ISEE Upper Level Quantitative Reasoning - How to find the area of a trapezoid

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 . 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

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

Trapezoidr

Find the area of the above trapezoid if $a=12\ cm$ , $b=14\ cm$ , and $h=10\ cm$ .

Figure not drawn to scale.

$130\ cm^2$

$120\ cm^2$

$110\ cm^2$

$100\ cm^2$

$90\ cm^2$

Explanation

The area of a trapezoid is given by

$Area=(\frac{b_{1}+b_{2}}{2})h$ ,

where , are the lengths of each base and is the altitude (height) of the trapezoid.

$b_{1}=a=12\ cm$

$b_{2}=b=14\ cm$

$h=10\ cm$

$Area=(\frac{b_{1}+b_{2}}{2})h=(\frac{12+14}{2})\times 10=13\times 10=130\ cm^2$

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.

Which is the greater quantity?

(a) The area of a trapezoid with bases 75 centimeters and 85 centimeters and height one meter.

(b) The area of a parallelogram with base 8 decimeters and height one meter.

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

Explanation

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 into the formula for area:

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

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

$A = \frac{1}{2} \cdot 160 \cdot 100$ '

square centimeters

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

$A = bh = 80 \cdot 100 = 8,000$ square centimeters

The figures have the same area.

A trapezoid has the base lengths of and . The area of the trapezoid is . Give the height of the trapezoid in terms of .

Explanation

The area of a trapezoid is given by

$Area=(\frac{b_{1}+b_{2}}{2})h$ ,

where $b_{}1$ , $b_{}2$ are the lengths of each base and is the altitude (height) of the trapezoid.

$Area=(\frac{b_{1}+b_{2}}{2})h=(\frac{3t+t}{2})\times h=8t^2\Rightarrow (2t)h=8t^2$

$\Rightarrow h=\frac{8t^2}{2t}=4t$

Trapezoid 1

Figure NOT drawn to scale.

The above figure depicts Trapezoid with midsegment $\overline{MN}$ . , and .

Give the area of Trapezoid .

Explanation

One way to calculate the area of a trapezoid is to multiply the length of its midsegment, which is 20, and its height, which here is

Midsegment $\overline{MN}$ bisects both legs of Trapezoid , in particular, $\overline{TP }$ . Since , $TP = 2 \cdot MT = 12$ .

Therefore, the area of the trapezoid is

$A = TP \cdot MN = 12 \cdot 20 = 240$

Note that the length of $\overline{RN}$ is irrelevant to the problem.

In the following trapezoid and . The area of the trapezoid is 54 square inches. Give the height of the trapezoid. Figure not drawn to scale.

Trap

$6\ in$

$5\ in$

$4\ in$

$7\ in$

$3\ in$

Explanation

The area of a trapezoid is given by

$Area=(\frac{b_{1}+b_{2}}{2})h$ ,

where $b_{}1$ , $b_{}2$ are the lengths of each base and is the altitude (height) of the trapezoid.

$b_{1}=a$

$b_{2}=b=2a$

Substitute these values into the area formula:

$Area=(\frac{b_{1}+b_{2}}{2})h=(\frac{a+2a}{2})a=54\Rightarrow \frac{3a^2}{2}=54\Rightarrow a^2=36\Rightarrow a=6\ in$

Trapezoid

Note: Figure NOT drawn to scale.

The above trapezoid has area . What is ?

Explanation

Substitute in the formula for the area of a trapezoid:

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

$\frac{1}{2} (16 + 12)h = 98$

$\frac{1}{2} (28)h = 98$

$14h \div 14= 98 \div 14$

Trapezoid A and Parallelogram B have the same height. Trapezoid A has bases 10 and 16; Parallelogram B has base 13. Which is the greater quantity?

(a) The area of Trapezoid A

(b) The area of Parallelogram B

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

Explanation

Let be the common height of the figures.

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

(b) The area of Parallelogram B is

The figures have the same area.

How to find the area of a trapezoid

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