help me to solve....rectangle??

Question

A rectangle has perimeter 36m. The square of the length of its diagonal is 164m2 (meter square).
what are the dimentions of the rectangle??

rec area - length * width
perimeter - 2 ( length+width)

Answer 1

On this problem we are going to use the pythagorrean theorem.

a^2+b^2=c^2.

We know that c^2=164

Now, we need to find a and b.

Let a=the length
Let b=the width

2a+2b=36
a^2+b^2=164

Now, let's use substitution. Let's solve the first equation for b.

b=(36-2a)/2
b=(18-a)

Plug this into the second equation.

a^2+(18-a)^2=164

Now, expand the equation using foil.

a^2+(324-36a+a^2)=164
Simplify
2a^2-36a+324=164
Subtract 164 from each side.
2a^2-36a+160=0
Factor out a 2.
2(a^2-18a+80)=0
Now, factor the equation.
2(a-8)(a-10)=0

So, a=8 or 10

Now, plug these answers into our first equation to find b.

2(8)+2b=36
2b=20
b=10

2(10)+2b=36
2b=16
b=8

So, b=8 or 10.

When a=8, b=10. When a=10, b=8. Therefore, we know that the dimensions of our rectangle must be 8m x 10m

I hope that this helps.

Answer 2

Using Pythagoras a^2 +b^2 = 164 and the perimeter =2(a+b)=36
a+b=18==> squaring a^2 +b^2 +2ab =324===> ab=80
b=80/a
so b= 18 -a and a( 18-a)= 80

a^2-18a +80=0 a= (18+-sqrtr(324-320))2 a=10 b=8 (this is the other solution taking minus)

Answer 3

The square on the hypotenuse = the sum of the squares of the other two sides.

The hypotenuse = 164 m. (Not m²)

All I can see is, that the base will be 10 m to give 10 x 10 = 100 m
and the perpendicular would be 8 m to give 8 x 8 = 64 m

(Probably a formula somewhere).
I guess the dimensions are therefore : 10 m x 8 m

Answer 4

l = length
w = width

perimeter = 36
=> l + w = 18
or l = 18 - w

diagonal squared = 164
=> l^2 + w^2 = 164
=> (18-w)^2 + w^2 = 164
=> 324 - 36w + w^2 + w^2 = 164
=> w^2 - 18w + 80 = 0
=> (w-10)(w-8) = 0

=> w = 10 and l = 8
or w = 8 and l = 10

Answer 5

2(l+b)=36
l+b=18
b=18-l
d^2=(l^2+b^2)
164 =l^2+(18-l)^2
solve

Answer 6

I think the width is 3m and the length is 12m

Answer 7

I dont know