Quadrilaterals - PSAT Math

Card 0 of 399

Question

If the perimeter of a square is equal to twice its area, what is the length of one of its sides?

Answer

Area of a square in terms of each of its sides:

Area = S x S

Perimeter of a square:

Perimeter = 4S

So if 'the perimeter of a square is equal to twice its area':

2 x Area = Perimeter

2 x \[S x S\] = \[4S\]; divide by 2:

S x S = 2S; divide by S:

S = 2

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Question

Rectangle ABCD is shown in the figure above. Points A and B lie on the graph of y = 64 – _x_2 , and points C and D lie on the graph of y = _x_2 – 36. Segments AD and BC are both parallel to the y-axis. The x-coordinates of points A and B are equal to –k and k, respectively. If the value of k changes from 2 to 4, by how much will the area of rectangle ABCD increase?

Answer

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Question

In Rhombus , $m \angle H = 45 ^{\circ }$ . If $\overline{RO }$ is constructed, which of the following is true about $\Delta RHO$ ?

Answer

The figure referenced is below.

The sides of a rhombus are congruent by definition, so $\overline{RH} \cong \overline{HO}$ , making $\Delta RHO$ isosceles. It is not equilateral, since $m \angle H = 45 ^{\circ }$ , and an equilateral triangle must have three $60 ^{\circ}$ angles.

Also, consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle HRM = 180^{\circ} - m \angle H = 180^{\circ} - 45 ^{\circ} = 135^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle HRO = \frac{1}{2} m \angle HRM = \frac{1}{2} \cdot 135 ^{\circ } = 67\frac{1}{2}^{\circ }$

Similarly, $m \angle HOR = 67\frac{1}{2}^{\circ }$

This makes $\Delta RHO$ acute.

The correct response is that $\Delta RHO$ is acute and isosceles, but not equilateral.

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Question

A rhombus has a side length of 5. Which of the following is NOT a possible value for its area?

Answer

The area of a rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

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Question

In Rhombus , $m \angle R = 60^{\circ }$ . If $\overline{HM}$ is constructed, which of the following is true about $\Delta RHM$ ?

Answer

The figure referenced is below.

Consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle RHO= 180^{\circ} - m \angle R = 180^{\circ} - 60^{\circ} = 120^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle RHM = \frac{1}{2} m \angle RHO= \frac{1}{2} \cdot 120^{\circ} = 60^{\circ}$

A similar argument proves that $m \angle RMH = 60^{\circ}$ .

Since all three angles of $\Delta RHM$ measure $60^{\circ}$ , the triangle is acute. It is also equiangular, and, subsequently, equilateral.

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Question

If the area of a rhombus is 24 and one diagonal length is 6, find the perimeter of the rhombus.

Answer

The area of a rhombus is found by

A = 1/2(_d_1)(_d_2)

where _d_1 and _d_2 are the lengths of the diagonals. Substituting for the given values yields

24 = 1/2(_d_1)(6)

24 = 3(_d_1)

8 = _d_1

Now, use the facts that diagonals are perpendicular in a rhombus, diagonals bisect each other in a rhombus, and the Pythagorean Theorem to determine that the two diagonals form 4 right triangles with leg lengths of 3 and 4. Since 32 + 42 = 52, each side length is 5, so the perimeter is 5(4) = 20.

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Question

Note: Figure NOT drawn to scale.

Calculate the perimeter of Quadrilateral in the above diagram if:

Answer

, so Quadrilateral is a rhombus. Its diagonals are therefore perpendicular to each other, and the four triangles they form are right triangles. Therefore, the Pythagorean theorem can be used to determine the common sidelength of Quadrilateral .

We focus on $\Delta AXB$ . The diagonals are also each other's bisector, so

$AX = \frac{1}{2} (AC) = \frac{1}{2} \cdot 48= 24$

$XB = \frac{1}{2} (BD) = \frac{1}{2} \cdot 20=10$

By the Pythagorean Theorem,

$AB = \sqrt{(AX)^{2}+ (XB)^{2}} = \sqrt{24^{2}+10^{2}} = \sqrt{576+100} = \sqrt{676} = 26$

26 is the common length of the four sides of Quadrilateral , so its perimeter is $4 \times 26 = 104$ .

