The length of the midsegment of a trapezoid is half sum of the lengths of the bases, so

\(\displaystyle XY = \frac{1}{2} (12 + 40) = \frac{1}{2} \cdot 52 = 26\).

Also, by definition, since Trapezoid \(\displaystyle TRAP\) is isosceles, \(\displaystyle TP = RA\). The midsegment divides both legs of Trapezoid \(\displaystyle TRAP\) into congruent segments; combining these facts:

\(\displaystyle TX = XP = \frac{1}{2} TP = \frac{1}{2} RA = RY = YA\)

\(\displaystyle TX = \frac{1}{2} XY = \frac{1}{2} \cdot 26 = 13\).

\(\displaystyle XP = YA= TX = 13\), so the perimeter of Trapezoid \(\displaystyle XYAP\) is 

\(\displaystyle XP+XY + YA + AP = 13+26+13+40 = 92\).

The midsegment of a trapezoid bisects both of its legs, so 

\(\displaystyle TX = XP\)  and  \(\displaystyle RY = YA\).

For reasons that will be apparent later, we will set

\(\displaystyle z = TX + RY = X P + YA\)

Also, the length of the midsegment is half sum of the lengths of the bases:

\(\displaystyle XY = \frac{1}{2} (12 + 36) = \frac{1}{2} \cdot 48= 24\).

The perimeter of  Trapezoid \(\displaystyle TRYX\) is 

\(\displaystyle TR + RY + YX + TX = TR + YX +( TX + RY) = 12+24+ z = 36 + z\)

Twice this is 

\(\displaystyle 2 (36 + z) = 72 + 2z\)

The  perimeter of  Trapezoid \(\displaystyle XYAP\) is

\(\displaystyle XY + YA + AP + XP = XY + AP + (XP+ YA) = 24 + 36 + z = 60 + z\)

\(\displaystyle 72 > 60\) and \(\displaystyle 2z > z\), so \(\displaystyle 72 + 2z > 60+ z\), making (a) the greater quantity.

Let \(\displaystyle h\) be the common height of the figures.

(a) The area of Trapezoid A is \(\displaystyle A = \frac{1}{2} (B+b) h = \frac{1}{2} (16+10) h = \frac{1}{2} (26) h= 13h\).

(b) The area of Parallelogram B is

\(\displaystyle A = bh = 13h\).

The figures have the same area.

\(\displaystyle \overline{PQ}\) divides the parallelogram into two trapezoids, each of which has the same height as the original parallelogram, which we will call \(\displaystyle h\). 

(a) The bases of Trapezoid \(\displaystyle APQD\) are \(\displaystyle \overline{AP}\) and \(\displaystyle \overline{DQ}\). \(\displaystyle AP = 13, DQ = 8\)

(b) The bases of Trapezoid \(\displaystyle BPQC\) are \(\displaystyle \overline{PB}\) and \(\displaystyle \overline{QC}\).

Opposite sides of a parallelogram are congruent, so since \(\displaystyle AB = 20\), \(\displaystyle DC= 20\) also.

\(\displaystyle PB = AB -AP = 20 - 13 = 7\)

\(\displaystyle QC = DC - DQ = 20-8 = 12\)

The sum of the bases of Trapezoid A is 21; the sum of those of Trapezoid B is 19. The two trapezoids have the same height. Thereforee, since the area is one-half times the height times the sum of the bases, Trapezoid A will have the greater area.

The easiet way to compare is to convert each measure to centimeters and calculate the areas in square centimeters. Both figures have height one meter, or 100 centimeters.

(a) Substitute \(\displaystyle B = 85, b = 75, h = 100\) into the formula for area:

\(\displaystyle A = \frac{1}{2} (B + b) h\)

\(\displaystyle A = \frac{1}{2} \cdot (85 + 75 ) \cdot 100\)

\(\displaystyle A = \frac{1}{2} \cdot 160 \cdot 100\)'

\(\displaystyle A = 8,000\) square centimeters

(b) 8 decimeters is equal to 80 centimeters, so multiply this base by a height of 100 centimeters:

\(\displaystyle A = bh = 80 \cdot 100 = 8,000\) square centimeters

The figures have the same area.

The easiest way to compare the areas might be to convert each of the dimensions to inches.

(a) The bases convert by multiplying the number of feet by twelve; the height is one yard, which is 36 inches.

