ISEE Upper Level Quantitative Reasoning - Geometry

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

In Parallelogram , and . Which of the following is greater?

(A)

(B)

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

(b) is the greater quantity

(a) is the greater quantity

(a) and (b) are equal

Explanation

In Parallelogram , $\overline{AB}$ and $\overline{BC }$ are adjoining sides; there is no specific rule for the relationship between their lengths. Therefore, no conclusion can be drawn of and , and no conclusion can be drawn of the relationship between and .

$\Delta ABC$ and $\Delta DEF$ are right triangles, with right angles $\angle B , \angle E$ , respectively. $m \angle A = m\angle D = 45^{\circ }$ and .

Which is the greater quantity?

(a)

(b)

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

Each right triangle is a $45 ^{\circ }-45 ^{\circ }-90 ^{\circ }$ triangle, making each triangle isosceles by the Converse of the Isosceles Triangle Theorem.

Since $\angle B$ and $\angle E$ are the right triangles, the legs are $\overline{AB}, \overline{BC}, \overline{DE}, \overline{EF}$ , and the hypotenuses are $\overline{AC}, \overline{DF}$ .

By the $45 ^{\circ }-45 ^{\circ }-90 ^{\circ }$ Theorem, $AB \sqrt{2} = AC$ and $EF \sqrt{2} = DF$ .

, so $AB \sqrt{2} > EF \sqrt{2}$ and subsequently, .

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:

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

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

(A) The perimeter of the pentagon

(B) The perimeter of the hexagon

(A) is greater

(A) and (B) are equal

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

(B) is greater

Explanation

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

Each side of the hexagon is 20% less than this length, or

The perimeter is five times this, or $6 \cdot 0.8s = 4.8s$ .

Since and is positive, , so the pentagon has greater perimeter, and (A) is greater.

Which is the greater quantity?

(a) The sum of the measures of the exterior angles of a thirty-sided polygon, one per vertex

(b) The sum of the measures of the exterior angles of a forty-sided polygon, one per vertex

(a) and (b) are equal

It is impossible to tell from the information given

(a) is greater

(b) is greater

Explanation

The Polygon Exterior-Angle Theorem states that the sum of the measures of the exterior angles of any polygon, one per vertex, is $360^{\circ }$ . This makes both quantities equal.

A wooden ball has a surface area of $900\pi m^2$ .

What is its radius?

Cannot be determined from the information provided

Explanation

A wooden ball has a surface area of $900\pi m^2$ .

What is its radius?

Begin with the formula for surface area of a sphere:

$SA_{sphere}=4\pi r^2$

Now, plug in our surface area and solve with algebra:

$900 \pi m^2=4\pi r^2$

Get rid of the pi

Divide by 4

Square root both sides to get our answer:

The circumferences of eight circles form an arithmetic sequence. The smallest circle has radius two inches; the second smallest circle has radius five inches. Give the radius of the largest circle.

1 foot, 11 inches

2 feet

2 feet, 1 inch

4 feet 2 inches

3 feet 10 inches

Explanation

The circumference of a circle can be determined by multiplying its radius by $2 \pi$ , so the circumferences of the two smallest circles are

$C_{1} = 2 \cdot 2 \pi = 4 \pi$

and

$C_{2} = 2 \cdot 5 \pi = 10 \pi$

The circumferences form an arithmetic sequence with common difference

$C_{2}- C_{1}= 10 \pi - 4 \pi = 6 \pi$

The circumference of a circle can therefore be found using the formula

$C_{n} = C_{1} + (n-1)d$

where $C_{1} = 4 \pi$ and $d = 6 \pi$ ; we are looking for that of the th smallest circle, so

$C_{8} = 4 \pi + (8-1)6 \pi$

$= 4 \pi + 7 \cdot 6 \pi$

$= 4 \pi + 42 \pi$

$= 46 \pi$

Since the radius of a circle is the circumference of the circle divided by $2 \pi$ , the radius of this eighth circle is

$r = \frac{46 \pi}{2 \pi} = 23$ inches, or 1 foot 11 inches.

What is the surface area of a cylinder of height in, with a radius of in?

$152\pi$

$122\pi$

$120\pi$

$136\pi$

$240\pi$

Explanation

Recall that to find the surface area of a cylinder, you need to find the surface area of its two bases and then the surface area of its "outer face." The first two are very easy since they are circles. The equation for one base is:

$A = \pi * r^2$

For our problem, this is:

$\pi * 4^2 = \pi * 16$

You need to double this for the two bases:

$2 * 16 * \pi = 32 * \pi$

The area of the "outer face" is a little bit trickier, but it is not impossible. It is actually a rectangle that has the height of the cylinder and a width equal to the circumference of the base; therefore, it is:

$2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 4 * 15 = 120 * \pi$

Therefore, the total surface area is:

$32\pi + 120\pi = 152\pi$

Icecreamcone

Refer to the above figure. The shaded region is a semicircle with area $16 \pi$ . Give the perimeter of $\bigtriangleup ABC$ .

$24 \sqrt{2}$

$24 \sqrt{3}$

$12 \sqrt{2}$

$12 \sqrt{3}$

Explanation

Given the radius of a semicircle, its area can be calculated using the formula

$A = \frac{1}{2} \pi r ^{2}$ .

Substituting $A = 16 \pi$ :

$\frac{1}{2} \pi r ^{2} = 16 \pi$

$\frac{2}{\pi} \cdot \frac{1}{2} \pi r ^{2} = \frac{2}{\pi} \cdot 16 \pi$

$r ^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}$

The diameter of this semicircle is twice this, which is $d = 2 \cdot 4\sqrt{2} = 8 \sqrt{2}$ ; this is also the length of $\overline{BC}$ .

$\bigtriangleup ABC$ has two angles of degree measure 60; its third angle must also have measure 60, making $\bigtriangleup ABC$ an equilateral triangle with sidelength $s = 8 \sqrt{2}$ . Its perimeter is three times this, or $3 \cdot 8 \sqrt{2} = 24 \sqrt{2}$

Geometry

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