Minimario's diagonal

As we see on the picture, there are two diagonals $AC = BD$ and $CX = DX.$ Let find the length of diagonal $AC$ and $CX$ by using Pythagorean Theorem

$AC^2 = AB^2 + BC^2 = 2^2 + 2^2 = 8$

$∴ AC = \sqrt{8}$

Let $B'$ be a point so $AB'$ and $B'X$ is perpendicular, then we found that $BB' = B'X = 1.$

$CX^2 = (AB + BB')^2 + B'X^2 = (2 + 1)^2 + 1^2 = 10$

$∴ CX = \sqrt{10}$

Since $AC < CX,$ then the longest diagonal is $CX = \sqrt{10}$ imply $x = 10.$

The longest diagonal of pentagon $AXBCD$ is the line $DX$ and $CX$ (Check this out about diagonal ), where they have same length. Then, we denote the length of them as $k$ . Finding $k$ , we get the new triagle $AXD$ or $BXC$ , in here, I assume that, I choose the triagle $BXC$ , actually both of $AXD$ and $BXC$ same.

Then, look the triangle $BXC$ , you get the side, $BC$ = $2$ , $BX$ = $\sqrt{2}$ , and $CX$ is the longest diagonal which we have to find it as $k$ . Get the $x$ using Law of Cosines . But, We have to know the angle of $CBX$ to get $k$ . We know that is $135^{\circ}$ . How? We know that from $\angle ABC + \angle ABX$ = $90^{\circ} + 45^{\circ}$ . Finally, using Law of Cosines :

$cos 135^{\circ}$ = $\frac{BX^{2} + BC^{2} - k^{2}}{2.BX.BC}$

You must get the $k$ as well as $CX$ , the longest diagonal of the pentagon is $\sqrt{10}$ , so $x$ = $10$ .

Let the mid point of AB is M. Then AM=M=1. Mid point of CD is N.Then CN =ND=1; so MN=2 & XM=1.hence the longest diagonal is XC or XD of length= XN^2+NC^2=10^1/2 . so,x=10.

Center Cartesian coordinates at A. Then D is at (0,2) and X is at (1,-1) since AXB is isosceles. Thus, DA has length $\sqrt{3^2 + 1^2} = \sqrt{10}$ .

The longest diagonal are DX and CX.

DX or CX= √(3²+1²)

=√10

the longest diagonal can be express as √x, therefore x=10

longest diagonal= line CX and line DX I'll choose line DX $DX^{2}=DA^{2}+AX^{2}-2 (DA) (AX) cos(DAX)$ $DA=2$ and $AX=\sqrt{2}$ let O=midpoint between A and B $AO=\frac{AB}{2}=\frac{2}{2}=1$ $OX^{2}=AX^{2}-AO{2}=(\sqrt{2})^{2}-1^{2}=2-1-1$ so $OX=1$ $OAX=arc tan \frac{OX}{AO}=arc tan(\frac{1}{1})=45$ degree $DAX=90 + OAX = 90 + 45 = 135$ degree $cos DAX= cos 135 = -\frac{1}{\sqrt{2}}$ $DX^{2}=DA^{2}+AX^{2}-2 (DA) (AX) cos(DAX)=2^{2}+(\sqrt{2})^{2}-4\sqrt{2}\frac{-1}{\sqrt{2}}$ $DX^{2}=4+2+4=10$ so the answer is $10$

• DX is the longest diagonal in pentagon AXBCD.
• We have AX= BX= √2 and AB=2, so ABX is the right-angled and isosceles triangle at X.
• Let E be the midpoint of AB and F be the midpoint of CD. => XF^2 = AX^2-AF^2 => XF=1.
• And EF is parallel with AD and BC, so AD=EF=2.
• We have EX=FX+EF=1+2=3.
• E is the midpoint of DC, so DE=DC/2=1.
• According to Pythagoras theory, we have DX^2 = XE^2 + DE^2 = 3^2 +1^2 = 10. => DX = √10

To begin with, let us notice a few things about the triangle $ABX$ . We know that the height $h$ of it drawn towards its side $AB$ (intersecting into a point $N$ ) divides $AB$ into $2$ equal parts, each having a value of $1$ . Now, we are aware that $BX=AX=\sqrt{2}$ . Thus, we can easily find the height of $ABX$ by using the Pythagorean Theorem, resulting in $h=1$ . This shows us that the triangles $ANX$ and $BNX$ are both isosceles, as well as right triangles. Therefore, it is obvious that both angles $ABX$ and $BAX$ equal $45^{\circ}$ .

