Hot answers tagged

9

By the Pythagorean theorem (the diagram can be rotated such that the interior segments are all horizontal/vertical), the diagonal $DA$ has length $\sqrt{12^2+5^2}=13$, so the square side length is $\frac{13}{\sqrt2}$ and the area $\frac{169}2$.


4

Your first way does not give a solution because the equality $$s=s-a=s-b=s-c$$ is impossible. By the way, by AM-GM $$\frac{A}{s^2}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s^2}\leq\frac{\sqrt{s\left(\frac{s-a+s-b+s-c}{3}\right)^3}}{s^2}=\frac{1}{3\sqrt3}.$$ The equality occurs for $$s-a=s-b=s-c$$ or $$a=b=c,$$ which says that we got a maximal value.


3

\begin{equation} x^3 = x^{\frac{1}{3}} \rightarrow x^9 = x \rightarrow x\left(x^8 - 1\right) = 0 \end{equation} And so we have $x = 0$ or $x^8 - 1 = 0$. For the later we employ the identity $$a^2 - b^2 = (a + b)(a - b)$$. Thus, \begin{equation} x^8 - 1 = 0 \rightarrow (x^4 + 1)(x^4 - 1) = 0 \end{equation} Assuming you are seeking Real Solutions only we see ...


3

Since $t^2\leq \sqrt{t}$ for $t\in [0,1]$, and $\sqrt{t}\leq t^2$ for $t\in [1,2]$; the area is given by $$\int_0^1 \big(\sqrt{t}-t^2\big)dt+\int_1^2 \big(t^2-\sqrt{t}\big)dt=\frac{1}{3}+\left(3-\frac{4\sqrt 2}{3}\right)=\frac{10-4\sqrt 2}{3}$$ For $t>1$ the area is enclosed by the two graphs and the line $x=2$.


3

Area of rectangle is $l\cdot b$ So, $$(x-3)(ax^2+bx+c) = 2x^3-13x^2+25x-12$$ Multiply and compare like powers of $x$


3

If the triangles $T$ and $T'$ are similar with simialrity factor $k$, then ratio of their areas is $k^2$. So since the sides of the starting triangle are twice as median triangle the result is ${1\over 4}$. Have I missed something? Edit. You probably meant what is the area of a triangle formed by medians? In that case, note that area of $BDG$ where $D$ ...


3

Without loss of generality, we can assume $r=1$. For convenience, position the circles vertically, with centers on the $y$-axis at the points $(0,h)$ and $(0,-h)$. Then the equation of the lower circle is $$x^2+(y+h)^2=1$$ Solving for $y$, and noting that $$h=1-\frac{s}{2}$$ we get that the upper half of the lower circle has the equation $$y=-1+\frac{s}{2}+...


2

The cross section of C by the plane $ x = t $ is a triangle. When $ t = 0 $ $(x = 0)$ , this is an isosceles right triangle with sides r, r and $ r \sqrt 2 $: cross section of cylinder by the plane x = 0. When $ t \neq 0 $, this is an isosceles right triangle with cathetus $ \sqrt {r ^ 2-t ^ 2} $: cross section of cylinder by the plane x = t. So, area of the ...


2

$SV$ is perpendicular to $WV$ and $SV=\sqrt{VT^2+ST^2}$. The area of the triangle is given by $\frac12 \cdot SV \cdot WV$.


2

$f(x)=-\frac{\sqrt2}{2}x^{\frac{3}{2}}$ When intersected with $g(x)=-2$ you get the point $x=2$ as you showed. So the area between $x=0$ , $y=-2$ and $f(x)$ is the area of $x=0$, $y=2$ and $-f(x)$. This area is the same as the area of $x=0$ , $x=2$ and $y=2$ minus the area of $-f(x)$ between $0$ and $2$. So $A=4-\int{f(x)dx}=4-\frac{8}{5}=\frac{12}{5}$


2

The area should be $$\int^2_0[f(x)-g(x)]dx$$ as $f(x)\ge g(x)$ in the given interval. So,$$\sqrt2\cdot2^{5/2} = 2^{1/2}2^{5/2} = 2^{3} = 8 \implies 4- \frac{8}{5} = \frac{12}{5}$$


2

Since $t^2 \leq \sqrt{t}$ for $t\in [0,1]$ and, $\sqrt{t} \leq t^2$ for $t \in [1,2]$, we can't just subtract the areas of the two functions, we need to incorporate the absolute value into our integrand. $$\displaystyle\int_0^2 |\sqrt{t}-t^2|dt$$ When we have an absolute value, we should use a piecewise function to solve the integral: \begin{cases} ...


1

Your Method 2 is fine and it gives the exact maximum value that you are looking for. Since $A=\sqrt{S(S-a)(S-b)(S-c)}$, the AM-GM inequality in Method 2 yields $$\frac{(S-a)+(S-b)+(S-c)}{3} \ge \sqrt[3]{(S-a)(S-b)(S-c)}$$ that is $$\frac{S^3}{3^3}\geq(S-a)(S-b)(S-c)=\frac{A^2}{S},$$ and therefore $$\frac{1}{3\sqrt{3}}\ge \frac{A}{S^2}.$$ Here equality holds ...


1

Let $AF$ and $CE$ be altitudes of $\Delta PAB$ and $\Delta PBC$ respectively. Thus, since $\Delta AFQ\sim\Delta CEQ$, we obtain: $$\frac{AQ}{QC}=\frac{AF}{CE}=\frac{\frac{1}{2}BP\cdot AF}{\frac{1}{2}BP\cdot CE}=\frac{S_{\Delta PAB}}{S_{\Delta PBC}}=\frac{2}{3}.$$ Now, you can get $S_{\Delta PQC}$ and $BP:PQ$. Can you end it now? I got $$\vec{OP}=\frac{3}...