\(\displaystyle 3 \frac{1}{2} \times 12 = \frac{7}{2} \times 12 = 42\) inches

\(\displaystyle 5 \frac{1}{4} \times 12 = \frac{21}{4} \times 12 = 63\) inches

Substitute into the formula for the area of a trapezoid, setting \(\displaystyle B = 63, b = 42, h = 36\):

\(\displaystyle A = \frac{1}{2} (B + b) h\)

\(\displaystyle A = \frac{1}{2} \cdot (63 + 42) \cdot 36\)

\(\displaystyle A = \frac{1}{2} \cdot 105 \cdot 36 = 1,890\) square inches

(b) The base of the parallelogram is 

\(\displaystyle 4 \frac{1}{3} \times 12 = \frac{13}{3} \times 12 = 52\).

Multiply this by the height:

\(\displaystyle A = bh = 52 \cdot 36 = 1,872\) square inches

The trapezoid has greater area.

The area of a trapezoid is half the product of its height, which here is \(\displaystyle x\), and the sum of the lengths of its bases, which here are \(\displaystyle x\) and \(\displaystyle 3x\):

\(\displaystyle A = \frac{1}{2} \cdot x \cdot (3x+x)\)

\(\displaystyle = \frac{1}{2} \cdot x \cdot (4x)\)

\(\displaystyle = \frac{1}{2} \cdot 4 \cdot x \cdot x\)

\(\displaystyle = 2x^{2}\)

The area of a square is the square of the length of a side, which here is \(\displaystyle 2x\):

\(\displaystyle A =( 2x) ^{2}= 2^{2}x^{2} = 4 x^{2}\)

The square has the greater area.

The area of a trapezoid is half the product of its height, which here is \(\displaystyle x\), and the sum of the lengths of its bases, which here are \(\displaystyle x\) and \(\displaystyle 3x\):

\(\displaystyle A = \frac{1}{2} \cdot x \cdot (3x+x)\)

\(\displaystyle = \frac{1}{2} \cdot x \cdot (4x)\)

\(\displaystyle = \frac{1}{2} \cdot 4 \cdot x \cdot x\)

\(\displaystyle = 2x^{2}\)

The area of a square, it being a rhombus, is half the product of the lengths of its diagonals, both of which are \(\displaystyle 2x\) here:

\(\displaystyle A = \frac{1}{2} \cdot 2x \cdot 2x\)

\(\displaystyle = \frac{1}{2} \cdot 2 \cdot 2 \cdot x \cdot x\)

\(\displaystyle = 2x^{2}\)

The trapezoid and the square have equal area.

Midsegment \(\displaystyle \overline{XY}\) divides Trapezoid \(\displaystyle TRAP\) into two trapezoids of the same height, which we will call \(\displaystyle h\); the length of the midsegment is half sum of the lengths of the bases:

\(\displaystyle XY = \frac{1}{2} (12 + 36) = \frac{1}{2} \cdot 48= 24\)

The area of a trapezoid is one half multiplied by its height multiplied by the sum of the lengths of its bases. Therefore, the area of Trapezoid \(\displaystyle TRYX\) - the shaded trapezoid - is

\(\displaystyle \frac{1}{2} \cdot h \cdot (TR+XY ) = \frac{1}{2} \cdot h \cdot (12+24) = \frac{1}{2} \cdot 36 \cdot h = 18 h\)

The area of Trapezoid \(\displaystyle TRAP\) is

\(\displaystyle \frac{1}{2} \cdot 2 h \cdot (TR+AP) = \frac{1}{2} \cdot 2 \cdot h \cdot (12+36) = \frac{1}{2} \cdot 2 \cdot 48 \cdot h = 48 h\)

The percent of Trapezoid \(\displaystyle TRAP\) that is shaded in is

\(\displaystyle \frac{18h}{48h} \cdot 100 \% = \frac{18}{48} \cdot 100 \% = 37 \frac{1}{2} \%\)

Midsegment \(\displaystyle \overline{XY}\) divides Trapezoid \(\displaystyle TRAP\) into two trapezoids of the same height, which we will call \(\displaystyle h\); the length of the midsegment is half sum of the lengths of the bases:

\(\displaystyle XY = \frac{1}{2} (12 + 40) = \frac{1}{2} \cdot 52 = 26\).

The area of a trapezoid is one half multiplied by its height multiplied by the sum of the lengths of its bases. Therefore, the area of Trapezoid \(\displaystyle TRYX\) is

\(\displaystyle \frac{1}{2} \cdot h \cdot (TR+XY ) = \frac{1}{2} \cdot h \cdot (12+26 ) = \frac{1}{2} \cdot 38 \cdot h = 19 h\)

The area of Trapezoid \(\displaystyle XYAP\) is

\(\displaystyle \frac{1}{2} \cdot h \cdot (XY +AP) = \frac{1}{2} \cdot h \cdot ( 26+40 ) = \frac{1}{2} \cdot 66 \cdot h = 33 h\)

The ratio of the areas is 

\(\displaystyle \frac{33 h}{19h} = \frac{33}{19}\), or 33 to 19.