Now let us turn our attention to the diagonals of the square $ABCD$ . We can see that $AC=BD=2\sqrt{2}$ . We also know that the angles $BAC$ and $ABD$ both equal $45^{\circ}$ as well, since the diagonals of a square split its angles exactly in half.

From all this we may also notice that the angles $XAC$ and $XBC$ are $90^{\circ}$ $(45^{\circ}+45^{\circ})$ .

The diagonals we are actually looking for are $XD$ and $XC$ , which both happen to be the hypotenuses of the right triangles $XAC$ and $XBD$ . Now, by using the Pythagorean Theorem once more we can easily find that $XD=XC=\sqrt{10}$ .

Since we know that $\sqrt{x}=\sqrt{10}$ we can easily find that $x=10$ , which is the correct answer to the problem.

The triangle ABC is rectangular. So  = 135°. By law off cosines: DX² = DA² + AX² - 2DA AX cos135 So, x = DX² = 10.

use a xy coordinate system taking origin at X. now find the coordinate of point D that is (-1,3). now from distance formula find the length of line XD. XD = [ (3-0)^2 + (-1-0)^2 ]^0.5 = (10)^0.5 = (x)^0.5 this implies x=10.

Obviously, the answer is the length of XD First let's calculate the perpendicular distance from X to AB: let F be a AB midpoint (XF = 1) since XA = XB therefore XF is perpendicualr on AB XF^2 = XA^2 - AF^2 = 2 - 1 XF = 1

Now we need to calculate XD: perpendicular distance from X to DC (V)= AF + AD = 1+2 = 3 perpendicular distance from X to AD (H)= AF = 1 XD^2 = V^2 + H^2 XD^2 = 10 XD equal square root 10

By assigning x and y coordinates to point D, A, and X, the distance formula can find X^(1/2). D can be the origin, (0,0), and since side DA is length 2, then A can be (0,-2). An altitude drawn from side AB is drawn from the midpoint of AB, which is length 2 again, so the x coordinate of X is 1. By Pythagorean's theorem, 1^2+1^2=c^2, and c=2^(1/2), so the y coordinate of X is A's -2-1, or -3. Then, by the distance formula, ((0-1)^2+(0-3)^2)^(1/2)=(1+9)^1/2=(10)^1/2. X is then 10.

By Pythagoras you get the diagonal of the square is sqrt(8)=2sqrt(2) which is double XA. By drawing a second square right below the one drawn, and by drawing the diagonal you are asked for (DX) a rectangle square is formed with sides 3 and 1.

Draw a line DX

DAB = 90 degree

Draw a line perpendicular to AB from X and label Z

XZ = 1

Since ABX is isosceles triangle , so , XAZ is 45 degree

@ = angle DAX

Use cosine formula as

c^2 = a^2 + b^2 - 2ab cos @

c^2 = (2)^2 + (sqrt 2)^2 - 2(2) (sqrt 2) cos @ = 10

it is easily see that XC or XD is the longest diagonal of the pentagon
from the side of XA, XB and AB, we found that XAB is a right-angled triangle ie. <AXB=90 ( $\sqrt (2 *\sqrt2$ ^2)\=2)
then we get <XBC=135
by using cosine rule, let XC=√x
x=2^2 +(\sqrt2) ^2 -2(2)(√2)cos 135 x=10

For this one, you will need to think outside the box, er, triangle. Using the information given, it is possible to determine that Triangle AXB is a 45-45-90 when split in half (we'll call that point y). The length of segment YB is 1 and so must be segment XY. We then add the length of CB and XY (making a segment down the pentagon with a length of 3). Finally we make a diagonal line from X to C. From there, all you do is use the Pythagorean Theorem to find the length of XC.

Let E be the midpoint AB. Triangle AEX is a right triangle with lengths 1, $sqrt(2)$ , and EX. Solving for EX (using Pythagorean theorem or special triangles) gives us a length of 1

The longest diagonal is DX ( or CX) because it is longer than the diagonal of a square.

Let F be the midpoint of DC. We form a right triangle DFX with lengths DF=1, FX=FE+EX=2+1=3, and DX.

Solving for DX (using Pythagorean theorem) gives us $sqrt(10)$ , thus x=10

It is clear that our longest diagonal is $DX = CX = \sqrt{1^2 + 3^2} = \sqrt{\boxed{10}}$ . $\blacksquare$