Trapezoids

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Trapezoid

In the above figure, $\overline{XY}$ is the midsegment of Trapezoid . What percent of Trapezoid has been shaded in?

$37 \frac{1}{2} %$

$33 \frac{1}{3} %$

Explanation

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

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

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 - the shaded trapezoid - is

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

The area of Trapezoid is

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

The percent of Trapezoid that is shaded in is

$\frac{18h}{48h} \cdot 100 % = \frac{18}{48} \cdot 100 % = 37 \frac{1}{2} %$

You recently bought a new bookshelf with a base in the shape of an isosceles trapezoid. If the small base is 2 feet, the large base is 3 feet, and the arms are 1.5 feet, what is the perimeter of the base of your new bookshelf?

Cannot be determined from the information provided.

Explanation

To find the perimeter of a bookshelf, we need to add up the lengths of the sides.

We know the two bases, we just need to add the lengths of the arms.

So, our answer is 8ft

Cannot be determined from the information provided.

Explanation

To find the perimeter of a bookshelf, we need to add up the lengths of the sides.

We know the two bases, we just need to add the lengths of the arms.

So, our answer is 8ft

Trapezoid

In the above figure, $\overline{XY}$ is the midsegment of Trapezoid . Give the ratio of the area of Trapezoid to that of Trapezoid .

33 to 19

10 to 3

13 to 6

20 to 13

Explanation

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

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

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 is

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

The area of Trapezoid is

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

The ratio of the areas is

$\frac{33 h}{19h} = \frac{33}{19}$ , or 33 to 19.

Trapezoid

In the above figure, $\overline{XY}$ is the midsegment of Trapezoid . Give the ratio of the area of Trapezoid to that of Trapezoid .

33 to 19

10 to 3

13 to 6

20 to 13

Explanation

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

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

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 is

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

The area of Trapezoid is

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

The ratio of the areas is

$\frac{33 h}{19h} = \frac{33}{19}$ , or 33 to 19.

Trapezoid

In the above figure, $\overline{XY}$ is the midsegment of Trapezoid .

Which is the greater quantity?

(a) Three times the area of Trapezoid

(b) Twice the area of Trapezoid

(b) is the greater quantity

(a) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

Explanation

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

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

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 is

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

Three times this is

$3 \cdot 18h = 54 h$ .

The area of Trapezoid is, similarly,

$\frac{1}{2} \cdot h \cdot (XY +AP ) = \frac{1}{2} \cdot h \cdot (24+36 ) = \frac{1}{2} \cdot 60 \cdot h = 30 h$

Twice this is

$2 \cdot 30 h = 60 h$ .

That makes (b) the greater quantity.

Trapezoid 1

The above figure depicts Trapezoid with midsegment $\overline{MN }$ . Express in terms of .

Explanation

The midsegment of a trapezoid has as its length half the sum of the lengths of the bases, which here are $\overline{TR}$ and $\overline{AP }$ :

$MN = \frac{1}{2} (TR + AP )$

$20 = \frac{1}{2} (x+ y)$

$20 \cdot 2 = \frac{1}{2} (x+ y) \cdot 2$

The correct choice is .

Trapezoid

The above diagram depicts trapezoid . Which is the greater quantity?

(a) $m \angle T + m \angle P$

(b) $m \angle R + m \angle A$

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

Explanation

$\overline{TR} \parallel \overline{PA}$ ; $\angle T$ and $\angle P$ are same-side interior angles, as are $\angle R$ and $\angle A$ .

The Same-Side Interior Angles Theorem states that if two parallel lines are crossed by a transversal, then the sum of the measures of a pair of same-side interior angles is always $180^{\circ}$ .

Therefore, $m \angle T + m \angle P = 180= m \angle R + m \angle A$ , making the two quantities equal.

Trapezoid

Figure NOT drawn to scale.

In the above figure, $\overline{XY}$ is the midsegment of isosceles Trapezoid . Also, $TX = \frac{1}{2} XY$ .

What is the perimeter of Trapezoid ?

Explanation

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

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

Also, by definition, since Trapezoid is isosceles, . The midsegment divides both legs of Trapezoid into congruent segments; combining these facts:

$TX = XP = \frac{1}{2} TP = \frac{1}{2} RA = RY = YA$

$TX = \frac{1}{2} XY = \frac{1}{2} \cdot 26 = 13$ .

, so the perimeter of Trapezoid is

Given Trapezoid , where $\overline{TR} \parallel \overline{PA}$ . Also, $m \angle A > m\angle P$

Which is the greater quantity?

(a) $m\angle T$

(b) $m \angle R$

(a) is greater

(b) is greater

(a) and (b) are equal

It is impossible to tell from the information given

Explanation

$\angle T$ and $\angle P$ are same-side interior angles, as are $\angle R$ and $\angle A$ .

The Same-Side Interior Angles Theorem states that if two parallel lines are crossed by a transversal, then the sum of the measures of a pair of same-side interior angles is always $180^{\circ}$ . Therefore,

$m \angle T + m \angle P = 180$ , or $m \angle P = 180 - m \angle T$