## marcoduuuh 2 years ago Can somebody please check my work? - Giving medals. (; (Picture below.)

1. marcoduuuh

2. Shane_B

I think you're a bit confused. The area of a right triangle is 1/2 base * height which you already know. The problem only gives you the hypotenuse and the base. You need to calculate the height before calculating the area. You can use the Pythagorean theorem to calculate the height pretty easily.

3. marcoduuuh

Both of them are wrong?

4. Shane_B

Yes, both are wrong. The height in the first one does come out to be 7.14 though...but you have then calculate the area.

5. marcoduuuh

Alright. So for the first one, A=1/2(7.14) times what? 15 or 9?

6. Shane_B

For the first one: $a^2+b^2=c^2$$7^2+b^2=10^2$$b^2=51$$b=7.14$ $A=\frac{1}{2}(7)(7.14)=?$The second one is answered the same way

7. marcoduuuh

Where do you get 7?

8. Shane_B

The first problem states that the leg (base) length is 7.

9. marcoduuuh

Nvm, I got confused. Sorry.

10. marcoduuuh

So for the first one: A=599.76

11. Shane_B

Did you put the above in a calculator? That's not even close :/

12. Shane_B

You should get an area of ~25

13. marcoduuuh

Oh wait! I put it in wrong, I got Area=24.99

14. Shane_B

That's correct :) Now just do the second one the same way

15. marcoduuuh

For the second one, I used the Pythagorean theorem and I got B=13.6015?

16. marcoduuuh

Btw thank for all the help and patience! @Shane_B

17. Shane_B

Neverming..windows calculator mistake. Yep, that's correct.

18. marcoduuuh

Okay so now, the answer should be A=142.8158?

19. marcoduuuh

@Shane_B

20. Shane_B

The base of the triangle is 16 and you calculated the height to be 13.6015. So the area should be:$A=\frac{1}{2}bh=\frac{1}{2}(16)(13.6015)=108.812$

21. marcoduuuh

Oh! That's what I did wrong. I used 21 instead of 16. Thank you so much.

22. Shane_B

np :)