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Question

A square has an area of 36. If all sides are doubled in value, what is the new area?

Answer

Let S be the original side length. SS would represent the original area. Doubling the side length would give you 2S2S, simplifying to 4(SS), giving a new area of 4x the original, or 144.

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Question

Quadrilateral ABCD contains four ninety-degree angles. Which of the following must be true?

I. Quadrilateral ABCD is a rectangle.

II. Quadrilateral ABCD is a rhombus.

III. Quadrilateral ABCD is a square.

Answer

Quadrilateral ABCD has four ninety-degree angles, which means that it has four right angles because every right angle measures ninety degrees. If a quadrilateral has four right angles, then it must be a rectangle by the definition of a rectangle. This means statement I is definitely true.

However, just because ABCD has four right angles doesn't mean that it is a rhombus. In order for a quadrilateral to be considered a rhombus, it must have four congruent sides. It's possible to have a rectangle whose sides are not all congruent. For example, if a rectangle has a width of 4 meters and a length of 8 meters, then not all of the sides of the rectangle would be congruent. In fact, in a rectangle, only opposite sides need be congruent. This means that ABCD is not necessarily a rhombus, and statement II does not have to be true.

A square is defined as a rhombus with four right angles. In a square, all of the sides must be congruent. In other words, a square is both a rectangle and a rhombus. However, we already established that ABCD doesn't have to be a rhombus. This means that ABCD need not be a square, because, as we said previously, not all of its sides must be congruent. Therefore, statement III isn't necessarily true either.

The only statement that has to be true is statement I.

The answer is I only.

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Question

A trapezoid has a base of length 4, another base of length s, and a height of length s. A square has sides of length s. What is the value of s such that the area of the trapezoid and the area of the square are equal?

Answer

In general, the formula for the area of a trapezoid is (1/2)(a + b)(h), where a and b are the lengths of the bases, and h is the length of the height. Thus, we can write the area for the trapezoid given in the problem as follows:

area of trapezoid = (1/2)(4 + s)(s)

Similarly, the area of a square with sides of length a is given by _a_2. Thus, the area of the square given in the problem is _s_2.

We now can set the area of the trapezoid equal to the area of the square and solve for s.

(1/2)(4 + s)(s) = _s_2

Multiply both sides by 2 to eliminate the 1/2.

(4 + s)(s) = 2_s_2

Distribute the s on the left.

4_s_ + _s_2 = 2_s_2

Subtract _s_2 from both sides.

4_s_ = _s_2

Because s must be a positive number, we can divide both sides by s.

4 = s

This means the value of s must be 4.

The answer is 4.

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Question

Note: Figure NOT drawn to scale.

The white region in the above diagram is a trapezoid. What percent of the above rectangle, rounded to the nearest whole percent, is blue?

Answer

The area of the entire rectangle is the product of its length and width, or

$100 \times 50 = 5,000$ .

The area of the white trapezoid is one half the product of its height and the sum of its base lengths, or

$\frac{1}{2} \times (64+100) \times 18 = 1,476$

Therefore, the blue polygon has area

.

This is

$\frac{3,524}{5,000} \times 100 = 70.48 %$ of the rectangle.

Rounded, this is 70%.

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Question

Refer to the above diagram. .

Give the area of Quadrilateral .

Answer

$\angle C \cong \angle BDX$ , since both are right; by the Corresponding Angles Theorem, $\overline{AC} || \overline{BD}$ , and Quadrilateral is a trapezoid.

By the Angle-Angle Similarity Postulate, since

$\angle C \cong \angle BDX$

and

$\angle X \cong \angle X$ (by reflexivity),

$\Delta ACX \sim \Delta BDX$ ,

and since corresponding sides of similar triangles are in proportion,

$\frac{AC}{CX} = \frac{BD}{DX}$

$\frac{AC}{24} = \frac{6}{9}$

$\frac{AC}{24} \times 24 = \frac{6}{9} \times 24$

, the larger base of the trapozoid;

The smaller base is .

, the height of the trapezoid.

