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?

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- Sep 7th 2008, 09:28 AMinneedofhelp!ariiefind 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? - Sep 7th 2008, 09:50 AMShyam
If you draw reactangle ABCD, AC=hypotenuse of triangle ADC=140 km

and CD=AB=40 km.

In triangle ADC,

$\displaystyle

AD^2 + CD^2 = AC^2$

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

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

$\displaystyle AD = \sqrt{19600 - 1600}$

$\displaystyle AD = \sqrt{18000}$

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

$\displaystyle AD = 60 \sqrt{5}$

$\displaystyle AD = 60 \times 2.236$

$\displaystyle AD = 134.164 \;km$

Area covered by triangle ADC;

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

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

$\displaystyle =2683.28\; km^2$ - Sep 7th 2008, 10:10 AMinneedofhelp!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 - Sep 7th 2008, 10:38 PMmr fantastic
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.