How to do more of these Pythagorean Theorem Geometry Questions?

`color(red)(4A-6)`

Let `Z` be the point where the boat changes direction

`XZY` is a rignt triangle

`XZ=10`

`ZY=10`

`XY^2=XZ^2+ZY^2=10^2+10^2`

`XY=10sqrt2`

The bearing of X from Y is `NW` which is `90+90+90+45=315`º

`color(red)(4A-7)`

It's the length of the longest diagonal

`l=sqrt(3^2+5^2+8^2)`

`=sqrt(9+25+64)`

`=sqrt98`

`=9.9cm`

`color(red)(4A-8)`

Driver A is at `X` after one hour (120km)

Driver B is at `Y` after one hour (135 km)

angle `hat(XCY)=90`º

The triangle `CXY` is rigt angle

Therefore,

`XY^2=XC^2+YC^2`

`=120^2+135^2`

`XY=sqrt(120^2+135^2)`

`=180.6 km`

`color(red)(4B-6)`

Angle `hat(ABC)=90`º

`AB=27`

`BC=21`

So,

`AC^2=AB^2+BC^2`

`=27^2+21^2`

`AC=sqrt(27^2+21^2)=34.21km`

`color(red)(4B-7)`

We have a right triangle

Distance of bottom of ladder to base of wall `=x`

Distance of top of ladder to the base of the wall `=2x`

Therefore,

`15^2=x^2+(2x)^2`

`225=x^2+4x^2`

`5x^2=225`

`x^2=225/5=45`

`x=sqrt45=6.71m`

Distance of top of ladder to the base of the wall `=2*6.71=13.42m`

`color(red)(4B-8)`

The longest diagonal is

`sqrt(8^2+7^2+3^2)`

`=sqrt122=11.05m`

`11.05 >11`

The answer is `YES`

`color(red)(4B-9)`

We have a right triangle of dimensions

`10, 1 and XY`

Therefore,

`XY^2=10^2+1^2`

`=100+1=101`

`XY=sqrt101=10.05cm`

`color(red)(3D-8)`

In triangle ABC

`AC^2=AB^2+BC^2`

`AC^2=100^2+100^2=2*100^2`

`AC=100sqrt2`

`AM=1/2AC=50sqrt2`

In triangle AME

`ME^2=AE^2-AM^2`

`=100^2-2*50^2`

`=50^2(4-2)`

`=2*50^5`

`AE =50sqrt2=70.71`

The height is `=70.71m`

`color(red)(4B-9)`

Let `a=`dimension of square base

The diagonal

`=sqrt(a^2*a^2)=sqrt2a^2=asqrt2`

Half this diagonal `=1/2*asqrt2=a/sqrt2`

In triangle `AME`

`AM=a/sqrt2`

`AE=15`

`EM=10`

Therefore,

`AE^2=AM^2+EM^2`

`15^2=a^2/2+10^2`

`225=100+a^2/2`

`a^2/2=225-100=125`

`a^2=125*2=250`

`a=sqrt(250)=15.81cm`

The side of the square is `=15.81cm`