WCAB is a quadrilateral with WA=WC and AB=BC. determine the perimeter of WCAB if AC= 12cm and BP = 6cm and BP:PW = 1:2. if necessary give your answer correct to one decimal place.

Answer 1

Since WA = WC, we can let x represent this side length.

Since AB = BC, we can let y represent this side length.

Given that AC = 12cm, we can use the Pythagorean theorem to find the value of x:

x² = (AC/2)² + BP²
x² = (12cm/2)² + 6cm²
x² = 36cm² + 36cm²
x² = 72cm²
x = √(72cm²)
x = 8.5cm (rounded to one decimal place)

Given that BP:PW = 1:2, we can find the value of PW.

BP = 6cm
PW = 2 * BP
PW = 2 * 6cm
PW = 12cm

Now, we can find the value of y using the Pythagorean theorem:

y² = (AC/2)² + WC²
y² = (12cm/2)² + (x - PW)²
y² = 36cm² + (8.5cm - 12cm)²
y² = 36cm² + (-3.5cm)²
y² = 36cm² + 12.25cm²
y² = 48.25cm²
y = √(48.25cm²)
y = 6.9cm (rounded to one decimal place)

Finally, we can calculate the perimeter of the WCAB quadrilateral:

Perimeter = WA + AB + BC + CW
Perimeter = x + y + y + x
Perimeter = 8.5cm + 6.9cm + 6.9cm + 8.5cm
Perimeter = 30.8cm (rounded to one decimal place)

Therefore, the perimeter of the WCAB quadrilateral is approximately 30.8cm.