The area of the trapezoid is

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

$A = \frac{1}{2} (BD + AC) \cdot CD$

$A = \frac{1}{2} (6+ 16) \cdot 15 = \frac{1}{2} (22) \cdot 15 = 165$

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Question

ABCD and EFGH are squares such that the perimeter of ABCD is 3 times that of EFGH. If the area of EFGH is 25, what is the area of ABCD?

Answer

Assign variables such that

One side of ABCD = a

and One side of EFGH = e

Note that all sides are the same in a square. Since the perimeter is the sum of all sides, according to the question:

4a = 3 x 4e = 12e or a = 3e

From that area of EFGH is 25,

e x e = 25 so e = 5

Substitute a = 3e so a = 15

We aren’t done. Since we were asked for the area of ABCD, this is a x a = 225.

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Question

A half circle has an area of $18\pi$ . What is the area of a square with sides that measure the same length as the diameter of the half circle?

Answer

If the area of the half circle is $18\pi$ , then the area of a full circle is twice that, or $36\pi$ .

Use the formula for the area of a circle to solve for the radius:

36π = πr2

r = 6

If the radius is 6, then the diameter is 12. We know that the sides of the square are the same length as the diameter, so each side has length 12.

Therefore the area of the square is 12 x 12 = 144.

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Question

Freddie is building a square pen for his pig. He plans to buy x feet of fencing to build the pen. This will result in a pen with an area of p square feet. Unfortunately, he only has enough money to buy one third of the planned amount of fencing. Which expression represents the area of the pen he can build with this limited amount of fencing?

Answer

If Freddie uses x feet of fencing makes a square, each side must be x/4 feet long. The area of this square is (x/4)2 = _x_2/16 = p square feet.

If Freddie uses one third of x feet of fencing makes a square, each side must be x/12 feet long. The area of this square is (x/12)2 = _x_2/144 = 1/9(_x_2/16) = 1/9(p) = p/9 square feet.

Alternate method:

The scale factor between the small perimeter and the larger perimeter = 1 : 3. Since we're comparing area, a two-dimensional measurement, we can square the scale factor and see that the ratio of the areas is 12 : 32 = 1 : 9.

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Question

If the diagonal of a square measures $16\sqrt{2} \ cm$ , what is the area of the square?

Answer

This is an isosceles right triangle, so the diagonal must equal $\sqrt{2}$ times the length of a side. Thus, one side of the square measures $16\ cm$ , and the area is equal to $(16 \ cm)^{2} = 256\ cm^{2}$

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Question

A square $A$ has side lengths of $z$ . A second square $B$ has side lengths of $2.25z$ . How many $A's$ can you fit in a single $B$ ?

Answer

The area of $A$ is $n$ , the area of $B$ is $5.0625n$ . Therefore, you can fit 5.06 $A's$ in $B$ .

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Question

The perimeter of a square is $12\ in.$ If the square is enlarged by a factor of three, what is the new area?

Answer

The perimeter of a square is given by $P=4s=12$ so the side length of the original square is $3\ in.$ The side of the new square is enlarged by a factor of 3 to give $s=9\ in.$

So the area of the new square is given by $A = s^{2} = (9)^{2} = 81 in^{2}$ .

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Question

Note: figure NOT drawn to scale.

Refer to the above figure.

$\textup{Rect }ABGF \sim \textup{Rect }ACDE$

, , .

Give the area of $\textup{Rect }ABGF$ .

.

Answer

$\textup{Rect }ABGF \sim \textup{Rect }ACDE$ , so the sides are in proportion - that is,

$\frac{AF}{AB}= \frac{AE}{AC}$

Set

, , and solve for :

$\frac{AF}{12}= \frac{10}{15}$

$AF = \frac{10}{15} \cdot 12 = 8$

$\textup{Rect }ABGF$ has area

$AB \cdot AF = 12 \cdot 8 = 96$

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Question

If the area of a square is units squared, what is the length of its diagonal?

Answer

The diagonal of a square creates two special 45-45-90 triangles, meaning that the diagonal of a square is just the length of one side of the square multiplied by the square root of 2.

In this problem, you can figure out the length of one side of the square by finding the square root of the area (which is equal to a side length), then multiplying that number by the square root of 2.

$\sqrt{81} = 9$

$9\times \sqrt{2} = 9\sqrt{2}$

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