Geometry lounge pack 1

Problem 1:

Three points A, B, C have the coordinates (3,4), (3,-2), (-5,-2) respectively

-Find which of the line segments joining the points is horizontal and what is the length?

-Which line segment is vertical and what is its length?

-What is the length of the 3rd segment and what is the equation of the line containing that segment?

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Problem 2:

A kite is flying with a length of the string already 60 meters. The string is forming a straight line making an angle of $75^{\circ}$ about the horizontal.

How high is it from the ground? Express your answer to the 10th of the meter.

Solution

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$\sin 75^{\circ}=\frac{x}{60} \Rightarrow x=60 \cdot \sin 75^{\circ}$

Finally the height $x$

$x=58.0\;meters$

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Problem 3:

An attic has a roof pitched at $60^{\circ}$ . A room has to be added per figure.

How far away from the sides must the walls be built?

What is the front area of the room?

Solution:

We have $AB=6 \;m$

$\tan 60^{\circ}=\frac{3}{AD}$

$AD=3 \cdot \cot 60^{\circ}=1.732 \;m$

Base of room:

$L=6-2 \times 1.732=2.536 \;m$

Area of the front:

$A_f=3(2.536)=7.608 \;m^{2}$

Problem 4:

A rectangular-shaped garden with sides of lengths 16 feet and 9 feet.

It has been decided that the garden has to be changed into a square design with the same design as the original rectangular-shaped garden.

What will be the measure of the side of the new square-shaped garden?

Solution:

If $x$ is the side of the square

$x^{2}=16 \times 9 \Rightarrow x=\sqrt{16 \times 9}$

$x=12 \;feet$

Problem 5:

The central angle of measure $45^{\circ}$ is subtended by a circular arc of length 6 meters.

What is the radius of the circle?

Solution:

The length of the arc:

$L=\alpha r$

In this problem $L=6$ and $\alpha=45^{\circ}=\frac{\pi}{4}$

$6=\frac{\pi}{4}r \Rightarrow r=\frac{6}{\frac{\pi}{4}}$

$r=\frac{24}{\pi}$

Problem 6:

A rectangular box with a base 2 inches by 6 inches is 10 inches tall and holds 12 ounces of breakfast cereal.

The manufacturer wants to use a new box with a base of 3 inches by 5 inches.

How many inches tall should the new box be in order to hold exactly the same volume as the original box?

Solution:

This is a classic problem that will come back in other packs.

It is important to understand the concept:

The two volumes are equal.

If $x$ is the new height:

We simply make the volumes identical

$2 \times 6 \times 10=3 \times 5 \times x$

$15x=120$

$x=8 inches$

Problem 7:

Circle c is centered at A and the small circle is tangent to circle c.

What is the value of the colored area if $AB=5$ units?

Solution:

From the graph we can depict that:

For the big circle: $R=10$

For the small circle: $r=5$

The colored area is simply the difference between the two areas

$A_c=10^{2} \pi-5^{2} \pi=100 \pi-25 \pi=75 \pi=235.62 \;square \;units$

Problem 8:

A 6-foot spruce tree is planted 15 feet from a lighted streetlight whose lamp is 18 feet above the ground.

How many feet long is the shadow of that tree?

Solution:

$\triangle BAC \sim \triangle EDC$ by $AA$ postulate

If $DC=x$

$\frac{x}{AC}=\frac{6}{18}$

But $AC=15+x$

$18x=6(15+x)$

$18x=90+6x$

$18x-6x=90$

$12x=90$

$x=\frac{90}{12}$

$x=7.5$

Problem 9:

Given DE=12, EF=7 and FG=10, What is the area of $\triangle DEG$ ?

Solution:

We have to find the difference between areas of $\triangle GFD$ and $\triangle GFE$

Colored area $\triangle GED$

$A_{GED}=A_{GFD}-A_{GFE}$ or simply $\triangle GED$ has a base of $12$ and a height of $10$

$A_{GED}=\frac{12 \times 10}{2}=60$

Problem 10:

Given the triangle ABC, what is $\tan C$ ?

