Review of Nuclear Battery Technology Essay

Review of Nuclear Battery Technology Essay Review of Nuclear Battery Technology Essay Example Review of Nuclear Battery Technology Essay Example The Role of Nuclear Battery for Smartphones In smart phones, batteries play a major role in providing power. Scientists as well as technology firms are continually looking for means to better the life of these batteries and their efficiency. The University of Missouri lately came up with a more efficient nuclear battery that is long lasting. This battery is capable of running several applications including a space flight. They also act as a reliable energy source in automobiles. Search for alternative sources of energy has made scientists indulge in extensive research in almost all fields to gather information on how to tackle the challenge of battery life of batteries used in various devices such as phones and laptops and come up with ones that are more efficient than the existing chemical batteries. The basic idea the researchers have been developing is that instead of consuming or utilizing the power in a battery`s chemical gradient, for instant in Lithium batteries, to employ the energy emitted by the decaying of isotopes of radioactive elements in a natural manner to generate energy. These batteries are referred to as nuclear batteries. Problem definition Nuclear batteries in smartphones emit radiation and in case of a leakage they can cause cancer and even death. For example, the gamma rays which have intense, penetrating power can only be checked with the introduction of a large lead lump; otherwise, cancer is inevitable. The casing is done to reduce radiation in smartphones, and it is done using the materials mentioned. Another possible disadvantage (though it is not common) is that terrorists may use the Strontium-90 to develop dirty bombs even though the substance is very expensive. Radiation protection principles presume that any radiation dose, no matter how small it might seem to be, can harm a person. Nuclear batteries are lighter than other ones, however, they can provide energy for smartphones, and they are much smaller in size as well as more efficient as other batteries. Nuclear batteries also have sufficient energy density. The radioisotope that is an example of nuclear energy can supply energy density that is approximately six orders of magnitude more than the batteries manufactured using chemical substances. Betavoltaic chipsets that are also nuclear batteries are commercially available and are of high demand due to low voltage. They are also amp products for the niche markets such as the military. Betavoltaic batteries generate power from beta radiations rather than photons. These beta radiations are high power electrons emitted by radioactive elements. Several commercial uses of nuclear technologies exist today, for example, fire control detectors and emergency exits in many buildings. Lithium-ion batteries have an aging problem, which becomes evident one year after the purchase and the manufacturers always go silent about this. It always happens whether the battery is in use or kept idle. Another disadvantage of lithium-ion batteries is that they are regarded as not completely mature since the chemicals and metals vary on a continuous basis. They need a protection circuit to preserve the voltage as well as the current within some safe limits . Natural Li is converted into isotopic clear 6Li. The merits of performing this task are the fact that thermal neutron combination of the cross-section is multiple of magnitude order larger than the natural Li cross-section implying that researchers can evaluate factual manufacturing methods and techniques. Functional requirements Terms such as atomic battery, nuclear battery and radioisotope and tritium generator are employed to depict devices that use energy produced from decaying of the radioactive isotope to generate electricity. Conversion method is divided into thermal and non-thermal one. Thermal converters consist of the thermionic and thermoelectric kinds of generators. Their output energy is always a function of a temperature difference. Non-thermal converters output of power is not a function of a temperature difference. It extracts a portion of power as it is degraded into the heat energy instead of using the thermal power to run electrons in the circle. Atomic batteries in most cases have an efficiency of between 0.1 to 5%. High efficiency beta voltaics have an efficiency of 6-8%. Thermal converters are classified into a thermionic converter that includes a hot electrode that emits electrons in a thermionic manner over a potential power barrier to a relatively cool form of an electrode and produce valuable electric energy output. Cesium vapor is employed to highly optimize electrode task function as well as present an ion supply to make sure the electron space charge has been neutralized. Beta voltaic is a battery that generates energy from radiation and scientists have studied the