Question

the perimeter p of a rectangle is found by the twice the sum of the length l and width w. then, solve for w. start with: p = 2(l + w) solve for w: the area of a trapezoid is \\(\frac{1}{2}\\) times the height h multiplied by the sum of the bases, \\(b_1\\) and \\(b_2\\). solve for \\(b_1\\). start with: \\(a = \frac{1}{2}h(b_1 + b_2)\\) solve for \\(b_1\\):

Explanation:

Part 1: Solve for \( w \) in \( P = 2(L + w) \)

Step 1: Divide both sides by 2

To isolate the term with \( w \), we divide both sides of the equation \( P = 2(L + w) \) by 2.
\( \frac{P}{2} = L + w \)

Step 2: Subtract \( L \) from both sides

Now, we subtract \( L \) from both sides to solve for \( w \).
\( w = \frac{P}{2} - L \)

Step 1: Multiply both sides by 2

To eliminate the fraction, we multiply both sides of the equation \( A = \frac{1}{2}h(b_1 + b_2) \) by 2.
\( 2A = h(b_1 + b_2) \)

Step 2: Divide both sides by \( h \)

Next, we divide both sides by \( h \) to isolate the term with \( b_1 \).
\( \frac{2A}{h} = b_1 + b_2 \)

Step 3: Subtract \( b_2 \) from both sides

Finally, we subtract \( b_2 \) from both sides to solve for \( b_1 \).
\( b_1 = \frac{2A}{h} - b_2 \)

Answer:

\( w = \frac{P}{2} - L \)

Part 2: Solve for \( b_1 \) in \( A = \frac{1}{2}h(b_1 + b_2) \)