Solution:

$tan C=\frac{AB}{BC}=\frac{12}{\sqrt{13^{2}-12^{2}}}=\frac{12}{5}$

Problem 11:

In the following figure:

$m\angle 2=(x^{2}-1)(x+1)$

$m\angle 8=185-x^{2}(x+1)$

Find $x$ and find the measure of all angles

Solution

Looking at the graph, $\angle 2$ and $\angle 6$ are corresponding angles and they are supplementary to $\angle 8$

That means:

$m \angle 2=180-m \angle 8$

We get:

$(x^{2}-1)(x+1)=180-(185-x^{2})(x+1))$

$(x^{2}-1)(x+1)=180-185+x^{2})(x+1))$

$(x^{2}-1)(x+1)-x^{2})(x+1)=180-185$

$(x+1)(x^{2}-1-x^{2})=-5$

$(x+1)(-1)=-5$

$x+1=5$

$x=5-1$

$x=4$

$m \angle 2=(4^{2}-1)(4+1)=(16-1)(5)=15 \times 5=75^{\circ}$

$m \angle 2=180-m \angle 8 \Rightarrow m \angle 8=180-m \angle 2=180-75=105^{\circ}$

Finally:

Angles: 2,3,6,and 7 all have same measure of $75^{\circ}$

Angles: 1,4,5,and 8 all have same measure of $105^{\circ}$

Problem 12:

In the following figure:

$m\angle 3=(x+1)(x+4)$

$m\angle 5=16(x+3)-(x^{2}-2)$

Find $x$ and find the measure of all angles

Solution

Looking at the graph, $\angle 5$ and $\angle 3$ are on the same side of transversal and are supplementary.

That means:

$m \angle 3+ m \angle 5=180$

We get:

$(x+1)(x+4)+16(x+3)-(x^{2}-2)=180$

$x^{2}+4x+x+4+16x+48-x^{2}+2=180$

$21x+54=180$

$21x=126$

$x=6$

$m \angle 3=(6+1)(6+4)=7 \times 10=70^{\circ}$

$m \angle 5=180-m \angle 3 \Rightarrow m \angle 5=180-70=110^{\circ}$

Finally:

Angles: 2,3,6,and 7 all have same measure of $70^{\circ}$

Angles: 1,4,5,and 8 all have same measure of $110^{\circ}$

Problem 13:

In the following figure:

$m\angle 4=2x^{2}-3x+6$

$m\angle 5=2x(x-1)-2$

Find $x$ and find the measure of all angles

Solution

$\angle 4$ and $\angle 5$ interior are alternate interior angles.

Alternate interior angles are congruent.

$m\angle 4=2x^{2}-3x+6$

$m\angle 5=2x(x-1)-2$

That yields:

$2x^{2}-3x+6=2x(x-1)-2$

$2x^{2}-3x+6=2x^{2}-2x-2$

$3x-2x=6+2$

$x=8$

We plug in:

$m\angle 4=2x^{2}-3x+6=2\times 8^{2}-3\times 8+6=128-24+6=110$

$m\angle 5=2x(x-1)-2=2\times 8(8-1)-2=16\times 7-2=112-2=110$

$m\angle 1,4,5\; or\; 8=110^{\circ}$
$m\angle 2,3,6\; or\; 7=70^{\circ}$

Problem 14:

In the following figure:

$m\angle 1=x^{2}+1$

$m\angle 8=8x-6$

Find $x$ and find the measure of all angles

Solution

$\angle 1$ and $\angle 8$ are alternate exterior angles and therefore congruent.

$m\angle 1=x^{2}+1$

$m\angle 8=8x-6$

That yields:

$x^{2}+1=8x-6$

$x^{2}-8x+7=0$

$\Delta=(-8)^{2}-4(1)(7)=64-28=36$

$\sqrt{\Delta}=6$

$x_1=\frac{8+6}{2}=7$

$x_2=\frac{8-6}{2}=1$

Case 1:

$m\angle 1=x^{2}+1=7^{2}+1=49+1=50$ and $m\angle 8=8x-6=8 \times 7-6=49+1=50$

$m\angle 1,4,5\; or\; 8=50^{\circ}$

$m\angle 2,3,6\; or\; 7=130^{\circ}$

Case 2:
$m\angle 1=x^{2}+1=1^{2}+1=1+1=2$ and $m\angle 8=8x-6=8 \times 1-6=8-6=2$

$m\angle 1,4,5\; or\; 8=2^{\circ}$

$m\angle 2,3,6\; or\; 7=178^{\circ}$

Problem 15:

