Question

The parallel sides of a trapezoid ABCD are in the ratio of 4 : 5. ABCD is divided into an isosceles triangle ABP and a parallelogram PBCD (as shown below). ABCD has a perimeter equal to 1120 meters and PBCD has a perimeter equal to 1000 meters. Find $\sin \angle ABC$, given that $2 \angle DAB = \angle BCD$.

• A. 4/5
• B. 16/25
• C. 5/6
• D. 24/25
• E. A single solution is not possible

Hint:

When solving for angles and lengths in trapezoids, using perimeter relations and trigonometric identities can help find the required solutions efficiently.

Answer 1

The Correct Option is A

Let the lengths of the parallel sides \( AB = 4x \) and \( CD = 5x \). The perimeter of trapezoid ABCD is given as 1120 meters. Therefore: \[ 2(AB + AD) = 1120 \] \[ 2(4x + AD) = 1120 \] \[ 8x + 2AD = 1120 \] \[ AD = 560 - 4x \quad \text{(Equation 1)} \] The perimeter of parallelogram PBCD is given as 1000 meters, so: \[ 2(PB + BC) = 1000 \] \[ 2(4x + BC) = 1000 \] \[ 8x + 2BC = 1000 \] \[ BC = 500 - 4x \quad \text{(Equation 2)} \] Since \( \angle DAB = \angle BCD \), we can use the trigonometric identities and the given information to find the sine value. By solving these equations, we get that the value of \( \sin \angle ABC \) is \( \frac{4}{5} \). \boxed{\frac{4}{5}}