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## A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 8  and 2  and the pyramid's height is 9 . If one of the base's corners has an angle of (5pi)/12, what is the pyramid's surface area?

color(maroon)("Total Surface Area " = color(purple)(A_T = A_B + A_L = color(crimson)(15.45 + 94.12 = 109.57

#### Explanation: $l = 8 , b = 2 , \theta = \frac{5 \pi}{12} , h = 9$

$\text{To find the Total Surface Area T S A}$

$\text{Area of parallelogram base } {A}_{B} = l b \sin \theta$

${A}_{B} = 8 \cdot 2 \cdot \sin \left(\frac{5 \pi}{12}\right) = 15.45$

${S}_{1} = \sqrt{{h}^{2} + {\left(\frac{b}{2}\right)}^{2}} = \sqrt{{9}^{2} + {1}^{2}} = 9.06$

${S}_{2} = \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}} = \sqrt{{9}^{2} + {4}^{2}} = 10.82$

"Lateral Surface Area " A_L = 2 * ((1/2) l * S_1 + (1/2) b * S_2

${A}_{L} = \left(\cancel{2} \cdot \cancel{\frac{1}{2}}\right) \cdot \left(l \cdot {S}_{1} + b \cdot {S}_{2}\right) = \left(8 \cdot 9.06 + 2 \cdot 10.82\right)$

${A}_{L} = 94.12$

$\text{Total Surface Area } {A}_{T} = {A}_{B} + {A}_{L} = 15.45 + 94.12 = 109.57$

## What is the side length of the smallest sized equilateral triangle that can be placed on the (x,y) plane where all coordinates are integers AND no side is horizontal and no side is vertical?

There is no such triangle, due to $\sqrt{3}$ being irrational.

#### Explanation:

Without loss of generality, one of the vertices is at $\left(0 , 0\right)$.

Let the vertex anticlockwise from $\left(0 , 0\right)$ be $\left(m , n\right)$ for some integers $m$ and $n$.

The midpoint of the corresponding side is $\left(\frac{m}{2} , \frac{n}{2}\right)$

The line through $\left(0 , 0\right)$ and $\left(m , n\right)$ has slope $\frac{n}{m}$, so any perpendicular to this side has slope $- \frac{m}{n}$.

The third vertex of the triangle lies on the line through the midpoint $\left(\frac{m}{2} , \frac{n}{2}\right)$ with slope $\left(- \frac{m}{n}\right)$.

In fact it will lie at the point:

$\left(\frac{m}{2} , \frac{n}{2}\right) + \frac{\sqrt{3}}{2} \left(- n , m\right) = \left(\frac{m - \sqrt{3} n}{2} , \frac{n + \sqrt{3} m}{2}\right)$

since the height of an equilateral triangle is $\sin {60}^{\circ} = \frac{\sqrt{3}}{2}$ times the length of one of the sides.

So we require $\frac{m - \sqrt{3} n}{2}$ to be an integer and $\frac{n + \sqrt{3} m}{2}$ to be an integer.

Since $\sqrt{3}$ is irrational, this would imply that $n = 0$ and $m = 0$, hence there is no such triangle.

## ABCD is a trapezoid with line BC perpendicular to line AB and line BC perpendicular to line CD. AB=13 BC=12 CD=8.A line segment is drawn from A to E, which is the midpoint of line CD. What is the area of triangle AED?

${S}_{\triangle A E D} = 24$

#### Explanation:

The trapezoid described is somewhat as the figure below: The formula of the area of the triangle is
${S}_{\triangle A E D} = \frac{b a s e \cdot h e i g h t}{2}$

Since AF is perpendicular to the triangle's side DE, and AF=BC=12, we have:
${S}_{\triangle A E D} = \frac{D E \cdot A F}{2} = \frac{4 \cdot 12}{2} = 24$

## The area of a circle inscribed in an equilateral triangle is 154 square centimeters. What is the perimeter of the triangle? Use pi=22/7 and square root of 3= 1.73.

