1. find the hyp.

rectangle a,b,c,d. a to b =40km a to c=140km what is distance a to d?
what is area covered by a,d,c?
402 +1402 =
1600+19600=
21200
√21200=
20√53
how do i work with this?

2. Originally Posted by inneedofhelp!ariie
rectangle a,b,c,d. a to b =40km a to c=140km what is distance a to d?
what is area covered by a,d,c?
402 +1402 =
1600+19600=
21200
√21200=
20√53
how do i work with this?
If you draw reactangle ABCD, AC=hypotenuse of triangle ADC=140 km
and CD=AB=40 km.

$

$AD^2 + 40^2 = 140^2$

$AD^2 = 140^2 - 40^2$

$AD = \sqrt{19600 - 1600}$

$AD = \sqrt{18000}$

$AD = \sqrt{5\times 3^2 \times 2^2 \times 10^2}$

$AD = 60 \sqrt{5}$

$AD = 60 \times 2.236$

$AD = 134.164 \;km$

$=\frac{base \times height}{2}$

$=\frac{40 \times 134.164}{2}$

$=2683.28\; km^2$

3. no the diagonal (hypotenuse) is A to D
A to B is 40km
A to C is 140km

the diagonal is always longest so cannot be 134

4. If your question is not clear you can't expect to get a good answer.

Let ad = x. Then x^2 + 40^2 = 140^2 => .....

By the way, don't ask for help in Live Chat. It clearly says that Live Chat is for small talk.

Originally Posted by inneedofhelp!ariie
rectangle a,b,c,d. a to b =40km a to c=140km what is distance a to d?

what is area covered by a,d,c?

402 +1402 =

1600+19600=

21200

√21200=

20√53

how do i work with this?
Originally Posted by Shyam
If you draw reactangle ABCD, AC=hypotenuse of triangle ADC=140 km

and CD=AB=40 km.

$

$AD^2 + 40^2 = 140^2$

$AD^2 = 140^2 - 40^2$

$AD = \sqrt{19600 - 1600}$

$AD = \sqrt{18000}$

$AD = \sqrt{5\times 3^2 \times 2^2 \times 10^2}$

$AD = 60 \sqrt{5}$

$AD = 60 \times 2.236$

$AD = 134.164 \;km$

$=\frac{base \times height}{2}$

$=\frac{40 \times 134.164}{2}$

$=2683.28\; km^2$
Originally Posted by inneedofhelp!ariie
no the diagonal (hypotenuse) is A to D
A to B is 40km
A to C is 140km

the diagonal is always longest so cannot be 134