Pentagons - ISEE Upper Level Quantitative Reasoning

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Question

The perimeters of six squares form an arithmetic sequence. The smallest square has area 9; the second smallest square has area 25. Give the perimeter of the largest of the six squares.

Answer

The two smallest squares have areas 9 and 25, so their sidelengths are the square roots of these, or, respectively, 3 and 5. Their perimeters are the sidelengths multiplied by four, or, respectively, 12 and 20. Therefore, in the arithmetic sequence,

$P_{1} = 12$

$P_{2} = 20$

and the common difference is .

The perimeter of the th smallest square is

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

Setting $P_{1} = 12, n = 6, d=8$ , the perimeter of the largest (or sixth-smallest) square is

$P_{6}=P_{1}+ (6-1)8 = 12+ 5 \cdot 8 = 12 + 40 = 52$ .

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Question

The area of a square is $36t ^{2} + 84t + 49$ .

Give the perimeter of the square.

Answer

The length of one side of a square is the square root of its area. The polynomial representing the area of the square can be recognized as a perfect square trinomial:

$36t ^{2} + 84t + 49$

$=( 6t )^{2} + 2\cdot 6t \cdot 7 + 7^{2}$

$= (6t+7)^{2}$

Therefore, the square root of the area is

$\sqrt{36t ^{2} + 84t + 49} = \sqrt{ (6t+7)^{2}} = |6t+7|$ ,

which is the length of one side.

The perimeter of the square is four times this length, or

.

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Question

The perimeters of six squares form an arithmetic sequence. The second-smallest square has sides that are two inches longer than those of the smallest square.

Which, if either, is the greater quantity?

(a) The perimeter of the third-smallest square

(b) The length of one side of the largest square

Answer

Let the length of one side of the first square be $s_{1}$ . Then the length of one side of the second-smallest square is $s_{2}= s_{1}+ 2$ , and the perimeters of the squares are

$P_{1}= 4s_{1}$

and

$P_{2} = 4 s_{2}= 4 (s_{1}+ 2) = 4s_{1} + 8 = P_{1}+ 8$

This makes the common difference of the perimeters 8 units.

The perimeters of the squares being in arithmetic progression, the perimeter of the th-smallest square is

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

Since , this becomes

$P_{n} = P_{1} + 8 (n-1)$

The perimeter of the third-smallest square is

$P_{3} = P_{1} + 8 (3-1) = P_{1}+ 8 \cdot 2 = P_{1} + 16$

The perimeter of the largest, or sixth-smallest, square is

$P_{6} = P_{1} + 8 (6-1) = P_{1}+ 8 \cdot 5= P_{1} + 40$

The length of one side of this square is one fourth of this, or

$s_{6} = \frac{ P_{6} }{4}= \frac{ P_{1} + 40}{4} = \frac{1}{4} P_{1} + 10$

Therefore, we are comparing $P_{1} + 16$ and $\frac{1}{4} P_{1} + 10$ .

Since a perimeter must be positive,

$\frac{1}{4} P_{1} < P_{1}$ ;

also, .

Therefore, regardless of the value of $P_{1}$ ,

$P_{1} + 16 > \frac{1}{4} P_{1} + 10$ ,

and

$P_{3}> s_{6}$ ,

making (a) the greater quantity.

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Question

The sidelength of a square is $2^{x}$ . Give its perimeter in terms of .

Answer

The perimeter of a square is four times the length of a side, which here is $2 ^{x }$ :

$P = 4 \cdot 2^{x} = 2 ^{2} \cdot 2^{x} =2 ^{x+2}$

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Question

A diagonal of a square has length $2^{x}$ . Give its perimeter.

Answer

The length of a side of a square can be determined by dividing the length of a diagonal by $\sqrt{2}$ - that is, $2 ^{\frac{1}{2}}$ . A diagonal has length $2^{x}$ , so the sidelength is

$2^{x} \div 2 ^{\frac{1}{2}}= 2 ^{x-\frac{1}{2}}$

Multiply this by four to get the perimeter:

$4 \cdot 2 ^{x-\frac{1}{2}} = 2^{2} \cdot 2 ^{x-\frac{1}{2}} = 2 ^{x-\frac{1}{2}+2} = 2 ^{x+\frac{3}{2}}$

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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

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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Question

Which of the following can be the measures of the four angles of a parallelogram?

Answer

Opposite angles of a parallelogram must have the same measure, so the correct choice must have two pairs, each of the same angle measure. We can therefore eliminate $38^{\circ }, 142 ^{\circ }, 28^{\circ }, 152 ^{\circ }$ and $102^{\circ }, 94 ^{\circ }, 86^{\circ }, 78 ^{\circ }$ as choices.

