Geometry - SAT Subject Test in Math I

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Question

The sides of a triangle have lengths 6 yards, 18 feet, and 216 inches. Which of the following is true about this triangle?

Answer

One yard is equal to 3 feet; it is also equal to 36 inches. Therefore:

18 feet is equal to $18 \div 3 = 6$ yards,

and

216 feet is equal to $216 \div 36 = 6$ yards.

The three sides are congruent, making the triangle equilateral - and all equilateral triangles are acute.

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Question

Figures not drawn to scale

The triangles above are similar. Given the measurements above, what is the length of side c?

Answer

You can find the length of c by first finding the length of the hypotenuse of the larger similar triangle and then setting up a ratio to find the hypotenuse of the smaller similar triangle.

$\small a{^{2}}+b{^{2}}=c{^{2}}$

$\small 6{^{2}}+8{^{2}}=c{^{2}}$

$\small 36+64=c{^{2}}$

$\small 100=c{^{2}}$

$\small c=10$

You also could have found 10 by recognizing this triangle is a form of a 3-4-5 triangle.

The hypotenuse of the bigger triangle is 10 inches.

Now that we know the length of the hypotenuse for the larger triangle, we can set up a ratio equation to find the hypotenuse of the smaller triangle.

$\small \frac{2.5}{6}=\frac{c}{10}$ cross multiply

$\small 6c=25$

$\small \frac{6c}{6}=\frac{25}{6}$

$\small c=4\frac{1}{6}$

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Question

If line 2 and line 3 eventually intersect when extended to the left which of the following could be true?

$\I.\ a = e \II.\ a = f \III.\ e+c = 180$

Answer

Read the question carefully and notice that the image is deceptive: these lines are not parallel. So we cannot apply any of our rules about parallel lines. So we cannot infer II or III, those are only trueif the lines are parallel. If we sketch line 2 and line 3 meeting we will form a triangle and it is possible to make a = e. One such solution is to make a and e 60 degrees.

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Question

What is the maximum number of distinct regions that can be created with 4 intersecting circles on a plane?

Answer

Try sketching it out.

Start with one circle and then keep adding circles like a venn diagram and start counting. A region is any portion of the figure that can be defined and has a boundary with another portion. Don't forget that the exterior (labeled 14) is a region that does not have exterior boundaries.

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Question

Note: Figure may not be drawn to scale

In rectangle has length and width and respectively. Point lies on line segment $\overline{BD}$ and point lies on line segment $\overline{EF}$ . Triangle has area , in terms of and what is the possible range of values for ?

Answer

Notice that the figure may not be to scale, and points and could lie anywhere on line segments $\overline{BD}$ and $\overline{EF}$ respectively.

Next, recall the formula for the area of a triangle:

$area = \frac{1}{2}bh$

To find the minimum area we need the smallest possible values for and .

To make smaller we can shift points and all the way to point . This will make triangle have a height of :

$area = \frac{1}{2}b(0)$

is the minimum possible value for the area.

To find the maximum value we need the largest possible values for and . If we shift point all the way to point then the base of the triangle is and the height is , which we can plug into the formula for the area of a triangle:

$area = \frac{1}{2}xy$

which is the maximum possible area of triangle

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Question

What is the area of the following kite?

Answer

The formula for the area of a kite:

$A = \frac{1}{2} (a)(b)$ ,

where represents the length of one diagonal and represents the length of the other diagonal.

Plugging in our values, we get:

$A = \frac{1}{2} (4: m)(5: m)$

$A = \frac{1}{2} (20: m^2) = 10: m^2$

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Question

Which of the following shapes is a kite?

Answer

A kite is a four-sided shape with straight sides that has two pairs of sides. Each pair of adjacent sides are equal in length. A square is also considered a kite.

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Question

The above figure depicts the rectangular swimming pool at an apartment. The apartment manager needs to purchase a tarp that will cover this pool completely. However, because of the cutting device the pool store uses, the length and the width of the tarp must each be a multiple of three yards. Also, the tarp must be at least one yard longer and one yard wider than the pool.

What will be the minimum area of the tarp the manager purchases?

Answer

Three feet make a yard, so the length and width of the pool are $\frac{50}{3} = 16 \frac{2}{3}$ yards and $\frac{35}{3} = 11 \frac{2}{3}$ yards, respectively. Since the dimensions of the tarp must exceed those of the pool by at least one yard, the tarp must be at least $17 \frac{2}{3}$ yards by $12 \frac{2}{3}$ yards; but since both dimensions must be multiples of three yards, we take the next multiple of three for each.

The tarp must be 18 yards by 15 yards, so the area of the tarp is the product of these dimensions, or

$18 \times 15 = 270$ square yards.

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Question

Note: Figure NOT drawn to scale.

Refer to the figure above, which shows a square inscribed inside a large triangle. What percent of the entire triangle has been shaded blue?

Answer

The shaded portion of the entire triangle is similar to the entire large triangle by the Angle-Angle postulate, so sides are in proportion. The short leg of the blue triangle has length 20; that of the large triangle, 30. Therefore, the similarity ratio is . The ratio of the areas is the square of this, or $3^{2}:2^{2}$ , or .

The blue triangle is therefore $\frac{4}{9}$ of the entire triangle, or $\frac{4}{9} \times 100 = 44 \frac{4}{9} %$ of it.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. What is the area of $\Delta ABC$ ?

