### All SAT Math Resources

## Example Questions

### Example Question #1 : How To Find If Right Triangles Are Similar

In the figure above, line segments *DC* and *AB* are parallel. What is the perimeter of quadrilateral *ABCD*?

**Possible Answers:**

95

75

85

80

90

**Correct answer:**

85

Because *DC* and *AB* are parallel, this means that angles *CDB* and *ABD* are equal. When two parallel lines are cut by a transversal line, alternate interior angles (such as *CDB* and *ABD*) are congruent.

Now, we can show that triangles *ABD* and *BDC* are similar. Both *ABD* and *BDC* are right triangles. This means that they have one angle that is the same—their right angle. Also, we just established that angles *CDB* and *ABD* are congruent. By the angle-angle similarity theorem, if two triangles have two angles that are congruent, they are similar. Thus triangles *ABD* and *BDC* are similar triangles.

We can use the similarity between triangles *ABD* and *BDC* to find the lengths of *BC* and *CD*. The length of *BC* is proportional to the length of *AD*, and the length of *CD* is proportional to the length of *DB*, because these sides correspond.

We don’t know the length of *DB*, but we can find it using the Pythagorean Theorem. Let *a*, *b*, and *c* represent the lengths of *AD*, *AB*, and *BD* respectively. According to the Pythagorean Theorem:

*a*^{2} + *b*^{2 }= *c*^{2}

15^{2} + 20^{2 }= *c*^{2}

625 = *c*^{2}

*c* = 25

The length of *BD* is 25.

We now have what we need to find the perimeter of the quadrilateral.

Perimeter = sum of the lengths of *AB*, *BC*, *CD*, and *DA*.

Perimeter = 20 + 18.75 + 31.25 + 15 = 85

The answer is 85.

### Example Question #112 : Sat Mathematics

A traffic light hangs *t* feet from the ground, over a street. A man standing the shadow of the traffic light is *h* feet tall, and his shadow is *s* feet long. How far is the man standing from the spot on the street directly under the traffic light?

**Possible Answers:**

**Correct answer:**

We can set this problem up like a set of similar triangles.

The first triangle is created by the three points: The Traffic light, the spot beneath the traffic light, and the spot where the man is standing (which is also the spot where the traffic light's shadow is).

The height of this Triangle is "T" as given in the question, and its base is the part that we are asked to solve for.

The second triangle is created by the top of the man's head, his feet, and the end of his shadow.

The height of this Triangle is "h" as given in the questions, and the base is "s".

We set up a proportion:

where X is the distance we are asked to find. Simply cross-multiply to solve.

### Example Question #1 : How To Find If Right Triangles Are Similar

An meteor crashed in the desert and created an oblong shaped crater. Scientists want to find the width of the crater as it is near their research facility. Line segments AC and DE intersect at B making the angles E and D the same. If AB is 2000 meters, BD is 1800 meters, DC is 600 meters and EB is 3600 meters, what is the width of the crater?

**Possible Answers:**

**Correct answer:**

To calculate the width of the crater, use the given information to establish that the image draws similar triangles. When triangles that have corresponding angles and a ratio to their side lengths they are considered to be similar triangles.

Identify the known information.

therefore,

and the bases of the triangles are parallel.

Also,

Set up the side ratios for this particular problem.

Looking at the only full ratio that is given, the scalar multiplier can be found.

Therefore, to find the width of the crater multiply by two.

### Example Question #91 : Triangles

In the given diagram, . Give the area of to the nearest whole number.

**Possible Answers:**

**Correct answer:**

By the Pythagorean Theorem,

Set , and solve for :

Take the positive square root of both sides:

, so corresponding sides are in proportion; specifically,

Set , and solve for :

A right triangle has as its area half the product of the length of its legs, so the area of is

To the nearest whole number, this rounds to 35.

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