The Right Triangles

Use similar triangles. ZY / 25 = 9 / ZY , ZY^2 = 225 , ZY = 15.

 We know XY as 25, so just add ZY with XY, which is 40.

let ZY be x..... and apply pythagorus theoram to rt. triangle ZWY, therefore, ZW^2=(x^2-81)

Now apply pyth. theoram to rt. triangle ZWX,

it gives, XZ^2=x^2+175

Now finally apply pyth. theoram to rt. triangle XZY, it gives XY^2=XZ^2+YZ^2 i.e. 25^2=x^2+175+x^2

YZ=x=15 so, XY+ZY=25+15=40 ANSWER

By angle chasing, triangle $XWZ$ and triangle $YWZ$ are both similar to triangle $XWZ$ . Thus $WZ^{2}$ = $(16)(9)$ = $144$ and $WZ=12$ . Therefore by Pythagoras Theorem we have $ZY=15$ and thus $ZY+XY=15+16+9=40$

The value of ZY is equal to the square root of the product of WY and XY. . . Substituting the values we acquire ZY as 15 so 15 + 25 = 40!

If xy & zy are viewed as vectors then the answer is zx ie. 20

Notice that it is a right triangle, and the hyp of it is side xy (which is 16+9=25)

Therefore.. a²+b²=c² ... So, c= 25 .. C²= 625.. There are only two numbers which satisfies the eq.. 20 and 15.. Since were looking for the shorter side.. Therefore , xy= 25 , equal 25+15 = 40

Here's how I solved it. When I saw this problem, the first thing I noticed was that 16 and 9, multiples of 4 and 3, added up to 25, a multiple of 5. 3-4-5... The simplest Pythagorean triple! I ended up disregarding all actual math and chasing after a solution of triples...

To find ZY, we need to utilize the Pythagorean Theorem. Before we can do THAT, we need to find a value for ZW that works for both triangles. The first thing I noticed was that ones of the legs of $\triangle ZXW$ was 16, which is a multiple of 4. $4*4=16$ , so I knew it was possible that the lengths of the sides might have been part of a Pythagorean triple created by multiplying 3-4-5 by 4. If this was true, that would mean that ZW would be 12. This once again opens up the possibility of a Pythagorean triple, but created by multiplying 3-4-5 by 3 instead of 4 (This is for $\triangle ZXY$ ). If this is true, that means the hypotenuse is 15 ( $5*3$ ...).

I can check my answer now by doing the same thing for ZX... ( $5*4=20$ ), causing the two legs of $\triangle ZYX$ to be equal to 15 and 20. Wait a minute... This seems to me to be yet ANOTHER Pythagorean triple, created by multiplying 3-4-5 by 5! This means that XY would be 25 since it is the hypotenuse... and sure enough, it is!

So now I've got ZY (15) and XY (25), and all I have to do is add them to get 40!

this can be done by using the similarities in the triangle .

we know that when any perpendicular is drawn from the the right angle of a triangle to its hypotenuse then all the triangles become similar .

so we can say that 16/zw = zw / 9 solving we get 144 = zw ^2 and so by solving it further we get the answer (40)

Letting XY=a and XZ=b

Now by use pythagorean theorem for [XYZ], we can see that a^2+b^2=625 ........ (i)

Use pythagorean from [XWZ] and [YWZ], gives

ZW^2 = b^2 -256 ZW^2 = a^2 -81

Equating thees two equations to obtain, b^2-256 = a^2 -81 a^2 -b^2 = -175, compare this into (i) gives,

2a^2 = 450 a^2 = 225, since a>0

So a=15, this implies ZY is 15 units

So, ZY+XY=15+25=40

Altitude(height) theorem

In a right triangle $\Delta ZXY$ the altitude $\overline {ZW}$ fulfills : $\overline {ZW}^2 = 16 \cdot 9 = 144 \Rightarrow \overline {ZW} = 12 \Rightarrow \overline {ZY}^2 = 144 + 81 = 225 \Rightarrow \overline {ZY} = 15$ ...

ZW is the geometric mean of XW and WY, and if you do that, you'll get 12 for ZW. From that, you use pythagorean's theorem, to get ZY of 15, and adding it to the known side of XY which is 25, you will get 40

XY=25 YZ=y XZ=p WZ=x p^2=x^2+〖16〗^2 y^2=x^2+9^2 p^2+y^2=2x^2+〖16〗^2+9^2 〖25〗^2=2x^2+〖16〗^2+9^2 x^2=9×16 x=12 p=20 y=15 XY+YX=25+15=40

ZW^2+16^2=ZX^2 ...1 ZW^2+9^2=ZY^2 ...2 adding 1 and 2 2ZW^2+256+81=ZX^2+ZY^2 2ZW^2+337=(16+9)^ (ZX^2+ZY^2=25^2) 2ZW^2=625-337 2ZW^2=288 ZW=144^1/2 ZW=12 ZW^2+WY^2=ZY^2 12^2+9^2=ZY^2 ZY+XY=15+25=40

|ZW|^2 + |XW|^2 = |ZX|^2

|ZW|^2 + |WY|^2 = |ZY|^2

|ZX|^2 + |ZY|^2 = |XY|^2

Add first to equations together and substitute |ZX|^2 + |ZY|^2 you get on the right side for |XY|^2. Solve for |ZW|, then plug the value in the second equation. You obtain |ZY|=15, add 25 and that's it

Triangles XZW and WYZ are similar with angle(ZXW)=angle(WZY). So if we let ZW=H: 9/H=H/16, so H^2=9*16=144, therefore H=12. Now we use pythagoras, with the 3,4,5 triangle to find that XZ=20 and ZY=15. Hence, ZY +XY=15+25=40.

in triangl WYZ : cos Y =9/ZY
intriangle XZY : cos Y =ZY/25
9/ZY =ZY/25
ZY power 2= 225
ZY =15
ZY + XY = 15 +25 =40

Let ZW=X , ZY²=9²+X²=81+X² , XZ²=16²+X²=256+X² , ZY²+XZ²=81+256+X²+X²=337+2X² , According to Pythagorus theorom , ZY²+XZ²=XY²=(16+9)²=25²=625 , 337+2X²=625 , 2X²=625-337=288 , X²=288/2=144 , X=√144=12 , ZW=12 , ZW²+WY²=ZY² , ZY²=12²+9²=144+81=225 , ZY=√225=15 , ZY+XY=15+25=40

(ZW)^=9*16=144 (ZW)=12 (ZY)^2=9^2+12^2 (ZY)^2=225 (ZY)=15 (ZY+XY)=15+(9+16) =40

perpendicular to hypotenuse is ZW so we can write ZW^2=16*9 ZW=12 and Zy^2=12^2+9^ ZY=15 therefore ZY+XY=15+25=40

XY = XW + WY = 16 + 9 = 25 Since ∆XZY is a right triangle, we can expect it to follow one of the right triangle triads: 3,4,5 Since XY = 25 = 5 x (5) ; ZY = 5 x (3) ; XZ = 5 x (4) (from the right triangle triad 3,4,5)

Therefore ZY = 15;

ZY + XY = 25 + 15 = 40

Appying Pythagoras Thm.
WZ^2 = XZ^2 - 16^2 WZ^2 = ZY^2 - 9^2 From Above Two Eq. We Get XZ^2 -ZY^2 = 175

Applying Pythagoras Thm Again we Get XZ^2 + ZY^2 = 625

Solving Eq. 3 nd 4 We Get XZ=20 YZ=15

So XY + YZ = 40