Plane Geometry

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ISEE Upper Level Quantitative Reasoning › Plane Geometry

Questions 1 - 10

Sector SOW has a central angle of $45^{\circ}$ . What percentage of the circle does it cover?

Explanation

Sector SOW has a central angle of $45^{\circ}$ . What percentage of the circle does it cover?

Recall that there is a total of 360 degrees in a circle. SOW occupies 45 of them. To find the percentage, simply do the following:

$\frac{45^{\circ}}{360^{\circ}}*100=.125*100=12.5%$

The perimeter of a square is one yard. Which is the greater quantity?

(a) The area of the square

(b) $\frac{1}{2}$ square foot

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell form the information given.

Explanation

One yard is equal to three feet, so the length of one side of a square with this perimeter is $\frac{3}{4}$ feet. The area of the square is $\frac{3}{4}\times \frac{3}{4} = \frac{9}{16}$ square feet. $\frac{9}{16} > \frac{1}{2}$ , making (a) greater.

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

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

Find the circumference of a circle with a radius of 4cm.

$8 \pi \text{ cm}$

$4\pi \text{ cm}$

$12\pi \text{cm}$

$2\pi \text{ cm}$

$16\pi \text{ cm}$

Explanation

To find the circumference of a circle, we will use the following formula:

$C = 2 \cdot \pi \cdot r$

where r is the radius of the circle.

Now, we know the radius of the circle is 4cm.

Knowing this, we can substitute into the formula. We get

$C = 2 \cdot \pi \cdot 4\text{cm}$

$C = 2\pi \cdot 4\text{cm}$

$C = 8\pi \text{ cm}$

Sector

Give the area of the white region of the above circle if $\overarc{AB}$ has length $12 \pi$ .

$567 \pi$

$81 \pi$

$729 \pi$

$243 \pi$

Explanation

If we let be the circumference of the circle, then the length of $\overarc{AB}$ is $\frac{80}{360} = \frac{2}{9 }$ of the circumference, so

$\frac{2}{9 } C = 12 \pi$

$\frac{9 }{2}\cdot \frac{2}{9 } C =\frac{9 }{2}\cdot 12 \pi$

$C =54\pi$

The radius is the circumference divided by $2 \pi$ :

$r= 54 \pi \div 2 \pi = 27$

Use the formula to find the area of the entire circle:

$A = \pi r ^{2} = \pi \left (27 \right )^{2} = 729 \pi$

The area of the white region is $1 - \frac{2}{9} = \frac{7}{9}$ of that of the circle, or

$\frac{7}{9} \cdot 729 \pi = 567 \pi$

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.

In Rhombus , and . Which is the greater quantity?

(A)

(B)

(a) is the greater quantity

(b) is the greater quantity

It cannot be determined which of (a) and (b) is greater

(a) and (b) are equal

Explanation

The four sides of a rhombus, by defintion, have equal length, so

$7x\div 7 = 8y \div 7$

$x = \frac{8}{7}y$

Since and are positive, .

One side of a regular pentagon is 20% longer than one side of a regular hexagon. Which is the greater quantity?

(A) The perimeter of the pentagon

(B) The perimeter of the hexagon

(A) and (B) are equal

(B) is greater

(A) is greater

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

Explanation

Let be the length of one side of the hexagon. Then its perimeter is .

Each side of the pentagon is 20% greater than this length, or

The perimeter is five times this, or $5 \cdot 1.2s = 6s$ .

The perimeters are the same.

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

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

A rectangle has perimeter 140 inches and area 1,200 square inches. Which is the greater quantity?

(A) The length of a diagonal of the rectangle.

(B) 4 feet

(A) is greater

(B) is greater

(A) and (B) are equal

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

Explanation

Let and be the dimensions of the rectangle. Then

and, subsequently,

Since the product of the length and width is the area, we are looking for two numbers whose sum is 70 and whose product is 1,200; through trial and error, they are found to be 30 and 40. We can assign either to be and the other to be since the result is the same.