -->}{\left(\text{5 in}\right)}^2\left(\text{3 in}\right)=75\mathrm{\pi <!--\; \pi \; -->}\text{ i}{\text{n}}^3$

Step 2: Find the volume of a can. The diameter is 3 inches, so the radius is 1.5 inches.

$V=\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h=\mathrm{\pi <!--\; \pi \; -->}{\left(\text{1}\text{.5 in}\right)}^2\left(\text{4 in}\right)=9\mathrm{\pi <!--\; \pi \; -->}\text{ i}{\text{n}}^3$

Step 3: Make a conversion factor using the volume of a can.

$75\mathrm{\pi <!--\; \pi \; -->}\cancel{\text{ inche}{\text{s}}^3}\left(\frac{1\text{ can}}{9\mathrm{\pi <!--\; \pi \; -->}\cancel{\text{inche}{\text{s}}^3}}\right)=\frac{25}{3}\text{ cans}=8\frac{1}{3}\text{ cans}$

You can add $8\frac{1}{3}$ cans of soup.

approximately 8 1/3 cans of soup

While we do not have the tools to truly answer this question, we can look at several examples and pick the best choice.

The circle gives the maximum area, but the question asks for the shape to be rectangular. Of the rectangular shapes, the square has the largest area at 6.25 feet by 6.25 feet.

6.25 ft by 6.25 ft

If all six sides are the same, then the ideal volume is $V=s^3$.

Find the surface area.

The rectangular base has an area $\text{B = LW = (1}\text{.26 ft)(1}\text{.26 ft) = 1}\text{.59 f}{\text{t}}^{\text{2}}$.

The perimeter of the base $\text{ = }p\text{ = 2(1}\text{.26 ft) + 2(1 }\text{.26 ft) = 5}\text{.04 ft}$.

The surface area $=2B+ph=2\left(1.59\text{ f}{\text{t}}^2\right)+\left(5.04\text{ ft}\right)\left(1.26\text{ ft}\right)=9.5304\text{ f}{\text{t}}^2$.

Make a conversion factor out of the cost of laminate.

$9.5304\cancel{\text{sq ft}}\left(\frac{\mathrm{\$<!--\; \$\; -->}10}{1\cancel{\text{sq ft}}}\right)\approx <!--\; \approx \; -->\mathrm{\$<!--\; \$\; -->}95.30$

The ideal dimensions are 1.26 feet wide by 1.26 feet long by 1.26 feet high at a cost of $95.30.

$1.26\text{ft}\text{wide}\times <!--\; \times \; -->1.26\text{ft}\text{long}\times <!--\; \times \; -->1.26\text{ft}\text{high},$ $95.26

The Pythagorean Theorem states $a^2+b^2=c^2$ where a and b are two sides (legs) of a right triangle and c is the hypotenuse.

$a=5$

1,140 ft

The distance from point $\left(x_1,y_1\right)$ to point $\left(x_2,y_2\right)$ is $d=\sqrt{{\left(x_2-<!--\; -\; -->x_1\right)}^2+{\left(y_2-<!--\; -\; -->y_1\right)}^2}$.

You go 7 south and 9 blocks west, where each block is 100 feet.

If (0,0) is where you end up, you are looking for the distance from (0,0) to (900, 700).

$d=\sqrt{{\left(900-<!--\; -\; -->0\right)}^2+{\left(700-<!--\; -\; -->0\right)}^2}$

$d=\sqrt{810,000+490,000}$

$d=\sqrt{1,300,000}$

$d\approx <!--\; \approx \; -->1,140$ feet

The distance is approximately 1,140 feet.

The Pythagorean Theorem states $a^2+b^2=c^2$ where a and b are two sides (legs) of a right triangle and c is the hypotenuse.

The slanted ramp must be at least 120.42 inches.

The slanted distance will be 120.4 inches.

The side lengths are $15,15\sqrt{3},30.$

The set of ratios in a 30°-60°-90° triangle are $1:\sqrt{3}:2$, or $x:x\sqrt{3}:2x$.

