How to do more of these Pythagorean Theorem Geometry Questions?

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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

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It's the length of the longest diagonal

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

#=sqrt(9+25+64)#

#=sqrt98#

#=9.9cm#

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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#

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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#

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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#

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The longest diagonal is

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

#=sqrt122=11.05m#

#11.05 >11#

The answer is #YES#

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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#

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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#

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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#