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SAT Math : Geometry

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Example Questions

Example Question #23 : Quadrilaterals

If the area of a square is 25 inches squared, what is the perimeter?

Possible Answers:

Not enough information

Correct answer:

Explanation:

The area of a square is equal to length times width or length squared (since length and width are equal on a square). Therefore, the length of one side is $l = \sqrt{25in^{2}}$ or $l=5 in.$ The perimeter of a square is the sum of the length of all 4 sides or $4 \times 5 in. =20 in.$

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Example Question #241 : Geometry

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.

Possible Answers:

II only

II and III only

I and II only

I only

I, II, and III

Correct answer:

I only

Explanation:

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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Example Question #242 : Geometry

If a polygon has 10 sides, what is the measure of each exterior angle?

Possible Answers:

$36^{\circ}$

$3.6^{\circ}$

$72^{\circ}$

$360^{\circ}$

$63^{\circ}$

Correct answer:

$36^{\circ}$

Explanation:

In order to figure this out, we need to remember the formula for finding exterior angles. $\frac{360^{\circ}}{n}$ , where is the number of sides of a polygon. Now we simply do the following calculation.

$\frac{360^{\circ}}{10}=36^{\circ}$

So the measure of each exterior angle is $36^{\circ}$ .

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Example Question #1 : How To Find The Area Of A Rhombus

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

Possible Answers:

Correct answer:

Explanation:

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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Example Question #271 : Plane Geometry

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

Possible Answers:

Correct answer:

Explanation:

The area of a rhombus is found by

A = 1/2(d₁)(d₂)

where d₁ and d₂ are the lengths of the diagonals. Substituting for the given values yields

24 = 1/2(d₁)(6)

24 = 3(d₁)

8 = d₁

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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Example Question #1 : Circles

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{CX}$ is twice as long as $\overline{AX}$ , BX = 20 , and DX = 10 .

Give the length of $\overline{AX}$ .

Possible Answers:

Insufficient information is given to find the length of $\overline{AX}$ .

$10 \sqrt{3}$

$10 \sqrt{2}$

$20\sqrt{2}$

Correct answer:

Insufficient information is given to find the length of $\overline{AX}$ .

Explanation:

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{CX}$ is twice this, or . The figure referenced is below:

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot 20 = 2t \cdot 10$

20 t = 20t

This statement is identically true. Therefore, without further information, we cannot determine the value of - the length of $\overline{AX}$ .

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Example Question #1 : Circles

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{BX}$ is 12 units longer than $\overline{AX}$ , CX = 8 , and DX = 10 .

Give the length of $\overline{AX}$ (nearest tenth, if applicable)

Possible Answers:

15.0

4.8

3.0

16.8

7.2

Correct answer:

4.8

Explanation:

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{BX}$ is t+12 . The figure referenced is below:

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot (t+12) = 8 \cdot 10$

$t \cdot t+t \cdot 12 = 8 0$

$t ^{2}+12 t = 8 0$

This quadratic equation can be solved by completing the square; since the coefficient of is 12, the square can be completed by adding

$\left (\frac{12}{2} \right ) ^{2} = 6 ^{2} = 36$

to both sides:

$t ^{2}+12 t + 36 = 8 0+ 36$

Restate the trinomial as the square of a binomial:

$(t+6 )^{2 } = 116$

Take the square root of both sides:

$t+6 = \pm \sqrt{116}$

$t+6 \approx - 10.8$ or $t+6 \approx 10.8$

Either

$t+6 \approx - 10.8$ ,

in which case

$t+6 - 6 \approx - 10.8 - 6$

$t \approx -16.8$ ,

$t+6 \approx 10.8$

in which case

$t+6 - 6 \approx 10.8 - 6$

$t \approx 4.8$ ,

Since is a length, we throw out the negative value; it follows that $t \approx 4.8$ , the correct length of $\overline{AX}$ .

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Example Question #1 : Chords

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at a point .

CD= 20, AX= 8

What is the diameter of the circle?

Possible Answers:

$20\frac{1}{2}$

$33\frac{1}{3}$

Insufficient information is given to answer the question.

Correct answer:

$20\frac{1}{2}$

Explanation:

In a circle, a diameter perpendicular to a chord bisects the chord. This makes the midpoint of $\overline{CD}$ ; consequently, $CX =DX = \frac{1}{2} CD = \frac{1}{2} \cdot 20 = 10$ .

The figure referenced is below:

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting AX= 8, CX =DX = 10 , and solving for :

$8 \cdot BX = 10 \cdot 10$

$8 \cdot BX = 100$

$8 \cdot BX \div 8 = 100 \div 8$

BX = 12.5

AB = AX + BX = 8 + 12.5 = 20.5 ,

the correct length.

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Example Question #1 : Circles

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point .

CX = AX + 6

BX = 12

DX = 10

Give the length of $\overline{AX}$ .

Possible Answers:

$3 \sqrt{14}$

$2 \sqrt{30}$

Insufficient information is given to answer the question.

Correct answer:

Explanation:

Let t = AX , in which case CX = t+ 6 ; the figure referenced is below (not drawn to scale).

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting AX= t, CX= t+ 6, BX = 12, DX = 10 , and solving for :

$AX \cdot BX = CX \cdot DX$

$t \cdot 12 = (t+6) \cdot 10$

12t = 10t + 60

12t- 10t = 10t + 60 - 10t

2t = 60

$\frac{2t}{2} = \frac{60}{2}$

t = 30 ,

which is the length of $\overline{AX}$ .

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Example Question #2 : Circles

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at point . AX = 12 and BX = 20 . Give the length of $\overline{CD}$ (nearest tenth, if applicable).

Possible Answers:

31.0

16.0

15.5

insufficient information is given to determine the length of $\overline{CD}$ .

32.0

Correct answer:

31.0

Explanation:

A diameter of a circle perpendicular to a chord bisects the chord. Therefore, the point of intersection is the midpoint of $\overline{CD}$ , and

$CX = DX = \frac{1}{2} CD$ .

Let stand for the common length of $\overline{CX}$ and $\overline{DX}$ ,

The figure referenced is below.

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$CX \cdot DX = AX \cdot BX$

Set AX = 12 and BX = 20 , and CX = DX = t ; substitute and solve for :

$t \cdot t = 12 \cdot 20$

$t^{2 } = 240$

$t = \sqrt{240} \approx 15.5$

This is the length of $\overline{CX}$ ; the length of $\overline{CD}$ is twice this, so

$CD = 2 \cdot t \approx 2 \cdot 15.5 = 31.0$

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SAT Math : Geometry

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #23 : Quadrilaterals

Example Question #241 : Geometry

Example Question #242 : Geometry

Example Question #1 : How To Find The Area Of A Rhombus

Example Question #271 : Plane Geometry

Example Question #1 : Circles

Example Question #1 : Circles

Example Question #1 : Chords

Example Question #1 : Circles

Example Question #2 : Circles

All SAT Math Resources

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