Since you are shown the right angle and the 60° angle, you know the unmarked angle is a 30° angle.

The side opposite the 30° angle is the shortest side, $a=x=15$.

$2x=2\left(15\right)=30$

$x\sqrt{3}=15\sqrt{3}$

The side lengths are 15, 30, $15\sqrt{3}$.

The set of ratios in a 30°-60°-90° triangle are $1:\sqrt{3}:2$, or $x:x\sqrt{3}:2x$.

Since you are shown the right angle are told about the 30° angle at the ground, you know the other angle is a 60° angle.

The ladder is the longest side, so $2x=24$ feet.

If $2x=24$, then $x=12$ and $b=12$ feet since it is the shorter leg, opposite the 30° angle

$x\sqrt{3}=12\sqrt{3}\approx <!--\; \approx \; -->20.8\text{ feet}$, the distance from the wall.

The ladder is 20.8 feet (or $12\sqrt{3}$ feet) from the wall and reaches 12 feet up the wall.

The ladder reaches 12 feet up the wall and sits $12\sqrt{3}\text{ feet}$ from the wall.

The set of ratios in a 45°-45°-90° triangle are $1:1:\sqrt{2}$, or $x:x:x\sqrt{2}$.

They show the 90° angle and two other angles of equal measure. That means the other two angles have a measure of 45° each.

Since 8 is the longest side, $8=x\sqrt{2}$.

Each side $x$ equals $x=4\sqrt{2}=\mathrm{5.66.}$

Find y.

The tangent of an angle is the opposite divided by the adjacent side: $\tan <!--\; \; -->{40}^{\circ <!--\; \circ \; -->}=\frac{\text{opp}}{\text{adj}}$.

$\tan <!--\; \; -->{40}^{\circ <!--\; \circ \; -->}=\frac{\text{y}}{\text{5}}$

$y=5\tan <!--\; \; -->{40}^{\circ <!--\; \circ \; -->}\approx <!--\; \approx \; -->4.2$

Find r.

The cosine of an angle is the adjacent side divided by the hypotenuse: $\cos <!--\; \; -->{40}^{\circ <!--\; \circ \; -->}=\frac{\text{adj}}{\text{hyp}}$.

$\cos <!--\; \; -->{40}^{\circ <!--\; \circ \; -->}=\frac{\text{5}}{r}$

$r\cos <!--\; \; -->{40}^{\circ <!--\; \circ \; -->}=5$

$r=\frac{5}{\cos <!--\; \; -->{40}^{\circ <!--\; \circ \; -->}}\approx <!--\; \approx \; -->6.53$

Note: You could have used the Pythagorean Theorem to find the second side. You can also use the Pythagorean Theorem to check your work. That is left to you.

$\begin{array}{l}r=6.53\\ y=4.2\end{array}$

Find $\mathrm{\alpha <!--\; \alpha \; -->}$.

The sine of an angle is the opposite side divided by the hypotenuse: $\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}=\frac{\text{opp}}{\text{hyp}}$.

$\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}=\frac{\text{6}}{8.3}$

$\mathrm{\alpha <!--\; \alpha \; -->}={\sin }^{-<!--\; -\; -->1}\left(\frac{\text{6}}{8.3}\right)$

Note: The inverse of the sine function is not 1 divided by the sine function. There is probably a second function associated with the sine key on your calculator. You might be entering something like this: [2^nd] [sine button] [left parentheses] [6] [÷] [8] [.] [3] [right parentheses] [EXE]

$\mathrm{\alpha <!--\; \alpha \; -->}\approx <!--\; \approx \; -->{46.3}^{\circ <!--\; \circ \; -->}$

Find $\mathrm{\beta <!--\; \beta \; -->}$.

Since this is a right triangle, you know $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$ add up to 90°.

$90-<!--\; -\; -->46.3={43.7}^{\circ <!--\; \circ \; -->}$.