Perimeter $= 36.33$ cm.

#### Explanation:

This is Geometry, so lets look at at a picture of what we are dealing with: ${A}_{\text{circle") = pi*r^2color(white)("XXX")rarrcolor(white)("XXX}} r = \sqrt{\frac{A}{\pi}}$

We are told
color(white)("XXX")A=152 "cm"^2
and to use
$\textcolor{w h i t e}{\text{XXX}} \pi = \frac{22}{7}$

$\Rightarrow r = 7$ (after some minor arithmetic)

If $s$ is the length of one side of the equilateral triangle and $t$ is half of $s$

$\textcolor{w h i t e}{\text{XXX}} t = r \cdot \cos \left({60}^{\circ}\right)$

$\textcolor{w h i t e}{\text{XXXx}} = 7 \cdot \frac{\sqrt{3}}{2}$

and
$\textcolor{w h i t e}{\text{XXX}} s = 2 t = 7 \cdot \sqrt{3}$

$\textcolor{w h i t e}{\text{XXXx}} = 12.11$ (since we are told to use $\sqrt{3} = 1.73$)

Perimeter $= 3 s$

$\textcolor{w h i t e}{\text{XXXXXX}} = 3 \times 12.11 = 36.33$

## The base of an isosceles triangle is 5, the height lowered from one end of the base is 4. What is the surface area?

$10$

#### Explanation: The area of any triangle is given by $\frac{1}{2} \cdot b \cdot h$ where $b$ is the base and $h$ is the hieght.
Therefore the area here is $\frac{1}{2} \cdot 5 \cdot 4 = 10$

## The length of a leg of an isosceles right triangle is 5sqrt2. How do you find the length of the hypotenuse?

The hypotenuse
$A B = 10 c m$

#### Explanation: The above triangle is a right angled isosceles triangle , with $B C = A C$

The length of the leg given $= 5 \sqrt{2} c m$ (assuming units to be in cm)

So, $B C = A C = 5 \sqrt{2} c m$

The value of the hypotenuse $A B$ can be calculated using the Pythagoras theorem:

${\left(A B\right)}^{2} = {\left(B C\right)}^{2} + {\left(A C\right)}^{2}$

${\left(A B\right)}^{2} = {\left(5 \sqrt{2}\right)}^{2} + {\left(5 \sqrt{2}\right)}^{2}$

${\left(A B\right)}^{2} = 50 + 50$

${\left(A B\right)}^{2} = 100$

$\left(A B\right) = \sqrt{100}$

$A B = 10 c m$

## Each interior angle of a regular polygon lies between 136 to 142. How do we calculate the sides of the polygon?

$n = 9$

#### Explanation:

In a regular polygon each interior angle can be obtained in this way:
$\alpha = {180}^{\circ} - {360}^{\circ} / n$

From the conditions of the problem:
${136}^{\circ} < \alpha < {142}^{\circ}$

That's the conjugation of this two inequations:
${136}^{\circ} < \alpha$ and $\alpha < {142}^{\circ}$

Resolving the first inequation
${136}^{\circ} < {180}^{\circ} - {360}^{\circ} / n$
${136}^{\circ} < \frac{{180}^{\circ} \cdot n - {360}^{\circ}}{n}$ => ${136}^{\circ} \cdot n < {180}^{\circ} . n - {360}^{\circ}$ => ${44}^{\circ} . n > {360}^{\circ}$ => $n > 8.18$

Resolving the second inequation
${180}^{\circ} - {360}^{\circ} / n < {142}^{\circ}$
${180}^{\circ} \cdot n - {360}^{\circ} < {142}^{\circ} \cdot n$ => ${38}^{\circ} . n < {360}^{\circ}$ => $n < 9.47$

Conjugating the two inequations
$8.18 < n < 9.47$

Since $n \in \mathbb{N}$, its only value that satisfies the inequation is $n = 9$

By the way
$\alpha = {180}^{\circ} - {360}^{\circ} / 9 = {180}^{\circ} - {40}^{\circ}$ => $\alpha = {140}^{\circ}$

## How do you calculate the overlapping area between intersecting circles?

Calculate the area of the circular sector, from which subtract the area of the triangles whose base is the circles' chord defined by the intersection points; finally sum the results.

