A dog is tied to circular pillar by a rope. Radius of this pillar is $1m$ and length of rope is $\pi m$. What is an area where dog can roam?   I tried to find the area of all semicircles and then to find its sum. It is easy to find an area at front side of the pillar. It is $\displaystyle\frac12\pi^2\pi=\frac{\pi^3}{2}$. Problem is how to find remaining area. I tried to write this area using compass and straightedge, but I couldn't. Then I wrote this in AutoCAD and it looks like this: Is it possible to find the exact value of this area?
122857
How do we find the area that the cow can graze?  The question goes as follows--  There is a circular barn house surrounded by a huge grazing field. A cow is tied to the rope ($AB$) at the end $A$ as shown. The length of the rope is half the circumference of the barn. Find the area that the cow can graze.  The left side of the area is obvious but i cannot get a hang of what is happening on the right side..all i can say is that the rope starts to wrap around the barn when the cow goes to the right.  The length of the rope is $16\pi$ units.
18824
From a square of unit length, pieces from the corners are removed to form a regular octagon. Find the value of the  removed area.  I got a weird answer, as I considered each side of the removed area will be $\dfrac{1}{3}$ and got $\dfrac{1}{18}$ for area of one triangle. Then I multiplied it by 4, to obtain  $\dfrac{2}{9}$ which I thought to be the answer but I was wrong.
115206
I have a thin Ring of mass $M$ and radius $R$, I have to find it's moment of Inertia about an axis passing through it's centre and at an angle of $\frac{\pi}{4}$ radians with the normal to the plane of the ring.  My approach:  I am trying to use perpendicular axes theorem. Suppose I place three mutually perpendicular axes on the centre so that one is the diameter, another one is perpendicular to it (also diameter), but on the plane of the ring, while the third is parallel to the normal, now I rotate the axes such that the one of them remains the diameter while the other two are mutually inclined at $\frac{\pi}{4}$ to the normal.  Now, as I know the moment of Inertia about a diameter ($\frac{MR^2}{2}$) So the required moment of Inertia (say $I$) must be :  $ I+I = \frac{MR^2}{2}$ from perpendicular axes theorem, So, $I = \frac{Mr^2}{4}$.  Is my way of thinking correct in this case?
52806
What is the area and circumference of a polygon with n equal sides? (triangle, square, pentagon all the way to a circle)  It doesn't matter if it's based on the radius (let's call it r) or the length n.  EDIT: I ment regular polygon.
123104
The area of ​​the portion of the sphere $ x^{2} + y^{2} +z^{2} = 1$ located inside of the cylinder $x = x^{2} + y^{2}$, and above the plane $z = 0$.  I'm stuck, so any tip will be helpful  Thanks in advance!
51856
Find the radius of the inscribed circle of a right triangle. The triangle's height is $\sqrt{6} + \sqrt{2}$ while the bisector of the right angle is 4.  This seemed like a generic similar triangles problem but it isn't that simple when I tried to solve it. Any help?  EDIT: Here is a picture of the problem
71195
Triangle $ABC$, $AB=4$, $BC=15$, $AC=13$. Two sides are tangents to the respective Parabolas. We have to find the area shaded. My approach-  I tried finding the area of the quadratures(Archimedes) formed but it doesn't help as- I have to find the area between the 2 intersecting Parabolas also which I can't find.
50835
Question:  Consider the following square sheet of metal (grey) with the side s = 10 cm. From it, we want to cut the parts of a cylindrical can with a lid and a bottom (bblack). Determine the radius and height of the can when the volume is maximal.   The rectangle and circles have to be cut in entire, singular pieces, so no trickery to minimize waste and no diagonal placement of the rectangle.  Solution has to use optimization / differentiation.  Attempted solution:  In order for this to work out the circumference of the circles and one side of the rectangle has to be identical. Otherwise they will not fit together in a cylindrical can.  This leads us to two cases that needs to be treated separately.  Case 1: The rectangle is standing upright.  Case 2: The rectangle is lying on its longest side.  There are two cases that needs to be considered separately.  Case 1:  The right-hand side of the above image must be the circumference of a circle, so  $$s_1 = 2r\pi$$  The bottom-side must be the height of the cylinder plus two radii of the circles:  $$s_1 = h + 2r$$  Putting these together becomes:  $$2r\pi = h + 2r = 10$$  Solving for h:  $$h = 10 - 2r$$  The volume of the can:  $$V = r^2\pi h$$  Adding in the earlier formula to get rid of h:  $$V = r^2\pi h = r^2\pi (10 - 2r) = 10r^2\pi - 2r^3\pi$$  Taking the derivative:  $$V' = 20\pi r - 6\pi r^2$$  Setting derivative equal to zero:  $$V' = 0 \Rightarrow r = \frac{10}{3}$$  (neglecting trivial r = 0 solution)  Height is then:  $$h = 10 - 2r = 10 - \frac{20}{3} = \frac{10}{3}$$  But this is not really possible given the side of the metal sheet cannot be more than 10 cm and $$2r\pi &gt; 10$$.  Case 2:  The bottom becomes:  $$2r + 2r\pi = 10 \Rightarrow r= \frac{5}{1+\pi}$$  The right side becomes:  $$h = 10$$  However, none of these answers match the solution:  $$r = \frac{5}{\pi}$$  $$h = 10(1-\frac{1}{\pi})$$  Did I get the starting point for the possible cases right? Where did I go wrong? How come it appears that no calculus is needed in the second step?
