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by jainrahul1985 » Sat Jul 25, 2009 9:38 pm
In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area of region
ABCDE ?
A. 76
B. 84
C. 92
D. 100
E. 108
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Source: — Problem Solving |

by truplayer256 » Sun Jul 26, 2009 6:49 am
Area of trapezoid ABDE= 1/2*h(base1+base2)= 1/2(8)(10)=40 square units

Area of triangle BCD= 1/2*b*h=1/2*(10)(12)=60 square units

60+40=100 square units D
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by hariharakarthi » Sun Jul 26, 2009 3:08 pm
@truplayer256

Can you explain how did you find the hight of the triangle BCD as 12?

Regards,
hhk.
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by truplayer256 » Sun Jul 26, 2009 4:40 pm
@truplayer256

Can you explain how did you find the hight of the triangle BCD as 12?

Regards,
hhk.
Sure. Drop a line across B and D to form triangle BCD. Now drop a perpendicular line in between B and D. Let's call the middle point between B and D, point M. Line segment MC is the height of the triangle and in order to find the heigh of the triangle, all we have to do is apply the pythagorean theorem. We have a right triangle with sides 5 and 13, so:

(MC)^(2)+ (5)^(2)=(13)^(2)

169-25=144

MC=sqrt(144)=12

We have to use the pythagoren theorem again to find the base of the triangle. If you look closely, you can see that the two legs of the triangle
that has the base of triangle BCD as the hypotenuse are 6 and 8, so:

6^2+8^2= (BD)^(2)

100=(BD)^(2)

BD=10

Area of triangle BCD= 1/2*base*height= 1/2*10*12=60 square units.
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by kaulnikhil » Mon Jul 27, 2009 2:15 am
join b and d ... u get bd as 10
use herons formulae to calculate area of triangle bdc .. which comes to 60
now u have a triangle and a rectangle remaning with areas 16 and 24 respectively
add them u get 60
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by maihuna » Tue Jul 28, 2009 12:04 pm
I also got 100 fig attached
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Charged up again to beat the beast :)
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