## anonymous one year ago CAN SOMEONE PLEASE HELP!! Just need to finish just 1 and I'm done!!

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1. anonymous

hello

2. anonymous

ummmmmm

3. anonymous

Calculate the area of triangle WXY with altitude YZ, given W (2,-1), X (6,3), Y (7,0) and Z (5,2) 8 square units 9.4 square units 7.7, square units 12 square units

4. anonymous

@mumustar78612

5. campbell_st

well I'd say you need the distance formula to find the base length and altitude the base would be WX, so find the distance from (2, -1) tp (6,3) and the altitude YZ is the distance from (7,0) to (5,2) then use the normal area of a triangle A =1/2bh

6. anonymous

I got 7.7 idk if that's right

7. campbell_st

8. anonymous

Is it 9.4?

9. campbell_st

$WX = \sqrt{(6 -2)^2 + (3 - (-1))^2}$ $YZ = \sqrt{(7 - 5)^2 + (0 - 2)^2}$ find those 2 values..

10. anonymous

I got 4

11. anonymous

Well 32/8

12. anonymous

I calculated everything

13. campbell_st

$WX = \sqrt{32} = 4\sqrt{2}~~~and~~~YZ = \sqrt{8} = 2\sqrt{2}$ then the area is $A = \frac{1}{2} \times WX \times YZ$ which is $A = \frac{1}{2} \times 4 \sqrt{2} \times 2 \sqrt{2}$

14. anonymous

12??

15. campbell_st

so you have $A = \frac{1}{2} \times 4 \sqrt{2} \times 2\sqrt{2} = 2 \times 2 \times \sqrt{2} \times \sqrt{2}$ and you need to know $\sqrt{a} \times \sqrt{a} = a$ hope that helps