battery since 1950 and regard it as a major source of nuclear energy. Day to day research is being pursued on nuclear batteries in various research institutions. Much of this task is centered on making the frontiers of these nuclear device technologies by using energy sources with the help of beta or alpha particle decay, which is based on the radioactive isotope emitted. The area of the beta voltaic, which is the most tackled by the researchers, is tritium. This is a hydrogen isotope that has a pair of neutrons and a single proton as well as electron inherent in its hydrogen form. It is a radioactive isotope with a half-life of 12.32 years during which it emits beta particle that is basically an electron. This makes it more preferred over other known solutions that emit dangerous gamma radiations . There are several other advantages of tritium like its weight; it is arguably the third lightest among the existing isotopes. It also has properties and reactivity similar to those of hydrogen. Researchers are well conversant with its production path, and they are also quite aware of its hazards. A specimen of Highly-Oriented Pyrolytic Graphite (HOPG) that is intercalated with some lithium so as to check loading before irradiation occurs. Nuclear energy sources when controlled arent inherently dangerous. These nuclear batteries employ radioactive isotopes referred to as strontium-90. The latter improves the electrochemical power in water-based solutions. An electrode consisting of nanostructured titanium dioxide and a coating made of platinum is responsible for converting the energy or the power into electrons . The water operates as a buffer. Surface Plasmon developed in the device emerges as a useful component since it improves the batterys efficiency. The Ionic solution, however, cannot be easily frozen at minimal temperatures. It could efficiently perform in a variety of applications, for instance, car batteries. Constructing a betavoltaic device, a silicon material inside two electrodes is wedged. By the time the radiation strikes the semiconductor there is a production of electrons flow, simply referred to as voltage electricity. Unfortunately, ancient materials were less suitable for enormous stacked arrays since the volume and the mass of the battery being developed would be large. Thinner and relatively lighter collectors and emitters were required for designing an array. Of late, developments in graphene are still to be correctly integrated into the architecture of this betavoltaic. When correct integration into these thin stacked kinds of betavoltaic arrays is completed, a wider utilization and efficient performance would be experienced. It is possible that betavoltaic energy can generate more power as compared to chemical batteries. The anticipated maximum efficiency of promethium and tritium batteries is 21% and 12% respectively. Factors leading to these efficiencies are the source construction and the secondary electron discharge as well as backscattering mainly from the collector. Experimentally, it was demonstrated that the efficiency of the tritium direct charge battery model with vacuum dielectrics and collectors with secondary electron emission suppression and backscattering coating reached 5.5%. This kind of battery has an activity of curies of approximately 108. The experiment also demonstrated a voltage of 5300 volts with short circuit current of 148 nanoamperes. However, the efficiency can be doubled with a double-sided source. A promethium-147 nuclear battery has an activity of above 2.6 curies. The experiment shows it generates a voltage of 60kv. The current for the short circuit is 6.0 nano-amperes reaching an efficiency of 15%. The effect of charge accumulation in dielectrics under mono-energetic electron beam irradiation was used for developing nuclear batteries. In this battery, the charge accumulated on the surface conducts electric current through an uncharged dielectric. A nuclear battery was fabricated and tested with a tritium source; taking into consideration that a dielectric layer is wider than the range of tritium beta elements and a metal collector is without a vacuum space, this model generated 0.4 microwatts of electricity. Natural radioactivity emits radiation that generates energy. Nuclear batteries also known as atomic batteries harness the energy. The power density of the final product and the application domains depend on a material employed to generate that energy. On the other hand, the output and the potential efficiency of the battery depend on the form of conversion employed. Thermal converter, that is a radioisotope generator, utilizes the thermal energy produced by radioisotope decay to generate electricity. Methods used for this process include thermocouple heating, a recognized charge accumulation effect found in the dielectrics. The nuclear batteries developed at the Missouri University consist of a platinum-coated titanium dioxide electrode that was with. Water was also incorporated in addition to