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ISEE Upper Level Quantitative : Geometry

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

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All ISEE Upper Level Quantitative Resources

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Example Questions

Example Question #1 : How To Find The Area Of A Trapezoid

Trapezoid

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

Which is the greater quantity?

(a) Three times the area of Trapezoid TRYX

(b) Twice the area of Trapezoid XYAP

Possible Answers:

(b) is the greater quantity

(a) and (b) are equal

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

(a) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

Midsegment $\overline{XY}$ divides Trapezoid TRAP 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 TRYX 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 XYAP 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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Example Question #9 : How To Find The Area Of A Trapezoid

Trapezoid 1

Figure NOT drawn to scale.

The above figure depicts Trapezoid TRAP with midsegment $\overline{MN}$ . MT = 6 , and RN = 9 .

Give the area of Trapezoid TRAP .

Possible Answers:

240

180

360

120

Correct answer:

240

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 TRAP , in particular, $\overline{TP }$ . Since MT = 6 , $TP = 2 \cdot MT = 12$ .

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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Example Question #1 : How To Find The Length Of The Side Of A Trapezoid

Trapezoid

In the above diagram, which depicts Trapezoid TRAP , TX = RY = 14 and XP = YA = 15 . Which is the greater quantity?

(a)

(b) 24

Possible Answers:

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

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

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 TX < XP and RY < YA , the midsegment, $\overline{MN }$ , is positioned as follows:

Trapezoid

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

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

XY < MN , so XY < 24 .

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Example Question #1 : How To Find The Length Of The Side Of A Trapezoid

Trapezoid 1

Figure NOT drawn to scale.

The above figure depicts Trapezoid TRAP with midsegment $\overline{MN}$ . MT = 6 , and RN = 9 .

Give the area of Trapezoid MNAP in terms of .

Possible Answers:

- 3x+ 180

3x- 60

3x + 180

3x+ 60

Correct answer:

- 3x+ 180

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 $\overline{AP }$ :

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

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

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

40 = x+ y

40 - x = y

Therefore, AP = 40 - x

The area of Trapezoid MNAP 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 MP = MT = 6 , and the area is

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

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

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

A =180 - 3x

A = - 3x+ 180

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Example Question #51 : Quadrilaterals

Trapezoid 1

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

Possible Answers:

y = 40 + x

y = 40 - x

y = 2 x + 20

y = 2x- 20

Correct answer:

y = 40 - x

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 $\overline{AP }$ :

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

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

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

40 = x+ y

40 - x = x+ y - x

40 - x = y

The correct choice is y = 40 - x .

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Example Question #52 : Quadrilaterals

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

Which is the greater quantity?

(a) $m\angle T$

(b) $m \angle R$

Possible Answers:

(a) is greater

It is impossible to tell from the information given

(b) is greater

(a) and (b) are equal

Correct answer:

(a) is greater

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

$m \angle R + m \angle A = 180$ , 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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Example Question #51 : Quadrilaterals

Consider trapezoid TRAP , 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) $m \angle R$

Possible Answers:

(a) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

(b) is greater.

Correct answer:

(b) is greater.

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 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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Example Question #53 : Quadrilaterals

Trapezoid

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

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

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

Possible Answers:

It is impossible to tell from the information given.

(b) is greater.

(a) is greater.

(a) and (b) are equal.

Correct answer:

(a) and (b) are equal.

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.

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Example Question #1 : How To Find The Area Of A Parallelogram

Parallelogram

In the above parallelogram, $\angle 1$ is acute. Which is the greater quantity?

(A) The area of the parallelogram

(B) 120 square inches

Possible Answers:

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

(B) is greater

(A) is greater

(A) and (B) are equal

Correct answer:

(B) is greater

Explanation:

Since $\angle 1$ is acute, a right triangle can be constructed with an altitude as one leg and a side as the hypotenuse, as is shown here. The height of the triangle must be less than its sidelength of 8 inches.

Parallelogram

The height of the parallelogram must be less than its sidelength of 8 inches.

The area of the parallelogram is the product of the base and the height - which is $A = 15 \cdot n$

Therefore,

n < 8

$15 \cdot n <15 \cdot 8$

A < 120

(B) is greater.

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Example Question #1 : Parallelograms

Parallelogram A is below:

Rhombus_1

Parallelogram B is below:

Parallelogram

Note: These figures are NOT drawn to scale.

Refer to the parallelograms above. Which is the greater quantity?

(A) The area of parallelogram A

(B) The area of parallelogram B

Possible Answers:

(A) and (B) are equal

(A) is greater

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

(B) is greater

Correct answer:

(A) and (B) are equal

Explanation:

The area of a parallelogram is the product of its height and its base; its slant length is irrelevant. Both parallelograms have the same height (8 inches) and the same base (1 foot, or 12 inches), so they have the same areas.

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ISEE Upper Level Quantitative : Geometry

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #1 : How To Find The Area Of A Trapezoid

Example Question #9 : How To Find The Area Of A Trapezoid

Example Question #1 : How To Find The Length Of The Side Of A Trapezoid

Example Question #1 : How To Find The Length Of The Side Of A Trapezoid

Example Question #51 : Quadrilaterals

Example Question #52 : Quadrilaterals

Example Question #51 : Quadrilaterals

Example Question #53 : Quadrilaterals

Example Question #1 : How To Find The Area Of A Parallelogram

Example Question #1 : Parallelograms

All ISEE Upper Level Quantitative Resources

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