Answer

If we see hypotenuse $\overline{AC}$ as the base of the large triangle, then we can look at the segment perpendicular to it, $\overline{BD}$ , as its altitude. Therefore, the area of $\Delta ABC$ is

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

.

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$

The area of $\Delta ABC$ is therefore

$A = \frac{1}{2} (AC) (BD) = \frac{1}{2} \cdot 30 \cdot 12 = 180$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the ratio of the area of $\Delta BDC$ to that of $\Delta ADB$ .

Answer

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$ .

The areas of $\Delta BDC$ and $\Delta ADB$ , each being right, are half the products of their legs, so:

The area of $\Delta ADB$ is $\frac{1}{2} \times (AD) (BD) = \frac{1}{2} \times 6 \times (BD)= 3(BD)$

The area of $\Delta BDC$ is $\frac{1}{2} \times (CD) (BD) = \frac{1}{2} \times 24 \times (BD)= 12(BD)$

The ratio of the areas is $\frac{12 (BD)}{3 (BD)} = \frac{12 }{3 } =\frac{4}{1}$ - that is, 4 to 1.

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Question

The above figure depicts the rectangular swimming pool at an apartment. The apartment manager needs to purchase a tarp that will cover this pool completely, but the store will only sell the material in multiples of ten square yards. How many square yards will the manager need to buy?

Answer

Three feet make a yard, so the length and width of the pool are $\frac{50}{3}$ yards and $\frac{35}{3}$ yards; the area of the pool, and that of the tarp needed to cover it, must be the product of these dimensions, or

$\frac{50}{3} \times \frac{35}{3}= \frac{1,750}{9} = 194 \frac{4}{9}$ square yards.

The manager will need to buy a number of square yards of tarp equal to the next highest multiple of ten, which is 200 square yards.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. In terms of area, $\Delta ADB$ is what fraction of $\Delta ABC$ ?

Answer

The area of $\Delta ADB$ , being right, is half the products of its legs, which is:

$\frac{1}{2} (AD) (BD) = \frac{1}{2} \times 6 (BD) = 3 (BD)$

The area of $\Delta ABC$ is one half the product of its base and height; we can use its hypotenuse as the base and as the height, so this area is

$\frac{1}{2} (AC) (BD) = \frac{1}{2} \times 30 (BD) = 15 (BD)$

Therefore, in terms of area, $\Delta ADB$ is $\frac{3 (BD)}{15 (BD)} = \frac{3}{15} = \frac{1}{5}$ of $\Delta ABC$ .

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Question

Note: figure NOT drawn to scale.

Refer to the above figure. Quadrilateral is a square. What is the area of Polygon ?

Answer

Polygon is a composite of $\Delta ABE$ and Square ; its area is the sum of the areas of the two figures.

$\Delta ABE$ is a right triangle; its area is half the product of its legs, which is

$A = \frac{1}{2} \cdot 5 \cdot 8 = 20$

$\overline{BE }$ is both one side of Square and the hypotenuse of $\Delta ABE$ ; its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

$BE = \sqrt{5^{2}+8^{2}} = \sqrt{25 + 64 } = \sqrt{89}$ .

Square has area the square of this, which is 89.

Polygon has as its area the sum of these two areas:

.

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Question

Find the area of a kite if the diagonal dimensions are and .

Answer

The area of the kite is given below. The FOIL method will need to be used to simplify the binomial.

$A=\frac{d1\cdot d2}{2}=\frac{(x+3)(x-6)}{2}= \frac{x^2-6x+3x-18}{2}$

$A= \frac{x^2-3x-18}{2}$

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Question

The diagonals of a kite are $\sqrt10:cm$ and $\sqrt5:cm$ . Find the area.

Answer

The formula for the area for a kite is

$A=\frac{d_1\cdot d_2}{2}$ , where and are the lengths of the kite's two diagonals. We are given the length of these diagonals in the problem, so we can substitute them into the formula and solve for the area:

$A = \frac{d1\cdot d2}{2} = \frac{\sqrt10 \cdot \sqrt5}{2} = \frac{\sqrt50}{2} = \frac{5\sqrt2}{2}:cm^2$

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Question

Find the area of a rhombus if the diagonals lengths are and .

Answer

Write the formula for the area of a rhombus:

$A=\frac{d_1 \cdot d_2}{2}$

Substitute the given lengths of the diagonals and solve:

$A=\frac{d1 \cdot d2}{2} = \frac{20:cm \cdot 40:cm}{2} =\frac{800:cm^2}{2} = 400:cm^2$

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Question

Find the area of a kite with diagonal lengths of and .

Answer

Write the formula for the area of a kite.

$A= \frac{d1\cdot d2}{2}$

Plug in the given diagonals.

$A= \frac{(a+b)\cdot (2a-2b)}{2}$

Pull out a common factor of two in and simplify.

$A= \frac{(a+b)\cdot2(a-b)}{2}$

Use the FOIL method to simplify.

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Question

Find the area of a rhombus if the diagonals lengths are and .

Answer

Write the formula for finding the area of a rhombus. Substitute the diagonals and evaluate.

$A=\frac{d1\cdot d2}{2}= \frac{2a \cdot 5a^2}{2}= 5a^3$

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Question

Find the area of a circle with a diameter of .

Answer

Write the formula for the area of a circle.

$A_{circle}= \pi r^2 = \frac{\pi d^2}{4}$

Substitute the diameter and solve.

$\frac{\pi (16)^2}{4} = \frac{\pi (16)(16)}{4}=\pi (16)(4)= 64\pi$

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