radioactive strontium-90. Sr-90 can decay radioactively with 28.79 years half-life. It generates an electron referred to as beta radiation; it also produces anti-neutrino as well as the isotope yttrium-90. This Y-90 has a half-life of 65 hours. This causes decay of additional electrons and anti-neutrinos. Stable Zirconium is also generated as a result of the decay. The wisest aspect of employing Sr-90 as a source of energy is the fact that it emits less or zero gamma radiations. Nuclear batteries are safe to handle and also very easy to use. Apart from being used in smartphones they are used extensively in health departments, for example, for cancer radiotherapy . Design concept Safety of radioactive substance is ensured by introducing an aluminum material between a human body part and the source of the rays. Thus, the safety of betavoltaic is checked in this way to avoid damage to people. One of the greatest advantages of nuclear batteries in smartphones is the fact that recharging will not be done as in the case with chemical batteries. As mentioned above, nuclear batteries with efficient packaging possess an energy density that is greater than in chemical ones. Additionally, the radioactive isotopes used to develop nuclear batteries are easily available at affordable market prices. Nuclear cells have a life span not less than ten years. This is an overwhelming term as they supply energy to equipment non-stop. Thus, the reliability and the longevity incorporated together may suffice a minor power needs for a decade. However, radiation safety standards need to be met. Incorporation of safety measures to ensure nuclear batteries are safe to handle. Devices, like smartphones batteries emit nuclear radiation that includes beta and gamma ray beams. This radiation is however kept in closed packages. Individuals worry that tritium in these batteries may diffuse due to the small size of the package and its mobility. They fear that it could diffuse through graphitic matrix, due to the complicated process of covering it . The worry for this is counterattacked by the fact that it is experimentally proven that the radiation would remain in the matrix as long as the temperatures remain below 627 degrees Celsius. The operating environment temperature that people live in is far much below this limit. The remaining challenge is the moisture. Nevertheless, the scientists are making use of a robust, hermetically fastened package. In less than three years to come, research companies, if funded adequately, will produce nuclear powered devices for general market. On this time framework, though, the researchers argue that it would depend on the regulatory framework. The addition of water was arguably the breakthrough of these batteries since that water can absorb a great amount of beta radiation since when in large quantities it can detriment to a betavoltaic semiconductor. However, beta radiation rips apart the molecules of water, generating free radicals as well as electricity. Comparative study with previous concept The cost of developing these nuclear batteries is relatively high. As for the case of most innovations, the starting cost is rather huge. However, as the innovation goes operational, these drawbacks varnish as the product is produced in bulk. Nuclear batteries for some specific applications like the size of laptop batteries may lead to some problems though it can be eliminated as time progresses; for instance the Xcel in laptops batteries is much more compared to the conventional one. Prospective commercial application of nuclear batteries in smartphones The aerospace firms would welcome smartphones recharging themselves. Oil and the gas companies are also potential commercial markets for the nuclear batteries due to their recharging factor. All these companies require some reliable energy sources in physical extremes for instant low temperatures and low pressure . The betavoltaic battery integrated into a flight data detector may signal to the searching squad for years rather than months. The odds of coming up with a commercially viable substance are reasonably perfect since the ultra-thin kind of collectors exists anyway. There is a growing global interest in the development of these thin beta-electron kinds of emitters. Applications Nuclear batteries are used widely due to their long life capability and high efficiency. This sort of innovation will undoubtedly change the current technology for the better and eliminate the power limitations brought about by chemical cells. In space applications, nuclear energy units are more significant as compared to the solar cells and the ordinary chemical batteries. Solar cells are easily destroyed when passing through radiation areas. The second reason is that the operations on planets such as Mars and the moon, where long phases of darkness need heavy batteries to provide power. Solar cells can only get energy from the sun. The third is that the missions conducted in space in an opaque atmosphere for instant on