Question

Four identical isosceles triangle AWB, BXC, CYD and DZE are arranged,
as shown, with points A, B, C, D and E lying on the same straight line. A
new triangle is formed with sides the same length as AX, AY and AZ. If
AZ = AE, the area of this new triangle in terms of x is euqal to
​

Answer 1

Answer:

x²√15

Step-by-step explanation:

AB = BC = CD = DE = x

Let say AW = BW = BX = CX = CY = DY = DZ = EZ = Y

and equal angle of isosceles = α

AX² = AB² + BY² - 2AB.BYCos(180 - α)

=> AX² = x² + y² + 2xyCosα    (Cos(180 - α) =  - cosα)

Similarly

AY² = AC² + CY² - 2AC.CYCos(180 - α)

=> AY² = (2x)² + y² + 2(2x)ycosα

=> AY² = 4x² + y² + 4xycosα

AZ² = 9x² + y² + 6xycosα

AZ = AE = 4x

16x² = 9x² + y² + 6xyCosα

AZ² = AE² + EZ² -2AE.EZCosα

=> (4x)² = (4x)² + y² - 2(4x)yCosα

=> y² = 8xyCosα

Using this value

16x² = 9x² + 8xyCosα + 6xyCosα

=> 7x² = 14xyCosα

=>  x² = 2xyCosα

y²/x² = 4

=> y² = 4x²

using  y² = 4x² & x² = 2xyCosα

AX² = x² + 4x² + x² = 6x²

AY² = 4x² + 4x² + 2x² = 10x²

AZ² = 9x² + 4x² + 3x² = 16x²

AX = x√6 , AY = x√10  & AZ = 4x

s = x(4 + √6 + √10)/2  

Using Hero's formula

Area = √s(s-AX)(s-AY)(s-AZ)

Solving this we get

= x²√15