$\mathrm{\beta <!--\; \beta \; -->}\approx <!--\; \approx \; -->{43.7}^{\circ <!--\; \circ \; -->}$.

Find x.

The Pythagorean Theorem states $a^2+b^2=c^2$ where a and b are two sides (legs) of a right triangle and c is the hypotenuse.

$\mathrm{\alpha <!--\; \alpha \; -->}={46.3}^{\circ <!--\; \circ \; -->},\mathrm{\beta <!--\; \beta \; -->}={43.7}^{\circ <!--\; \circ \; -->}$, $x=5.73$

First, convert one mile to 5,280 feet.

The sine of an angle is the opposite side divided by the hypotenuse: $\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}=\frac{\text{opp}}{\text{hyp}}$.

$\sin <!--\; \; -->{23}^{\circ <!--\; \circ \; -->}=\frac{y}{5,280}$

$y=5,280\left(\sin <!--\; \; -->{23}^{\circ <!--\; \circ \; -->}\right)$

$y\approx <!--\; \approx \; -->2,241$ feet

The plane is approximately 2,241 feet in the airport when it passes over the peak.

Note: $2,241-<!--\; -\; -->1,500=741$ feet, so the plane is 741 feet over the peak.

2,241 ft

Find c.

The Pythagorean Theorem states $a^2+b^2=c^2$ where a and b are two sides (legs) of a right triangle and c is the hypotenuse.

Find the angle opposite 7.

The tangent of an angle is the opposite divided by the adjacent side: $\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{\text{opp}}{\text{adj}}$.

Find the angle opposite 4.

Since this is a right triangle, you know the sum of the measure of the other two angles adds up to 90°.

$90-<!--\; -\; -->60.3={29.7}^{\circ <!--\; \circ \; -->}$.

The angle opposite the 4 side has a measure of 29.7°.

The side c has a measure of 8.06. The angles have measures of 60.3° and 29.7°.

$c=8.06$, one angle is ${60}^{\circ <!--\; \circ \; -->}$, and the other angle is ${30}^{\circ <!--\; \circ \; -->}$.

The angle at the ground in the drawing will have the same measure as the angle of depression since alternate interior angles have the same measure.

The tangent of an angle is the opposite divided by the adjacent side: $\tan <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}=\frac{\text{opp}}{\text{adj}}$.

${11.9}^{\circ <!--\; \circ \; -->}$

The sine of an angle is the opposite side divided by the hypotenuse: $\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}=\frac{\text{opp}}{\text{hyp}}$.

$\sin <!--\; \; -->{50}^{\circ <!--\; \circ \; -->}=\frac{y}{60}$

$y=60\left(\sin <!--\; \; -->{50}^{\circ <!--\; \circ \; -->}\right)$

$y\approx <!--\; \approx \; -->46$ feet

The kite is approximately 46 feet above the ground.

46 ft

The line containing points C and B is $\stackrel{\leftrightarrow <!--\; \leftrightarrow \; -->}{CB}$, or you could label it $\stackrel{\leftrightarrow <!--\; \leftrightarrow \; -->}{BC}$.

The line containing points $C$ and $B$ is a straight line that extends infinitely in both directions and contains points $C$ and $B$.

You are looking for the union of line segment $\stackrel{¯<!--\; ¯\; -->}{AB}$ and line segment $\stackrel{¯<!--\; ¯\; -->}{BD}$.

Sketch the two parts separately above the line.

Union includes all the elements that are in both drawings.

The union is the line segment $\stackrel{¯<!--\; ¯\; -->}{AD}$.

$\stackrel{¯<!--\; ¯\; -->}{AB}\cup <!--\; \cup \; -->\stackrel{¯<!--\; ¯\; -->}{BD}=\stackrel{¯<!--\; ¯\; -->}{AD}$

$\stackrel{¯<!--\; ¯\; -->}{AB}\cup <!--\; \cup \; -->\stackrel{¯<!--\; ¯\; -->}{BD}=\stackrel{¯<!--\; ¯\; -->}{AD}$. The union of line segment $\stackrel{¯<!--\; ¯\; -->}{AB}$ and the line segment $\stackrel{¯<!--\; ¯\; -->}{BD}$ contains all points in each line segment combined.

