In the figure, ABC and ADE are right-angled isosceles triangles. BD=CE=2cm. The area of the shaded part is 18cm². Find the length of AB.

I’m trying to find a method to solve this question other than guess and check. Can anyone help me?

In the figure, ABC and ADE are right-angled isosceles triangles. BD=CE=2cm. The area of the shaded part is 18cm². Find the length of AB.

I’m trying to find a method to solve this question other than guess and check. Can anyone help me?

mr.tan

Solution 1:Consider the blue rectangle BFGA. It has the same height and width as parallelogram BFEC. So they both have the same area, 2k. The triangle DBF has area 2 x 2 x 0.5 = 2. Orange strip has area 18 = DBF + BFEC = 2 + 2k, which implies k = 8. Therefore AB is 8cm. Check: 0.5 x 10 x 10 - 0.5 x 8 x 8= 50 - 32 = 18

See: https://www.geogebra.org/classic/rrztmrbv

Solution 2: (if you know algebra)Area ADE = 0.5 x (k+2)**2 = 0.5 (k**2 + 2k + 4) = 0.5 k**2 + 2k + 2 Area ABC = 0.5 x k**2 So orange strip has area 18 = ADE - ABC = 0.5 k**2 + 2k + 2 - 0.5 x k**2 = 2k + 2, which implies k = 8. So AB is 8cm. The equation is essentially the same 2k+2 = 8 in both solutions. The algebra is interpreted as area in Solution 1.

colemanmaths

thanks

mr.tan

You are welcomed!

Here is a simpler solution which involves drawing 1 auxiliary line:

Area of orange triangle = 1/2 x 2 x (2+k) = 2 + k Area of blue triangle = 1/2 x 2 x k = k So 2 + k + k = 18 and hence k = 8.

waterflow

Extend line BC and draw another horizontal line FD to form parallelogram CEFD as shown in diagram

Let AB=AC=k and

Area of shape BCDE = Area of Parallelogram CEFD – Area of triangle FBD

18 = 2(k+2) – ½*2*2

18 = 2k +4 -2

2k = 16

K = 8

So length of AB is 8cm