Jupiter. There is no light there, thus solar cell are useless there. The nuclear source of power would be useful in space. Nuclear batteries would also eliminate the necessity of heating electronics in areas where temperatures are -245 degrees Celsius, for instance in space. These incredible advantages would ensure the nuclear batteries will easily replace current chemical sources of power. All applications including the phones that require large powers and a high lifetime and not forgetting a definite design over density will automatically prefer the nuclear source. The other application is the use of these batteries in mobile devices. A nuclear-powered battery for a laptop or a phone can provide supply approximate 8,000 times the life of the ordinary laptop or phone battery. Nuclear brings about forgetting the tedious process of recharging and replacing batteries. A nuclear battery through research has been found that it can endure a minimum of five years. The Xcel-N has never been switched off since it started its operation. It has been working for eight months in a row, without using any external energy supply. Low energy electronics are ending up being versatile. Therefore, these kinds of batteries are nowadays becoming commercially relevant. They act as power sources for machinery that ought to function unattended for a long time, like satellites. Also, if it packed correctly, it can be applied to spaceship and pacemakers. These batteries can provide energy to a variety of objects from the tiny sensors to enormous systems. The plans of these nuclear batteries A proof-of-principle form of analysis starts with an emitter. Irradiation of the high-energy grapheme-based kind of beta emitters is necessary. When this is fully optimized, then invention and development of nuclear powered cells is quite possible. The key hurdles are experienced in the transportation of these devices and their handling. It is advisable to collaborate with Defense Advanced Research Project Agency (DARPA) in developing the geometry, and in field-testing of these devices . Feasibility assessment The above-mentioned researches concerning the nuclear batteries present adequate hope in the supply of power and energy in future for devices and applications. Upon implementation of these technologies, feasibilities and credibility of devices such as smartphones will be elevated. This calls for keen observation of all standards while producing nuclear batteries so as to avoid the leakage of radioactive substances. Economic feasibility will be dictated by advantages and its applications. With a variety of features being added to these researchers, nuclear batteries will undoubtedly be one of the greatest inventions made in human history. Dose calculator Since we live in a radioactive universe where radiation is a part of the natural environment, its essential to measure the radiation dose. The unit used to measure is known as the millirem (mrem). The regarded annual dose in every person should be around 350mrems, whether it comes from a natural or a man-made source. It is nor desirable for any individual to receive more than that dose annually. Absorbed dose refers to a quantity of radiation experienced by a person in the body. The absorbed dose units are (rad) and gray (Gy). Dose equivalent adds together the radiation quantity that is absorbed with the medical effects of that radiation type. For the beta and the gamma rays found in smartphones have the same dose equivalent as the absorbed dose. The dose equivalent for these rays is much higher than those of the neutron and the alpha. This is because these types are more harmful to human body. The dose equivalent units are the roentgen man (rem) and the sievert (Sv). The biological equivalent of the dose is estimated in 1/1000th of a rem, which is known as millirem. For practical purpose, 1R (exposure) = 1rad (absorbed dose) = 1 rem or 1000mrem (dose equivalent). A measure presented as Ci shows substances radioactivity. A measure in rem or mrem indicates the energy amount that is deposited in living tissues by a radioactive substance. Nuclear energy source will replace conventional cells as well as the adaptors; hence, the future will be full of exciting innovations with new ways of powering the portable devices. Although automobiles are in the first phase of their development, it is a clear indication of how nuclear energy is being employed. It is highly promising that the nuclear cells will definitely find a niche in automobiles and issues like running out of fuel or the battery life will come to an end. Though they pose a negative effect, the advantages brought about by nuclear batteries outweigh the disadvantages. The good thing is that these demerits are controllable. In future, the world of science will continue to use electric power from indispensable radioisotope. The scientific world argues that small devices ought to use small batteries to supply them with power. The urge for extra power arises as technology improves.