In the following figure:

$AC=10\;feet$ and $BC=12\;feet$

1. Find the values of $x$ and $y$

2. Find $\angle B$

3.Find $b$ and $a$

4. Find the area of figure $ABCD$

Solution:

From $\triangle CDA$

$x+100+2x-y=180$

$3x-y=180-100$

$3x-y=80$ $(1)$

From transversal $AD$ :

$3x+2y+x+100=180$

$4x+2y=180-100$

$4x+2y=80$

$2x+y=40$ $(2)$

Adding $1$ and $2$ :

$5x=120$

$x=\frac{120}{5}$

From $2$

$2x+y=40$

$y=40-2x$

$y=40-2 \times 24$

$y=40-48$

$y=-8$

Finding $\angle CAB$

$m\angle CAB=3x+2y=3 \times 24-2 \times 8=72-16=56^{\circ}$

$m\angle DAC=x=24^{\circ}$

$m\angle DCA=2x-y=2\times 24 -(-8)=48+8=56^{\circ}$ . This verifies the sum of $180^{\circ}$

$m\angle ACB=180-56-43.7=80.3^{\circ}$

From $\triangle CAB$ :

$\frac{\sin B}{10}=\frac{\sin 56}{12} \Rightarrow \sin B=\frac{10}{12}\sin 56 \Leftarrow m\angle B \approx 43.7^{\circ}$

Finding the height $h$

$\sin B=frac{h}{12}\Rightarrow h=12\times \sin 43.7=8.3 \; feet$

Finding $a$ and $b$ :

$\frac{10}{\sin 43.7}=\frac{b}{\sin 80.3} \Rightarrow b=14.3 \;feet$

$\frac{10}{\sin 100}=\frac{a}{\sin 24} \Rightarrow a=4.1 \;feet$

$Area ABCD=\frac{14.3+4.1}{2}8.3=76.36$

Problem 16

Solve the triangle shown for

$A=42^{\circ}$ and $c=36$

Solution:

Solve for $\triangle ABC$

We have to make sure that all 6 elements have been calculated.

ANGLES
$A=42^{\circ}$
$C=90^{\circ}$
$B=?$

SIDES:
$a=?$
$b=?$
$c=36$

This is the right triangle geometry:

$\sin A=frac{a}{c} \Rightarrow a=c \sin A=36 \sin 42=36 \times 0.66913=24.08868$

$m\angle B=90-m\angle A=90-42=48^{\circ}$

Now let’s calculate $b$ and check it:

$\sin B=frac{b}{c} \Rightarrow b=c \sin B=36 \sin 48=36 \times 0.74314=26.75304$

Now let’s verify:

$\sqrt{a^{2}+ b^{2}}=\sqrt{(24.08868)^{2}+ (26.75304)^{2}}=35.99986 \approx=36$
This verifies the length of $c$

Answer:
ANGLES
$A=42^{\circ}$
$C=90^{\circ}$
$B=42^{\circ}$

SIDES:
$a=24.08868$
$b=26.75304$
$c=36.00000$

Problem 17

Solve the triangle shown for

$a=7.12$ and $c=8.44$

SOLUTION

ANGLES
$C=90^{\circ}$
$A=?$
$B=?$

SIDES:
$a=7.12$
$c=8.44$
$b=?$

Here we have to find $m\angle A$ , $m\angle B$ and length side $b$ .
We are lucky it is a right triangle, easy.

$\sin A=\frac{a}{c}=\frac{7.12}{8.44}=0.84360 \Rightarrow A=\arcsin (0.84360)=57.52245^{\circ}$

Now $B=90-A=90-57.52245=32.47755^{\circ}$

We used the $sine$ function for we knew that since the sum of the two angles was 90, none would exceed 90.

Finding the last element $b$

$\sin B=\frac{b}{c} \Rightarrow b=c \sin B=8.44 \sin 32.47755=8.44 \times 0.53697=4.53203$

Now let’s verify:

$\sqrt{a^{2}+ b^{2}}=\sqrt{(7.12)^{2}+ (4.53203)^{2}}\approx 8.44$
This verifies the length of $c$

Answer:
ANGLES
$A=57.52245^{\circ}$
$B=32.47755^{\circ}$
$C=90^{\circ}$

SIDES:
$a=7.12000$
$b=4.53203$
$c=8.44000$

Problem 18

An observer on the earth observes an airplane flying overhead that subtends an angle of $0.38^{\circ}$ .