1

Let $k\in DC$ such that $FK||BD$. Thus, $DK:KC=2:3.$ Let $DC=5x$. Thus, $DK=2x$ and $AD=10x$, which gives $$\frac{AP}{PF}=\frac{AD}{DK}=\frac{10x}{2x}=5.$$ Similarly, let $M\in EC$ such that $FM||BE.$ Thus, $EM:MC=2:3,$ which gives $EM=4x$ and $$\frac{AQ}{QF}=\frac{AE}{EM}=\frac{5x}{4x}=\frac{5}{4}.$$ From here we obtain: $$AQ:QP:PF=10:5:3.$$ Now, $$S_{\...


1

The branch of $\sqrt{|x|}+\sqrt{|y|}=1$ in the first quadrant is $$ y=(1-\sqrt{x})^2=1-2\sqrt{x}+x $$ The area under it is $$ \int_0^1(1-2\sqrt{x}+x)\,dx $$ The area under the line $x+y=1$ (from $x=0$ to $x=1$) is… Subtract and multiply by $4$.


1

Hint: Since $$x\geq 0$$ we get $$\sqrt{x}+\sqrt{|y|}=1$$ and $$x+|y|=1$$ and you can compute $y$


1

Hints: For a cylinder or prism, volume is base area multiplied by height This implies base area would be volume divided by height These calculations are often easier if you use the same base unit throughout (except in the land of the acre-foot)


1

Solve the equation $x^3=\sqrt[3]{x}$: $$ x^3=x^{\frac13}\implies\\ \left(x^3\right)^3=\left(x^{\frac13}\right)^3\implies\\ x^9=x. $$ Divide both sides by $x$ and observe that $x=0$ is a solution: $$ \frac{x^9}{x}=1\implies\\ x^{9-1}=1\implies\\ x^8=1\implies\\ x=\pm1. $$ So, the solution set consists of three elements: $\{-1,0,1\}$. And those are the $x$ ...


1

To find the intersection between the graph, just find the solutions of $x^3=x^{1/3}$. $$x^3=x^{1/3} \Rightarrow x^9-x=x(x-1)(x+1)(x^2+1)(x^4+1)=0 \Rightarrow x=-1,0,1$$ Since $x^3\geq x^{1/3}$ for $x\in[-1,0]$, and $x^{1/3}\geq x^3$ for $x\in[0,1]$, the area between the two graphs is given by $$\int_{-1}^0 (x^3-x^{1/3})dx+\int_0^1 (x^{1/3}-x^3)dx$$


1

By AM-GM and by your work we obtain: $$9=s(s-a)(s-b)(s-c)\leq s\left(\frac{s-a+s-b+s-c}{3}\right)^3=\frac{s^4}{27}.$$ Can you end it now?


1

As you have discovered, $\triangle ACT$ is isosceles. $$AC = AT = 24$$ Mark $D$ on $AB$ where $AB\perp CD$. Consider two triangles $\triangle ACD$ and $\triangle BCD$. $\triangle ACD$ is a $60^\circ-30^\circ-90^\circ$ triangle. $$AD = 12, CD = 12\sqrt 3$$ $\triangle BCD$ is a $45^\circ-45^\circ-90^\circ$ triangle. $$BD = CD = 12\sqrt 3$$ The area is ...


1

Since $$\measuredangle CTA=45^{\circ}+30^{\circ}=75^{\circ}=\measuredangle C,$$ we obtain: $$AC=AT=24.$$ Now, let $TK$ and $TN$ be altitudes of $\Delta ABT$ and $\Delta ACT$ respectively. Thus, $$TK=TN=\frac{1}{2}AT=12$$ and $$S_{\Delta ATC}=\frac{12\cdot 24}{2}=144.$$ Also, $$S_{\Delta ABT}=\frac{12\cdot(12+\sqrt{24^2-12^2})}{2}.$$ Can you end it now? I ...


1

When you are applying the quadratic equation, you are solving for $x^2$, not $x$ since this is a fourth degree polynomial equation which is quadratic in form. The first solution is not possible because $x_1^2 $cannot be negative. So that leaves the second solution $x_2^2=a^2$. So you get two solutions for $x$. You get $x=\pm a$. So to get the area ...


1

It's $$\frac{24\cdot\sqrt{15^2+8^2}}{2}.$$


1

Consider a half of the lens shaped lune cut along a central vertical line. Sector angle at center of either circle is $2 \theta$ $$ \frac{2 \theta}{2 \pi} .\pi r^2 -\frac12 r \sin \theta. r \cos\theta = \pi r^2/4$$ Simplifying $$ \theta - \sin \theta \cos \theta =\frac{\pi}{4} $$ A transcendental equation this is.. can be solved by numerical methods e.g., ...


1

Colorado is not perfectly rectangular, and its northern border (assuming that it runs along a parallel) is shorter than the southern one. To see this it is sufficient to consider the Earth as an ideal sphere, in which case: $$ L_{E/W}=R|\theta_N-\theta_S|,\quad L_{S/N}=R|\phi_W-\phi_E|\cos\theta_{S/N},\quad A=R^2|(\phi_W-\phi_E)(\sin\theta_N-\sin\theta_S)|, $...


1

We know that if the area of a circle with radius 1 is $\pi$, then the area of a circle with any radius $r$ is $\pi r^2$ so it suffices to show that the the area of a circle with radius 1 is $\pi$. Here's a visual proof that the area of a circle with radius 1 is $\pi$. Alternatively, you can prove that the area of any circle is $\pi r^2$ entirely by ...


Only top voted, non community-wiki answers of a minimum length are eligible