You are looking for the intersection of ray $\stackrel{\leftarrow <!--\; \leftarrow \; -->}{BD}$ and line segment $\stackrel{¯<!--\; ¯\; -->}{BC}$.

Sketch the two parts separately above the line.

Intersection includes only the elements that are common to both drawings.

The intersection includes only line segment $\stackrel{¯<!--\; ¯\; -->}{BC}$.

$\stackrel{\leftarrow <!--\; \leftarrow \; -->}{BD}\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{BC}=\stackrel{¯<!--\; ¯\; -->}{BC}$

$\stackrel{\to <!--\; \to \; -->}{BD}\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{BC}=\stackrel{¯<!--\; ¯\; -->}{BC}$. The intersection of the ray $\stackrel{\to <!--\; \to \; -->}{BD}$ and the line segment $\stackrel{¯<!--\; ¯\; -->}{BC}$ contains only the points common to each set, $\stackrel{¯<!--\; ¯\; -->}{BC}$.

Straight angles measure exactly 180°.

This is a straight angle.

straight

Obtuse angles have a measure between 90° and 180°, not inclusively.

This angle is obtuse.

obtuse

Right angles measure exactly 90°.

This angle is a right angle.

right

Acute angles have a measure between 0° and 90°, not including 90°.

This angle is acute.

acute

Angle 1 and the 31° angle are supplementary, so their measures add up to 180°.

m∡1 = 180° – 31° = 149°

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}1={149}^{\circ <!--\; \circ \; -->}$ by supplementary angles with $\mathrm{\measuredangle <!--\; \measuredangle \; -->}{31}^{\circ <!--\; \circ \; -->}.$

Angle 3 and the 31° angle are vertical angles. Vertical angles have equal measures.

m∡3 = 31°

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}3={31}^{\circ <!--\; \circ \; -->}$ by vertical angles with the angle measuring ${31}^{\circ <!--\; \circ \; -->}.$

Angle 1 and the 31° angle are supplementary, so their measures add up to 180°.

The measure of angle 1 = 180° – 31° = 149°

Angle 5 and angle 1 are corresponding angles, so they have equal measures.

m∡5 = 149°

$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}5={149}^{\circ <!--\; \circ \; -->}$ by corresponding angles with $\mathrm{\measuredangle <!--\; \measuredangle \; -->}1.$

The sum of the angle measures must be 180°.

The angle measures 89°.

$x={89}^{\circ <!--\; \circ \; -->}$

The sum of the angle measures must be 180°. A right angle measures 90°.

The angle measures 67°.

$x={67}^{\circ <!--\; \circ \; -->}$

In an isosceles triangle, at least two angles have equal measures. You know the angles at the bottom have equal measure.

$x={77}^{\circ <!--\; \circ \; -->}$

The sum of the angle measures must be 180°. A right angle measures 90°.

$x={77}^{\circ <!--\; \circ \; -->}$ and $y={26}^{\circ <!--\; \circ \; -->}$

Decide which triangle’s side to put in the numerator. It does not matter which you pick to go where, if you are consistent.

In this exercise, let’s always put the smaller triangle in the numerator.

Triangle ABE is similar to triangle ACD.

To help you set up your proportions, you can use the order of the letters in the triangles. For instance, make a proportion of the first two letters in each triangle.

$\frac{AB}{AC}=\frac{3}{3+4}$

Make a proportion of the second and third letters.

$\frac{BE}{CD}=\frac{2}{x}$

Because these are similar triangles, the proportions are equal.

Make a proportion of the first and third letters.

$\frac{AE}{AD}=\frac{4}{4+y}$

Because these are similar triangles, the proportions are equal. Use one of the early proportions.

These are similar triangles, so we can solve using proportions. Then, $x=\frac{14}{3}$ and $y=\frac{16}{3}$.