Question

figure abcd is a parallelogram. points x and y are placed so that (overline{bx}congoverline{dy}) and (overline{cd}perpoverline{xy}). the area of bxyc is 71.5 square units. what is the area of abcd? 99 square units 110 square units 126 square units 143 square units

Explanation:

Step1: Find the length of \( BX \)

Use the trapezoid area formula \( A=\frac{(a + b)h}{2} \) for \( BXYC \). Let \( BX=x \). Here \( a = x \), \( b = 4 \), \( h = 11 \), and \( A=71.5 \).

$$71.5=\frac{(x + 4)\times11}{2}$$

Multiply both sides by \( 2 \): \( 143=(x + 4)\times11 \).
Divide both sides by \( 11 \): \( x+4 = 13 \), so \( x=BX = 9 \).

Step2: Find the length of \( AB \)

Since \( ABCD \) is a parallelogram, \( AB=CD \). And \( CD=DY + YC \), \( BX\cong DY \), so \( DY = 9 \), \( YC = 4 \), then \( CD=9 + 4=13 \).

Step3: Calculate the area of parallelogram \( ABCD \)

Use the parallelogram area formula \( A = base\times height \). The base \( CD = 13 \) and the height (distance between \( AB \) and \( CD \)) is \( 11 \).

$$A=13\times11=143$$

Answer:

143 square units