area of the quadi;lateral

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.

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 could not understand why are you dividing by 2?

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.

stop@800 wrote:

parallel_chase wrote:IMO 100.
base or BC/2 = 10/2 = 5

I could not understand why are you dividing by 2?

I used heroes formula and got the answer but would like to know this method also.

Thanks!!

Well I divided BC not by 2 but I divided into half which more or less means the same

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.

6 - 8 - 10
5 - 12 - 13

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

parallel_chase wrote:

stop@800 wrote:

parallel_chase wrote:IMO 100.
base or BC/2 = 10/2 = 5

I could not understand why are you dividing by 2?

I used heroes formula and got the answer but would like to know this method also.

Thanks!!

Well I divided BC not by 2 but I divided into half which more or less means the same

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 agree
I did not knew this bisection logic.
Thanks a lot.