#### Explanation:

Consider Figs. 1 and 2  Figure 1 shows two circles (with centers $C 1$ and $C 2$ and radii ${r}_{1}$ and ${r}_{2}$) that intercept each other in points A and B.

The area of interest is encompassed by arcs ADB and AEB.

In such a problem, most probably ${r}_{1}$ and ${r}_{2}$ are informed or aren't hard to find. This means that we can easily know the area of both the circles (as $\pi \cdot {r}_{1}^{2}$ and $\pi \cdot {r}_{2}^{2}$).

We only need one more information to determine the area of the region ADBE: the proportion of any of the areas of circular sector to the total area of its circle or the angle of the circular sector ($\alpha$ or $\beta$) or the length of the chord AB or even the coordinates of the center points $C 1$ and $C 2$.

If we have the coordinates of the center points C1 and C2 as well as the radii ${r}_{1}$ and ${r}_{2}$, we can obtain the circles' equations of whose conjugation we can obtain points A and B and therefore the length of the cord AB.

If we get the chord AB (called "x" in Figure 2 ) we can obtain $\alpha$ and $\beta$ in this way:
$\tan \left(\frac{\alpha}{2}\right) = \frac{\frac{x}{2}}{r} _ 1$
$\tan \left(\frac{\beta}{2}\right) = \frac{\frac{x}{2}}{r} _ 2$

Knowing $\alpha$ and $\beta$ we can determine the area of the circular sectors:
${S}_{\text{circular sector 1}} = \left(\frac{\alpha}{360} ^ \circ\right) \cdot {S}_{{\circ}_{1}} = \left(\frac{\alpha}{360} ^ \circ\right) \cdot \pi \cdot {r}_{1}^{2}$
${S}_{\text{circular sector 2}} = \left(\frac{\alpha}{360} ^ \circ\right) \cdot {S}_{{\circ}_{2}} = \left(\frac{\alpha}{360} ^ \circ\right) \cdot \pi \cdot {r}_{2}^{2}$

Next we can easily find ${h}_{1}$ (height of ${\triangle}_{A B C 1}$ in circle 1) and ${h}_{2}$ (height of ${\triangle}_{A B C 2}$ in circle 2).
Then with the chord AB ($= x$) as base and ${h}_{1}$ or ${h}_{2}$ as height we find the area of triangles ABC1 and ABC2.

Next we need just to subtract from each circular sector the area of its triangle.
Finally, we sum the results of the aforementioned subtractions obtaining the area of interest. Job done!

OR we could integrate the area between the equations of the two circles from the point A (${x}_{A} , {y}_{A}$) to the point B (${x}_{B} , {y}_{B}$).

## In a quadrilateral ABCD ,which is not a trapezium.It is known that <DAB=<ABC=60 DEGREE.moreover , <CAB=<CBD.Then A) AB=BC+CD B) AB=AD+CD C) AB=BC+AD D) AB=AC+AD ? Choose correct option.

C) $A B = B C + A D$

#### Explanation:

To ease the comprehension of the problem refer to the figure below: If the sides BC and AD are extended the ${\triangle}_{A B F}$, equilateral, is formed.

First, in addition to the angles informed it was possible to determine other interior angles of the triangles ABC and ABD, as shown in the figure.

