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anonymous
 5 years ago
What is the area of the trapezoid given X=13 , y=9 and A = 55 degree?
A.)164.5cm^2
B.)210.3cm^3
C.)133.9cm^2
D.)267.8cm^2
*sn the picture is there when you click on the picture. Please help thanks
anonymous
 5 years ago
What is the area of the trapezoid given X=13 , y=9 and A = 55 degree? A.)164.5cm^2 B.)210.3cm^3 C.)133.9cm^2 D.)267.8cm^2 *sn the picture is there when you click on the picture. Please help thanks

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Height of the trapezoid can be found be taking the sin 55 Sin is opposite/hypothenuse Sin 55 = h/19 Sin 55 = h 7.37 = h You also need to know the two bases.. The one base is x which is 13. The other base is 13 + whatever 2 of those extra lengths are. The extra lengths are the cos 55.. cos 55 = L/9 9 cos 55 = L 5.16 = L So the other base is 5.16 + 13 + 5.16 = 23.32 The formula for area of a trapezoid is 1/2h(b1 + b2) A = 1/2 * 7.37(13 + 23.32) A = 133.9 cm^2
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