How To Do These Pythagorean Theorem Math Questions?

Answer to ex. 9c:

Let C be the intersection of MN with AB:
the triangles AMC and NBC are similar

(#hatM=hatN=90°# and #BhatCN=MhatCA# since they are opposite angles);

let #x=CN# and #5-x=MC#, then it is:

#3/1=(5-x)/x#

that's

#3x=5-x#

#4x=5#

#x=5/4 ->CN=5/4#

#MC=5-x=5-5/4=15/4#

Then AB=AC+BC and, by the Pythagoras theorem, it is:

#AB=sqrt(3^2+(15/4)^2)+sqrt((5/4)^2+1^2)#

#=sqrt(9+225/16)+sqrt(25/16+1=#

#=sqrt(369/16)+sqrt(41/16)#

#=3/4sqrt(41)+1/4sqrt(41)#

#=sqrt41#

Answer to ex.3

Let x the lower side of the rectangle, then the greater one will be 2x, then let's solve the equation:

#x^2+(2x)^2=20^2# (Pythagoras theorem)

#x^2+4x^2=400#

#5x^2=400#

#x^2=80#

#x=sqrt(80)=4sqrt5# (the lower side)

#2x=8sqrt5# (the second side)

Then the perimeter is:

#2*4sqrt5+2*8sqrt5=8sqrt5+16sqrt5=24sqrt5#

and the area is:

#4sqrt5*8sqrt5=32*5=160#

Answer to ex. 5

In a square the relation between side and diagonal is:

diagonal=side#*sqrt2#

Then side=diagonal#/sqrt2=10/sqrt2=10sqrt2/2=5sqrt2#

Please see the diagram below for reference.

Here we have extended #BN# to #P# and draw #AP# perpendicular to #BP#.

Now as #APNM# is a rectangle, #MA=NP=3# #m# and hence #BP=1+3=4# #m#

Similarly #AP=MN=5# #m#

Now as #DeltaABP# is a right angled triangle,

#AB^2=BP^2+AP^2=4^2+5^2=16+25=41#

and #AB=sqrt41=6.403#

(3) Let #x=# length of smaller side

Length of other side is #=2x#

Therefore,

#x^2+(2x)^2=20^2#

#x^2+4x^2=400#

#5x^2=400#

#x^2=400/5=80#

#x=sqrt80=sqrt(16*5)=4sqrt5#

Perimeter #=2x+4x=6x=6*4sqrt5=24sqrt5 cm#

Area #=x*2x=2x^2=2*80=160 cm^2#

(5)Let #a=# length of the side of the square

#a^2+a^2=10^2#

#2a^2=100#

#a^2=50#

#a=sqrt50=sqrt(25*2)=5sqrt2#

(9c) Easier way to calculate #AB#

#AB^2=MN^2+(MA+NB)^2#

#AB^2=5^2+(3+1)^2#

#AB^2=25+16=41#

#AB=sqrt41#

Let #O# be the intersection of AB and MN

Let #MO=x#, then #ON=5-x#

Triangles OMA and ONB are similar

Therefore,

#x/3=(5-x)/1#

#x=15-3x#

#4x=15#

#x=15/4#

#AO^2=MA^2+MO^2#

#AO^2=3^2+(15/4)^2=9+225/16=(144+225)/16#

#AO=sqrt369/4#

#NO=5-15/4=5/4#

#OB^2=ON^2+BN^2#

#=(5/4)^2+1=25/16+1=41/16#

#OB=sqrt41/4#

Finally,

#AB=AO+OB=sqrt369/4+sqrt41/4#

I hope that this will help!!!