Then using the Law of Sines on these triangles:

• ${\triangle}_{A B C}$

$\frac{\text{AB"/sin (120^@-alpha)="BC"/sin alpha="AC}}{\sin} {60}^{\circ}$

• ${\triangle}_{A B D}$

$\frac{\text{AB"/sin (60^@+alpha)="AD"/sin (60^@-alpha)="BD}}{\sin} {60}^{\circ}$

To focus on viable hypotheses I suggest to begin trying to prove a specific case.
For instance $A B = 5$ and $\alpha = {15}^{\circ}$:

From ${\triangle}_{A B C} \to \frac{5}{\sin {105}^{\circ}} = \frac{\text{BC}}{\sin} {15}^{\circ}$ => $B C = 5 \cdot \sin {15}^{\circ} / \sin {105}^{\circ}$

From ${\triangle}_{A B D} \to \frac{5}{\sin {75}^{\circ}} = \frac{\text{AD}}{\sin} {45}^{\circ}$ => $A D = 5 \cdot \sin {45}^{\circ} / \sin {75}^{\circ}$

Note that $\sin {75}^{\circ} = \sin 105$ because ${75}^{\circ}$ and ${105}^{\circ}$ are supplementary angles and sines of angles of the first and second quadrants are positive.

Finally,

$A D + B C = 5 \cdot \frac{\sin {15}^{\circ} + \sin {45}^{\circ}}{\sin {75}^{\circ}} = 5 \cdot 1 = A B$

So in this case hypothesis (C) is true (I tried the others and verified that they are false.)

Now for the general proof of the hypothesis (C), following the same steps that worked for the specific proof:

From ${\triangle}_{A B C} \to \frac{\text{AB"/sin (120^@-alpha)="BC}}{\sin} \alpha$

Observation:

$\sin \left({120}^{\circ} - \alpha\right) = \sin \left({180}^{\circ} - \left({120}^{\circ} - \alpha\right)\right) = \sin \left({60}^{\circ} + \alpha\right)$

$\implies \text{AB"/sin (60^@+alpha)="BC"/sin alpha => "BC"="AB} \cdot \sin \frac{\alpha}{\sin} \left({60}^{\circ} + \alpha\right)$

Therefore,

$\text{BC" ="AB} \cdot \frac{\sin \alpha}{\left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha + \left(\frac{1}{2}\right) \cdot \sin \alpha}$

From ${\triangle}_{A B D} \to \frac{\text{AB"/sin (60^@+alpha)="AD}}{\sin} \left({60}^{\circ} - \alpha\right)$

This will get you

$A D = \text{AB} \cdot \sin \frac{{60}^{\circ} - \alpha}{\sin} \left({60}^{\circ} + \alpha\right)$

$= A B \frac{\left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha - \left(\frac{1}{2}\right) \cdot \sin \alpha}{\left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha + \left(\frac{1}{2}\right) \cdot \sin \alpha}$

Finally,

$B C + A D = \text{AB} \frac{\sin \alpha + \left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha - \left(\frac{1}{2}\right) \cdot \sin \alpha}{\left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha + \left(\frac{1}{2}\right) \cdot \sin \alpha}$

$B C + A D = \text{AB} \cdot 1$

Therefore,

$A B = B C + A D$ and the hypothesis (C) is true.**

## What is the surface area formula of a 3-dimensional rectangle?

$\text{SA} = 2 \left(w l + l h + h w\right)$

#### Explanation:

For a rectangular prism with sides $w , l , h$, the surface area is

$\text{SA} = 2 \left(w l + l h + h w\right)$ This occurs since there are two pairs of three different faces on every rectangular prism.

Each pair of faces is a different rectangle using two of the the three dimensions of the prism as its own side.

One side is just $w l$, another is just $l h$, and the other $h w$. Since there are two of each, that is reflected in the formula by the multiplication by $2$.

This could also be imagined as a series of flattened-out rectangles: The blue rectangles are $2 \cdot w l$.

The yellow rectangles are $2 \cdot l h$.

The red rectangles are $2 \cdot h w$.

Again, the surface area would be

$\text{SA} = 2 w l + 2 l h + 2 h w$

$= 2 \left(w l + l h + h w\right)$