Plane Geometry - ISEE Upper Level Quantitative Reasoning

Card 0 of 1116

Question

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

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

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

Answer

$\overline{TR} \parallel \overline{PA}$ ; $\angle T$ 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.

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Question

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) $m \angle R$

Answer

$\angle T$ 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 $m \angle P = 180 - m \angle T$

$m \angle R + m \angle A = 180$ , or $m \angle A = 180 - m \angle R$

Substitute:

$m \angle A > m\angle P$

$180 - m \angle R > 180 - m\angle T$

$180 - m \angle R -180 > 180 - m\angle T-180$

$- m \angle R > - m\angle T$

$- (- m \angle R )<-\left ( - m\angle T \right )$

$m \angle R < m\angle T$

(a) is the greater quantity

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Question

Consider trapezoid , where $\overline{TR} \parallel \overline{PA}$ . Also, $\angle A$ is acute and $\angle P$ is obtuse.

Which is the greater quantity?

(a) $m \angle T$

(b) $m \angle R$

Answer

$\angle T$ 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 same-side interior angles are supplementary. A pair of supplementary angles comprises either two right angles, or one acute angle and one obtuse angle. Since $\angle A$ is acute and $\angle P$ is obtuse, $\angle R$ is obtuse and $\angle T$ is acute. Therefore $\angle R$ the greater measure of the two, making (b) greater.

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Question

Which quantity is greater?

(a) The perimeter of the above trapezoid

(b) The perimeter of a rectangle with length and width and , respectively.

Answer

The perimeter of a rectangle is twice the sum of its length and its width:

Since the height of the trapezoid in the figure is , both of its legs must have length greater than or equal to . But for a leg to be of length, it must be perpendicular to the bases. Since perpendicularity of both legs would make the trapezoid a rectangle - which it cannot be - it follows that both legs cannot be of length . Therefore, the perimeter of the trapezoid is:

The perimeter of the trapezoid must be greater than that of the rectangle.

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Question

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 ?

Answer

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

.

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Question

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

Which is the greater quantity?

(a) Twice the perimeter of Trapezoid

(b) The perimeter of Trapezoid

Answer

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

and .

For reasons that will be apparent later, we will set

Also, 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 perimeter of Trapezoid is

Twice this is

The perimeter of Trapezoid is

and , so , making (a) the greater quantity.

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Question

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

Answer

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.

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Question

On Parallelogram , , locate point on $\overline{AB}$ such that ; locate point on $\overline{CD}$ such that . Draw $\overline{PQ}$ .

Which is the greater quantity?

(a) The area of Quadrilateral

(b) The area of Quadrilateral

Answer

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

(a) The bases of Trapezoid are $\overline{AP}$ and $\overline{DQ}$ .

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

Opposite sides of a parallelogram are congruent, so since , also.

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.

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Question

Which is the greater quantity?

(a) The area of a trapezoid with bases $3 \frac{1}{2}$ feet and $5 \frac{1}{4}$ feet and height one yard.

(b) The area of a parallelogram with base $4 \frac{1}{3}$ feet and height one yard.

Answer

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.

$3 \frac{1}{2} \times 12 = \frac{7}{2} \times 12 = 42$ inches

$5 \frac{1}{4} \times 12 = \frac{21}{4} \times 12 = 63$ inches

Substitute into the formula for the area of a trapezoid, setting :

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

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

$A = \frac{1}{2} \cdot 105 \cdot 36 = 1,890$ square inches

(b) The base of the parallelogram is

$4 \frac{1}{3} \times 12 = \frac{13}{3} \times 12 = 52$ .

Multiply this by the height:

$A = bh = 52 \cdot 36 = 1,872$ square inches

The trapezoid has greater area.

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Question

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.

Answer

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.

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Question

Which quantity is greater?

(a) The area of the above trapezoid

(b) The area of a square with sides of length

Answer

The area of a trapezoid is half the product of its height, which here is , and the sum of the lengths of its bases, which here are and :

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

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

$= \frac{1}{2} \cdot 4 \cdot x \cdot x$

$= 2x^{2}$

The area of a square is the square of the length of a side, which here is :

$A =( 2x) ^{2}= 2^{2}x^{2} = 4 x^{2}$

The square has the greater area.

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Question

Which quantity is greater?

(a) The area of the above trapezoid

(b) The area of a square with diagonals of length

Answer

The area of a trapezoid is half the product of its height, which here is , and the sum of the lengths of its bases, which here are and :

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

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

$= \frac{1}{2} \cdot 4 \cdot x \cdot x$

$= 2x^{2}$

The area of a square, it being a rhombus, is half the product of the lengths of its diagonals, both of which are here:

$A = \frac{1}{2} \cdot 2x \cdot 2x$

$= \frac{1}{2} \cdot 2 \cdot 2 \cdot x \cdot x$

$= 2x^{2}$

The trapezoid and the square have equal area.

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Question

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

Answer

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

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

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Question

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

Answer

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.

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Question

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

Answer

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.

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Question

Figure NOT drawn to scale.

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

Give the area of Trapezoid .

Answer

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.

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Question

In the above diagram, which depicts Trapezoid , and . Which is the greater quantity?

(a)

(b) 24

Answer

To see that (b) is the greater quantity of the two, it suffices to construct the midsegment of the trapezoid - the segment which has as its endpoints the midpoints of legs $\overline{TP }$ and $\overline{RA }$ . Since and , the midsegment, $\overline{MN }$ , is positioned as follows:

The length of the midsegment is half the sum of the bases, so

$MN = \frac{1}{2} (TR + AP )= \frac{1}{2} \cdot (12 + 36 ) = 24$

, so .

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Question

Figure NOT drawn to scale.

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

Give the area of Trapezoid in terms of .

Answer

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

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

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

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

Therefore,

The area of Trapezoid is one half multiplied by the height, , multiplied by the sum of the lengths of the bases, and . The midsegment of a trapezoid bisects both legs, so , and the area is

$A = \frac{1}{2} \cdot MP \cdot (MN + AP)$

$A = \frac{1}{2} \cdot 6 \cdot (20 + 40 - x )$

$A =3 \cdot (60 - x )$

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Question

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

Answer

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 .

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Question

Refer to the above figure, which shows a parallelogram. What is equal to?

Answer

The sum of two consecutive angles of a parallelogram is $180^{\circ }$ .

157 is the correct choice.

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