If the plane is known to have a length of $250\; feet$ , what is its altitude to nearest hundred feet?

See figure.

SOLUTION

Here we can use two methods but it does not matter at that distance since the arc and the straight line representing the length of the plane are the same

$L=\alpha r \Rightarrow r=\frac{L}{\alpha}=\frac{250}{\frac{0.38\pi}{180}}=250\times \frac{180}{0.38\pi}\approx 37700 \;feet$ rounded the 100 feet accuracy.

Problem 19

A curve along a highway is an arc of a circle with 250-meter radius. If the curve corresponds to a central angle of 2 radians, what is the length of the highway along the curve?

SOLUTION

The curve:
$c=\alpha r=2 \times 250=500$

Problem 20:

Given:

$A=31^{\circ}$
$a=22$
$b=9$
Find the other elements if the triangle is possible.

SOLUTION

ANGLES
$A=31^{\circ}$
$B=?$ C=?

SIDES:
$a=22$
$c=9$
$b=?$

We notice that side $a>b$ this means that $m\angle B< m\angle A$

We can safely use the LAW OF SINES:

$\frac{\sin A}{a}=\frac{\sin B}{b}\Rightarrow \sin B=\frac{b}{a}\sin A=\frac{9}{22}\sin 31=0.21070$
$m\angle B=\arcsin (0.21070)=12.16322^{\circ}$

$m\angle C=180-(31+12.16322)=136.83678^{\circ}$

For side c:
$\frac{a}{\sin A}=\frac{a}{\sin A}\Rightarrow c=\frac{a \sin C}{\sin A}=\frac{22 \sin (136.83678)}{\sin 31}=29.22063$

Answer:

ANGLES
$A=31.00000^{\circ}$
$B=12.16322^{\circ}$
$C=136.83678^{\circ}$

SIDES:
$a=22.00000$
$b=9.00000$
$c=29.22063$

We also verify that $a+b>c$ because $22+9>29.22063$

Problem 21:

$B=160^{\circ}$
$a=19$
$b=28$
Find the other elements if the triangle is possible.

SOLUTION

ANGLES

$A=?$

$B=160^{\circ}$

$C=?$

SIDES:
$a=19$
$b=28$
$c=?$

We notice the the sum of the measures of $\angle A$ and $\angle C$ is only $20^{\circ}$

Let’s try to calculate:

$\frac{a}{\sin A}=\frac{b}{\sin B} \Rightarrow \sin A=\frac{a}{b}\sin B=\frac{19}{28} \sin 160=0.23209 \Leftarrow m\angle A=13.42015$

$m \angle C=180-(160+13.42015)=6.57985$

$\frac{a}{\sin A}=\frac{c}{\sin C} \Rightarrow c=\frac{0.11459}{0.23209} 19=9.38089$

Answer:

ANGLES
$A=13.42015^{\circ}$
$B=160.00000^{\circ}$
$C=6.57985^{\circ}$

SIDES:
$a=19.00000$
$b=28.00000$
$c=9.38089$

Problem 22:

$A=68^{\circ}$
$b=8$
$c=5$
Find the other elements if the triangle is possible.

SOLUTION

ANGLES
$A=68^{\circ}$
$B=?$ C=?

SIDES:
$a=?$
$b=8$
$c=5$

We notice that side $c>b$ this means that $m\angle C< m\angle B$

But in this case we don’t know if one angle is obtuse. We use the law of cosines.

Side $a$

$a^{2}=b^{2}+c^{2}-2bc \cos A$

$a^{2}=8^{2}+5^{2}-2\times 5 \times 8 \times cos 68=64+25-80 \times 0.37461=59.0312 \Rightarrow a=7.68318$

For $m\angle B:$

$b^{2}=a^{2}+c^{2}-2ac \cos B \Rightarrow \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}=\frac{7.68318^{2}+5^{2}-8^{2}}{2 \times 7.68318 \times 5}=0.26072 \Leftarrow m\angle B=74.88721$

$m\angle C=180-(68+74.88721)=37.11279$

Answer:

ANGLES
$A=68.00000^{\circ}$
$B=74.88721^{\circ}$
$C=37.11279^{\circ}$

SIDES:
$a=7.68318$
$b=8.00000$
$c=5.00000$

Problem 23:

A circle of center $O(0,0)$ has been inscribed in a square ABCD, of side 16 meters.

