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SAT Math : Right Triangles

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Example Question #91 : Right Triangles

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:

$\frac{hs}{T}$

hsT

$\frac{Th}{s}$

$\frac{sT}{h}$

$\frac{h}{sT}$

Correct answer:

$\frac{sT}{h}$

Explanation:

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:

$\frac{height}{base} = \frac {T}{X} = \frac{h}{s}$

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

$X = \frac{sT}{h}$

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Example Question #91 : Right Triangles

Screen shot 2016 02 12 at 8.29.40 am

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:

$1800\text{ m}$

$1200\text{ m}$

$1400\text{ m}$

$800\text{ m}$

$1600\text{ m}$

Correct answer:

$1200\text{ m}$

Explanation:

Screen shot 2016 02 12 at 8.29.40 am

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.

$\\ \angle E=\angle D \\\angle B=\angle B$

therefore,

$\angle A=\angle C$

and the bases of the triangles are parallel.

Also,

$\\\overline{AB}=1800 \text{ m} \\\overline{EB}=3600\text{ m} \\\overline{BD}=1800 \text{ m} \\\overline{CD}=600\text{ m}$

Set up the side ratios for this particular problem.

$\frac{AB}{BC}; \frac{AE}{CD}; \frac{EB}{BD}$

$\frac{1800}{BC}; \frac{x}{600}; \frac{3600}{1800}$

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

$\frac{3600}{1800}=2$

Therefore, to find the width of the crater multiply $\overline{CD}$ by two.

$\\\frac{x}{600}=2 \\\\x=2\cdot 600 \\x=1200\text{ m}$

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Example Question #4 : How To Find If Right Triangles Are Similar

Triangles

In the given diagram, $\bigtriangleup ABC \sim \bigtriangleup DCA$ . Give the area of $\bigtriangleup DCA$ to the nearest whole number.

Possible Answers:

173

Correct answer:

Explanation:

By the Pythagorean Theorem,

$(AC)^{2}= (AB)^{2} + (BC)^{2}$

Set AB = 5, BC= 13 , and solve for :

$(AC)^{2}= (AB)^{2} + (BC)^{2}$

$(AC)^{2}= 5^{2} + 12^{2}$

$(AC)^{2}= 25+144$

$(AC)^{2}=169$

Take the positive square root of both sides:

$AC = \sqrt{169} = 13$

$\bigtriangleup ABC \sim \bigtriangleup DCA$ , so corresponding sides are in proportion; specifically,

$\frac{DC}{AB} = \frac{AC}{CB}$

Set AB= 5, CB = 12, AC = 13 , and solve for :

$\frac{DC}{5} = \frac{13}{12}$

$\frac{DC}{5} \cdot 5 = \frac{13}{12} \cdot 5$

$DC = \frac{65}{12} = 5 \frac{5}{12}$

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

$\frac{1}{2} \cdot AC \cdot DC$

$= \frac{1}{2} \cdot 13 \cdot 5 \frac{5}{12}$

$= 35 \frac{5}{24}$

To the nearest whole number, this rounds to 35.

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SAT Math : Right Triangles

Study concepts, example questions & explanations for SAT Math

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Example Question #91 : Right Triangles

Example Question #91 : Right Triangles

Example Question #4 : How To Find If Right Triangles Are Similar

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