Triangles and Polygons on SAT Math Strategies and Practice Questions for Geometry

Triangles and Polygons on SAT Math Strategies and Practice Questions for Geometry SAT / ACT Prep Online Guides and Tips 25 to 30% of the SAT math section will involve geometry, and the majority of those questions will deal with polygons in some form or another. Polygons come in many shapes and sizes and you will have to know your way around them with confidence in order to ace those SAT questions on test day. Luckily, despite their variety, polygons are often less complex than they look, and a few simple rules and strategies will have you breezing through those geometry questions in no time. This will be your complete guide to SAT polygons- the rules and formulas for various polygons, the kinds of questions you’ll be asked about them, and the best approach for solving these types of questions. What is a Polygon? Before we talk about polygon formulas, let’s look at what exactly a polygon is. A polygon is any flat, enclosed shape that is made up of straight lines. To be â€Å"enclosed† means that the lines must all connect, and no side of the polygon can be curved. Polygons NOT Polygons Polygons come in two broad categories- regular and irregular. A regular polygon has all equal sides and all equal angles, while irregular polygons do not. Regular Polygons Irregular Polygons (Note: most all of the polygons on the SAT that are made up of five sides or more will be regular polygons, but always double-check this! You will be told in the question whether the shape is "regular" or "irregular.") The different types of polygons are named after their number of sides and angles. A triangle is made of three sides and three angles (â€Å"tri† meaning three), a quadrilateral is made of four sides (â€Å"quad† meaning four), a pentagon is made of five sides (â€Å"penta† meaning five), and so on. Most of the polygons you’ll see on the SAT (though not all) will either be triangles or some sort of quadrilateral. Triangles in all their forms are covered in our complete guide to SAT triangles, so let’s look at the various types of quadrilaterals you’ll see on the test. With polygons, you may notice that many definitions will fit inside other definitions. Quadrilaterals There are many different types of quadrilaterals, most of which are subcategories of one another. Parallelogram A parallelogram is a quadrilateral in which each set of opposite sides is both parallel and congruent (equal) with one another. The length may be different than the width, but both widths will be equal and both lengths will be equal. Parallelograms are peculiar in that their opposite angles will be equal and their adjacent angles will be supplementary (meaning any two adjacent angles will add up to 180 degrees). Rectangle A rectangle is a special kind of parallelogram in which each angle is 90 degrees. The rectangle’s length and width can either be equal or different from one another. Square If a rectangle has an equal length and width, it is called a square. This means that a square is a type of rectangle (which in turn is a type of parallelogram), but NOT all rectangles are squares. Rhombus A rhombus is a type of parallelogram in which all four sides are equal and the angles can be any measure (so long as their adjacents add up to 180 degrees and their opposite angles are equal). Just as a square is a type of rectangle, but not all rectangles are squares, a rhombus is a type of parallelogram (but not all parallelograms are rhombuses). Trapezoid A trapezoid is a quadrilateral that has only one set of parallel sides. The other two sides are non-parallel. Kite A kite is a quadrilateral that has two pairs of equal sides that meet one another. And here come the formulas- mwahaha! Polygon Formulas Though there are many different types of polygons, their rules and formulas build off of a few simple basic ideas. Let’s go through the list. Area Formulas Most polygon questions on the SAT will ask you to find the area or the perimeter of a figure. These will be the most important area formulas for you to remember on the test. Area of a Triangle $$(1/2)bh$$ The area of a triangle will always be half the amount of the base times the height. In a right triangle, the height will be equal to one of the legs. In any other type of triangle, you must drop down your own height, perpendicular from the vertex of the triangle to the base. Area of a Square $$l^2 \or {lw}$$ Because each side of a square is equal, you can find the area by either multiplying the length times the width or simply by squaring one of the sides. Area of a Rectangle $$lw$$ For any rectangle that is not a square, you must always multiply the base times the height to find the area. Area of a