In the following figure, a smaller circle of center $L$ and tangent of Circle $\odot O$ at $F$ has been inscribed per figure.

1. What is the equation of $\odot O$ ?
2. What is the equation of the small circle $\odot L$ ?

3.What is the area of the colored area?

(All calculations to be to the thousandth)

For video solution

For text:

The circle center $O(0,0)$

General Equation of a circle of center $(x_0, y_0)$ is:

$(x-x_0)^{2}+(y-y_0)^{2}=R^{2}$

We can see that the radius is half of the width of the square.

$R=8$

$(x)^{2}+(y)^{2}=8^{2}$

Now let’s draw aline passing through points $O$ and $D$

Points $D$ coordinates $D=(8,8)$

Distance $OD$ :

$OD^{2}=8^2+8^2 \Rightarrow OD=8 \sqrt{2}$

Distance $FD$

$FD=OD-OF=8 \sqrt{2}-8=8(\sqrt{2}-1)$

The line passing through $O \& D$

$y=x$

It’s perpendicular line $k$ has a slope of $m'=-1$ .

That line passes through the tangency point $F$ .

Coordinates of $F$

$\cos \frac{\pi}{4}=\frac{x_F}{R} \Rightarrow x=R \cos \frac{\pi}{4}=\frac{8 \sqrt{2}}{2}=4 \sqrt{2}$

For point $F$ :

$F=(4 \sqrt{2}, 4 \sqrt{2})$

Line $k$ passes through $F$ with a slope of $-1$ . We use the general form:

$y-y_0=m(x-x_0)$

$y-4\sqrt{2}=-1(x-4\sqrt{2})$

$y=-x+4\sqrt{2}+4\sqrt{2}$

$y=-x+8\sqrt{2}$

This is the equation of line $k$ .

$x+y=8\sqrt{2}$

Line $k$ meets the square in two points $G$ and $H$

For point $G$ :

Point $G$ is the intersection of lines $y=8$ and line $k$ or $y=-x+8\sqrt{2}$

We get at $G$ :

$-x+8\sqrt{2}=8$

$x=8\sqrt{2}-8$

$x=8(\sqrt{2}-1)$

Finally:

$G=(8(\sqrt{2}-1), 8)$

Or:

$G=(3.314, 8)$

For point $G$ :

Point $H$ is the intersection of lines $x=8$ and line $k$ or $y=-x+8\sqrt{2}$

Let’s plug in the value of $x$ :

For point $G$ :

$y=-x+8\sqrt{2}=-8+8\sqrt{2}=8(\sqrt{2}-1)$

So:

$H=(8,8(\sqrt{2}-1))$

Or:

$H=(8,3.314)$

Now we can see the triangle $\triangle GDH$

Circle $\odot L$ is inscribed in $\triangle GDH$ with $L$ being the incerter.

Lines from any vertex bisects that vertex.

We see that angle:

$m\angle FGD=\frac{\pi}{4}$

The line from vertex $G$ to $L$ will make an angle of $-\frac{\pi}{8}$

Let’s find it’s tangent or the slope.

$\tan (-\frac{\pi}{8})=-\tan \frac{\pi}{8}$

From our formula page:

$\tan \theta=\frac{1-\cos 2\theta}{\sin 2\theta}$

$\tan \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}$

$\tan \frac{\pi}{8}=\frac{1-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=\sqrt{2}-1$

$\tan (-\frac{\pi}{8})=-(\sqrt{2}-1)$ . This is the slope.

The line $m$ passes through $G$

$G=(8(\sqrt{2}-1), 8)$

We use the standard equation:

$y-y_G=m(x-x_G)$

$y-8=-(\sqrt{2}-1)(x-8(\sqrt{2}-1))$

$y=-(\sqrt{2}-1)x+8(2-2\sqrt{2}+1)+8$

$y=-(\sqrt{2}-1)x+8(2-2\sqrt{2}+1+1)$

$y=-(\sqrt{2}-1)x+16(2-\sqrt{2})$

The lines $y=x$ and $y=-(\sqrt{2}-1)x+16(2-\sqrt{2})$ intersect at the incenter $L$ , center of $\odot L$

