IMO 100.
Kindly refer the below image.
AX = 2
XB = 8-2 = 6
Area of Rectangle AXDE = 8*2 = 16
Area of triangle XBD = 1/2 * XB * XD = 1/2 *6* 8 = 24
Length of BD
BD^2 = XB^ + DX^2
Either you can solve it by long method above or if you just notice that this forms one of the pythagorean triplets
base = 6, height = 8, hypoteneuse = 10
BD = 10
Height of traingle BCD
CD = 13
base or BC/2 = 10/2 = 5
Using the pythagoreous theorem or we could again use the pythegorean triplets
Base = 5
height = 12
hypotenuese = 13
area triangle BCD = 1/2 * height* BC = 1/2 * 12 * 10 = 60
Total area = 60 + 16 + 24 = 100.
Hope this helps.
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area of the quadi;lateral
Source: Beat The GMAT — Problem Solving |
I could not understand why are you dividing by 2?parallel_chase wrote:IMO 100.
Kindly refer the below image.
AX = 2
XB = 8-2 = 6
Area of Rectangle AXDE = 8*2 = 16
Area of triangle XBD = 1/2 * XB * XD = 1/2 *6* 8 = 24
Length of BD
BD^2 = XB^ + DX^2
Either you can solve it by long method above or if you just notice that this forms one of the pythagorean triplets
base = 6, height = 8, hypoteneuse = 10
BD = 10
Height of traingle BCD
CD = 13
base or BC/2 = 10/2 = 5
I used heroes formula and got the answer but would like to know this method also.
Thanks!!
Using the pythagoreous theorem or we could again use the pythegorean triplets
Base = 5
height = 12
hypotenuese = 13
area triangle BCD = 1/2 * height* BC = 1/2 * 12 * 10 = 60
Total area = 60 + 16 + 24 = 100.
Hope this helps.
Well I divided BC not by 2 but I divided into half which more or less means the samestop@800 wrote:I could not understand why are you dividing by 2?parallel_chase wrote:IMO 100.
base or BC/2 = 10/2 = 5
I used heroes formula and got the answer but would like to know this method also.
Thanks!!
the height (which is perpendicular bisector drawn from angle C) divides the base into two equal parts. This way you can use the Pythagoras theorem to find the height and hence find the area.
Hope this clears your doubt.
Also, I am not a big fan of Heron's formula especially on this one, since I was able to this in less than 1 minute without touching the pen and just using the pythagorean triplets.
No rest for the Wicked....
Also we can find out the area of trapezium BADE -1 /2 (sum of parallel sides * distance between them )+ area of the triangle BCD( sq rt s ( s-a)(s-b)(s-c) ) where s is a+ b+ c /2
I agreeparallel_chase wrote:Well I divided BC not by 2 but I divided into half which more or less means the samestop@800 wrote:I could not understand why are you dividing by 2?parallel_chase wrote:IMO 100.
base or BC/2 = 10/2 = 5
I used heroes formula and got the answer but would like to know this method also.
Thanks!!
the height (which is perpendicular bisector drawn from angle C) divides the base into two equal parts. This way you can use the Pythagoras theorem to find the height and hence find the area.
Hope this clears your doubt.
Also, I am not a big fan of Heron's formula especially on this one, since I was able to this in less than 1 minute without touching the pen and just using the pythagorean triplets.
I did not knew this bisection logic.
Thanks a lot.