Parallelogram $$bh$$ Finding the area of a parallelogram is exactly the same as finding the area of a rectangle. Because a parallelogram may slant to the side, we say we must use its base and its height (instead of its length and width), but the principle is the same. You can see why the two actions are equal if you were to transform your parallelogram into a rectangle by dropping down straight heights and shifting the base. Area of a Trapezoid $$[(l_1+l_2)/2]h$$ In order to find the area of a trapezoid, you must find the average of the two parallel bases and multiply this by the height of the trapezoid. Now let's look at an example: In the figure, WXYZ is a rectangle with $\ov{WA} = \ov{BZ} = 4$. The area of the shaded region is 32. What is the length of $\ov{XY}$? [Note: figure not to scale] A. 6B. 8C. 12D. 16E. 20 First, let us fill in our given information. Our shaded figure is a trapezoid, so let us use the formula for finding the area of a trapezoid. area $=[(l_1+l_2)/2]h$ Now if we call the longest base q, the shortest base will be $q−4−4$, or $q−8$. (Why? Because the shortest leg is equal to the longest leg minus our two given lengths of 4). This means we can now plug in our values for the leg lengths. In addition, we are also given a height and an area, so we can plug all of our values into the formula in order to find the length of our longest side, q. $32=[(q+(q−8))/2]2$ $32=(2q+2q−16)/2$ $64=4q−16$ $80=4q$ $20=q$ The length of $\ov{XY}$ (which we designated $q$) is 20. Our final answer is E, 20. In general, the best way to find the area of different kinds of polygons is to transform the polygon into smaller and more manageable shapes. This will also help you if you forget your formulas come test day. For example, if you forget the formula for the area of a trapezoid, turn your trapezoid into a rectangle and two triangles and find the area for each. Let us look to how to solve the above problem using this method instead. We are told that the area of the trapezoid is 32. We also know that we can find the area of a triangle by using the formula ${1/2}bh$. So let us find the areas for both our triangles. ${1/2}bh$ ${1/2}(4)(2)$ ${1/2}8$ $4$ Each triangle is worth 8, so together, both triangles will be: $4+4$ $8$ Now if we add the area of our triangles to our given area of the trapezoid, we can see that the area of our full rectangle is: $32+8$ $40$ Finally, we know that we find the area of a rectangle by multiplying the length times the width. We have a given width of 2, so the length will be: $40=lw$ $40=2l$ 20=l The length of the rectangle (line $\ov{XY}$) will be 20. Again, our final answer is E, 20. Always remember that there are many different ways to find what you need, so don’t be afraid to use your shortcuts! Whichever solving path you choose depends on how you like to work best. Angle Formulas Whether your polygon is regular or irregular, the sum of its interior degrees will always follow the rules of that particular polygon. Every polygon has a different degree sum, but this sum will be consistent, no matter how irregular the polygon. For example, the interior angles of a triangle will always equal 180 degrees (to see more on this, be sure to check out our guide to SAT triangles), whether the triangle is equilateral (a regular polygon), isosceles, acute, or obtuse. All of these triangles will have a total interior degree measure of 180 degrees. So by that same notion, the interior angles of a quadrilateral- whether kite, square, trapezoid, or other- will always add up to be 360 degrees. Why? Because a quadrilateral is made up of two triangles. For example: One interior angle of a parallelogram is 65 degrees. If the remaining angles have measures of $a$, $b$ and $c$, what is the value of $a+b+c$? All quadrilaterals have an interior degree sum of 360, so: $a+b+c+65=360$ $a+b+c=295$ The sum of $\bi a, \bi b$, and $\bi c$ is 295. Interior Angle Sum You will always be able to find the sum of a polygon’s interior angles in one of two ways- by memorizing the interior angle formula, or by dividing your polygon into a series of triangles. Method 1: Interior Angle Formula $$(n−2)180$$ If you have an $n$ number of sides in your polygon, you can always find the interior degree sum by the formula $(n−2)$ times 180 degrees. If you picture starting from one angle and drawing connecting lines to every other angle to make triangles, you can see why this formula has an $n−2$. The reason being that you cannot make a triangle by using the immediate two connecting sides that make up the angle- each would simply be a straight line. To see this in action, let us look at our second method. Method 2: Dividing Your Polygon Into Triangles The reason the above formula works is because you are