We get:

$x=-(\sqrt{2}-1)x+16(2-\sqrt{2})$

$x(1-\sqrt{2}-1)=16(2-\sqrt{2})$

$x\sqrt{2}=16(2-\sqrt{2})$

$x=\frac{16(2-\sqrt{2})}{\sqrt{2}}$

$x=\frac{16\sqrt{2}(2-\sqrt{2})}{2}$

$x=8(2\sqrt{2}-2)$

$x=16(\sqrt{2}-1)$

Since the point $L$ is on $y=x$ :

$L=(16(\sqrt{2}-1), 16(\sqrt{2}-1))$

The length $FL$ is the radius of $\odot L$

Let’s calculate that distance:

$F=(4 \sqrt{2}, 4 \sqrt{2})$

$L=(16(\sqrt{2}-1), 16(\sqrt{2}-1))$

$FL=\sqrt{(16\sqrt{2}-16-4\sqrt{2})^{2}+(16\sqrt{2}-16-4\sqrt{2})^{2}}$

$FL=\sqrt{2(16\sqrt{2}-16-4\sqrt{2})^{2}}$

$FL=\sqrt{2}(12\sqrt{2}-16)$

$FL=24-16\sqrt{2}$

If $r$ is the radius of $\odot L$ :

$r=24 -16 \sqrt{2}$

Let’s use decimals to simplify:

$r=1.373$

Equation of the circle:

$(x-16(\sqrt{2}-1))^{2}+(x-16(\sqrt{2}-1))^{2}=( 24-16\sqrt{2})^{2}$

Or in decimals:

$(x-6.627)^{2}+(y-6.627)^{2}=(1.373)^{2}$

Now for the final question we have to take a look at the graph:

The shaded is simply the difference between the areas of $\triangle JID$ and circle segment of arc $IJ$

Let’s calculate that difference.

Triangle $\triangle LJI$ is a right triangle.

The circle segment intercepts an arc with a central angle of $\frac{\pi}{2}$

$Area\;of\;segment=\frac{1}{2}r^{2}(\alpha-\sin \alpha)$ with $\alpha=\frac{\pi}{2}$ in our case.

$Area\;of\;segment=\frac{1}{2}(1.373)^{2}(\frac{\pi}{2}-\sin \frac{\pi}{2})$

$Area\;of\;segment=0.942(\frac{\pi}{2}-1)=0.942(0.571)$

$Area\;of\;segment=0.538$

Area of $\triangle JID$

Let’s calculate the length of segment $LN$

The height of $\triangle JID$ is:

$DN=OD-ON$ with $ON=OF+FL+LN=R+r+LN$

We can see that : $DN=OD-R-r-LN=8\sqrt{2}-8-24+16\sqrt{2}-LN$

$JI=r \sqrt{2}=(24-16\sqrt{2})\sqrt{2}=24 \sqrt{2}-32=1.941$

$NI=\frac{1}{2}JI=0.971$

In the right triangle $\triangle LNI$

$r^2=(NI)^{2}+(LN)^{2}$

$LN=\sqrt{r^{2}-(NI)^{2}{2}}=\sqrt{1.884-(0.971)^{2}}$

$LN=\sqrt{1.884-0.943}=\sqrt{0.941}=0.970$

We can also say that $LN=ND=0.971$ without approximation for $N$ is the intersection of the diagonals of the quadrilateral $DJLI$

Finally:

$DN=8\sqrt{2}-8-24+16\sqrt{2}-0.970=11.314-8-24+22.627-0.970=0.971$

$DN=0.971$

Area of $\triangle JID$ is:

$Area\; \triangle JID=\frac{1}{2}JI \cdot DN=\frac{1}{2}1.941 \times 0.971=0.942\; m^{2}$

Shaded Area:

$Shaded\; Area=0.942-0.538=0.404$

Answer: $Shaded\; Area=0.404\; m^{2}$

Problem 24:

The isosceles trapezoid, ABCD, shown in the following figure has the parallel sides measuring 8 meters and 18 meters.

1. Find the area of the inscribed circle.

2. Find the value of the shaded area inside the trapezoid.

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Problem 25:

Find the value of the blue shaded area in the following design. The area is inside a square with a side of 8 meters. The four holes are circles with 600 millimeters as diameter each.

Round your result to the hundredth.