20504
Suppose, for the sake of illustration, that $AF=2$, where $F\mathbb{'s}$ coordinates are $(2,0)$ and $A$ is the point of origin; $E=(0,1)$ and circle $A$ has $R=1.5$  How do I solve the area of the intersection of the circle and rectangle by integration?  If we let $f(x)=1$, then the area of the rectangle is defined by: $$\int_0^Ff(x)dx$$   If we let $g(x)=\sqrt{R^2-x^2}$, then the area of the circle is defined by: $$4\int_0^Rg(x)dx$$ Now my best attempt to find the area is through: $$\int_0^Rg(x)dx-\int_0^{\sqrt{R^2-1}}(g(x)-1)dx$$  The first integral gives me a quarter of the area circle $R$, the second gives me the area of the circle that is outside the rectangle, because the circle and the top line intersects at $x=\sqrt{R^2-1}$  Is my solution correct?
70038
A Triangle has 2 sides of length 4 and 9. The largest circle than can be drawn in the triangle so as to touch all 3 sides has radius 1 cm. Find the length of the 3rd side.
22172
Prove that: area of triangle $S_{ABC}=p \cdot R$, where p is half perimeter of triangle $DEF$. $AD,CF, BE$ are altitudes. Is there a simple way?
102178
I have coordinates of 3d triangle and I need to calculate its area. I know how to do it in 2D, but don't know how to calculate area in 3d. I have developed data as follows.  (119.91227722167969, 122.7717056274414, 39.3568115234375),  (119.8951187133789, 122.7717056274414, 39.38057327270508),  (121.11941528320312, 123.2818832397461, 38.41301345825195)
121601
This question arose while I was answering this question, (). First the original question:  $ABCD$ is a quadrilateral. A line through $D$ parallel to $AC$ meets $BC$ produced at $P$ we need to show $$Area(\Delta APD)=Area(ABCD)$$    Its easy to see that the OP must have meant to prove, $Area(\Delta ABP)=Area(ABCD)$, the proof of which is given in my answer. However, it set me thinking when actually $Area(\Delta ADP)=Area(ABCD)$ is true? I have not been able to derive any good conclusion (I mean, in terms of the elements (diagonals, angles or sides) of $ABCD$). Can anyone help?
82630
A rectangle is inscribed in a semi circle with radius $r$ with one of its sides at the diameter of the semi circle. Find the dimensions of the rectangle so that its area is a maximum.    My Try:   Let length of the side be $x$, Then the length of the other side is $2\sqrt{r^2 -x^2}$, as shown in the image.    Then the area function is   $$A(x) = 2x\sqrt{r^2-x^2}$$  $$\begin{align}A'(x) &amp;= 2\sqrt{r^2-x^2}-\frac{4x}{\sqrt{r^2-x^2}}\\ &amp;=\frac{2}{\sqrt{r^2-x^2}} (r^2 - 2x -x^2)\end{align}$$  setting $A'(x) = 0$,  $$\implies x^2 +2x -r^2 = 0$$  Solving, I obtained:  $$x = -1 \pm \sqrt{1+r^2}$$  That however is not the correct answer, I cannot see where I've gone wrong? Can someone point out any errors and guide me the correct direction. I have a feeling that I have erred in the differentiation.   Also how do I show that area obtained is a maximum, because the double derivative test here is long and tedious.   Thanks!
101746
In arbitrary triangle $ABC$, let $G$ be centroid and $AH$ be a height.We extend $HG$ beyond $G$ to intersect circumcircle at $K$.Prove that $AK||BC$       I connected midpoint of $BC$ to $K$ and $A$ to show $MA=MK$ but failed...
92342
Let both ends of a string of length $L$ be tied to a stick of length $S$. Among all plane regions enclosed by this contraption, it achieves maximum when the string forms a circular arc.  It is noted that if $S=0$, this theorem is reduced to the isoperimetric theorem, which states, for the length $L$ of a closed curve and the area $A$ of the planar region that it encloses,   $4\pi A \le L^2$,  and that equality holds if and only if the curve is a circle.  Your comments and ideas are greatly appreciated!
92530
Consider the red path from A that zigzags to B, which takes $n$ even steps of length $w$. The path length of the route $P_n$ will be equal to:  $ P_n = P_x + P_y = \frac{n}{2}\times w + \frac{n}{2}\times w = n \times w $  But $\frac{n}{2}\times w = 1$ beacuse it is the length of one of the sides of the triangle so:  $P_n = 2$  Which will be true no matter how many steps you take. However in the limit $n \to \infty, w \to 0$ the parth length $P_\infty$ suddenly becomes:  $P_\infty = \sqrt{1^2 + 1^2} =  \sqrt{2}$  Due to Pythagoras. Why is this the case? It seems the path length suddenly decreases by 0.59!