Also, the sum of the measures of the angles of any quadrilateral must be $360 ^{\circ }$ , so we add the angle measures of the remaining choices:

$88^{\circ }, 97 ^{\circ }, 88^{\circ }, 97 ^{\circ }$ :

$88^{\circ } + 97 ^{\circ } + 88^{\circ }+ 97 ^{\circ } = 370$ , so we can eliminate this choice.

$77^{\circ }, 133 ^{\circ }, 77^{\circ }, 133 ^{\circ }$ :

$77^{\circ } + 133 ^{\circ } + 77^{\circ }+ 133 ^{\circ } =420$ , so we can eliminate this choice.

$57^{\circ }, 123 ^{\circ }, 57^{\circ }, 123 ^{\circ }$

$57^{\circ } +123 ^{\circ }+57^{\circ }+123 ^{\circ } = 360$ ; this is the correct choice.

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Question

In Parallelogram , $m \angle A = (3x)^{\circ }$ and $m \angle C = (2y)^{\circ }$ .

Which is the greater quantity?

(a)

(b)

Answer

In Parallelogram , $\angle A$ and $\angle C$ are opposite angles and are therefore congruent. This means that

$m \angle A = m \angle C$

$3x \div 3 = 2y \div 3$

$x = \frac{2}{3}y$

Both are positive, so .

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Question

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

(A) The perimeter of the parallelogram

(B) 46 inches

Answer

The measure of $\angle 1$ is actually irrelevant. The perimeter of the parallelogram is the sum of its four sides; since opposite sides of a parallelogram have the same length, the perimeter is

inches,

making the quantities equal.

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Question

Parallelogram A is below:

Parallelogram B is below:

Note: These figures are NOT drawn to scale.

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

(A) The perimeter of parallelogram A

(B) The perimeter of parallelogram B

Answer

The perimeter of a parallelogram is the sum of its sidelengths; its height is irrelevant. Also, opposite sides of a parallelogram are congruent.

The perimeter of parallelogram A is

inches;

The perimeter of parallelogram B is

inches.

(A) is greater.

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Question

Figure NOT drawn to scale.

The above figure depicts Rhombus with and .

Give the perimeter of Rhombus .

Answer

All four sides of a rhombus have the same length, so we can find the perimeter of Rhombus by taking the length of one side and multiplying it by four. Since , the perimeter is four times this, or $4 \cdot AB = 4 \cdot 32 = 128$ .

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

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Question

Rhombus has two diagonals that intersect at point ; $\angle RHO = 75 ^{\circ }$ . Which is the greater quantity?

(a) $m \angle RPH$

(b) $m \angle HPO$

Answer

The diagonals of a rhombus always intersect at right angles, so $m \angle RPH = m \angle HPO = 90 ^{\circ }$ . The measures of the interior angles of the rhombus are irrelevant.

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Question

A rhombus has diagonals of length two and one-half feet and six feet. Which is the greater quantity?

(A) The perimeter of the rhombus

(B) Four yards

Answer

It will be easier to look at these measurements as inches for the time being:

$2 \frac{1}{2} \times 12 = 30$ and $6 \times 12 = 72$ , so these are the lengths of the diagonals in inches.

The diagonals of a rhombus are each other's perpendicular bisector, so, as can be seen in the diagram below, one side of a rhombus and one half of each diagonal form a right triangle. If we let be the length of one side of the rhombus, then this is the hypotenuse of that right triangle; its legs are one-half the lengths of the diagonals, or 15 and 36 inches.

By the Pythagorean Theorem,

$s = \sqrt{ 15 ^{2} + 36^{2}} = \sqrt{ 225 + 1,296} = \sqrt{ 1,521} = 39$

Each side of the rhombus measures 39 inches, and its perimeter is

$4 \times 39 = 156$ inches.

Four yards is equal to $4 \times 36 = 144$ inches, so (A) is greater.

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Question

A rectangle has a width of 2_x_. If the length is five more than 150% of the width, what is the perimeter of the rectangle?

Answer

Given that w = 2_x_ and l = 1.5_w_ + 5, a substitution will show that l = 1.5(2_x_) + 5 = 3_x_ + 5.

P = 2_w_ + 2_l_ = 2(2_x_) + 2(3_x_ + 5) = 4_x_ + 6_x_ + 10 = 10_x_ + 10 = 10(x + 1)

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