essentially dividing your polygon into a series of triangles. Because a triangle is always 180 degrees, you can multiply the number of triangles by 180 to find the interior degree sum of your polygon, whether your polygon is regular or irregular. Individual Interior Angles If your polygon is regular, you will also be able to find the individual degree measure of each interior angle by dividing the degree sum by the number of angles. (Note: $n$ can be used for both the number of sides and the number of angles; the number of sides and angles in a polygon will always be equal.) $${(n−2)180}/n$$ Again, you can choose to either use the formula or the triangle dividing method by dividing your interior sum by the number of angles. Angles, angler fish...same thing, right? Side Formulas As we saw earlier, a regular polygon will have all equal side lengths. And if your polygon is regular, you can find the number of sides by using the reverse of the formula for finding angle measures. A regular polygon with n sides has equal angles of 120 degrees. How many sides does the figure have? 3 4 5 6 7 For this question, it will be quickest for us to use our answers and work backwards in order to find the number of sides in our polygon. (For more on how to use the plugging in answers technique, check out our guide to plugging in answers). Let us start at the middle with answer choice C. We know from our angle formula (or by making triangles out of our polygons) that a five sided figure will have: $(n−2)180$ $(5−2)180$ $(3)180$ $540$ degrees. Or again, you can always find your degree sum by making triangles out of your polygon. This way you will still end up with $(3)180=540$ degrees. Now, we also know that this is a regular polygon, so each interior angle will be this same. This means we can find the individual angles by dividing the total by the number of sides/angles. So let us find the individual degree measures by dividing that sum by the number of angles. $540/5=108$ Answer choice C was too small. And we also know that the more sides a figure has, the larger each individual angle will be. This means we can cross off answer choices A and B (60 degrees and 90 degrees, respectively), as those answers would be even smaller. Now let us try answer choice D. $(n−2)180$ $(6−2)180$ $(4)180$ $720$ Or you could find your internal degree sum by once again making triangles from your polygons. Which would again give you $(4)180=720$ degrees. Now let’s divide the degree sum by the number of sides. $720/6=120$ We have found our answer. The figure has 6 sides. Our final answer is D, 6. Luckily for us, the SAT is predictable. You don't need a psychic to figure out what you're likely to see come test day. Typical Polygon Questions Now that we’ve been through all of our polygon rules and formulas, let’s look at a few different types of polygon questions you’ll see on the SAT. Almost all polygon questions will involve a diagram in some way (especially if the question involves any polygon with four or more sides). The few problems that do not use a diagram will generally be simple word problems involving rectangles. Typically, you will be asked to find one of three things in a polygon question: #1: The measure of an angle (or the sum of two or more angles)#2: The perimeter of a figure#3: The area of a figure Let’s look at a few real SAT math examples of these different types of questions. The Measure of an Angle: Because this hexagon is regular, we can find the degree measure of each of its interior angles. We saw earlier that we can find this degree measure by either using our interior angle formula or by dividing our figure into triangles. A hexagon can be split into 4 triangles, so $180 °*4=720$ degrees. There are 6 interior angles in a hexagon, and in a regular hexagon, these will all be equal. So: $720/6=120$ Now the line BO is at the center of the figure, so it bisects the interior angle CBA. The angle CBA is 120, which means that angle $x$ will be: $120/2=60$ Angle $x$ is 60 degrees. Our final answer is B, 60. The Perimeter of a Figure: We are told that ABCE is a square with the area of 1. We know that we find the area of a square by multiplying the length and the width (or by squaring one side), which means that: $lw=1$ This means that: $l=1$ And, $w=1$ We also know that every side is equal in a square. This means that $\ov{AB}, \ov{BC}, \ov{CE}, and \ov{AE}$ are ALL equal to 1. We are also told that CED is an equilateral triangle, which means that each side length is equal. Since we know that $\ov{CE} = 1$, we know that $\ov{CD}$ and $\ov{DE}$ both equal 1 as well. So the perimeter of the polygon as a whole- which is made of lines $\ov{AB}, \ov{BC}, \ov{CD}, \ov{DE}, and \ov{EA}$- is equal to: $1+1+1+1+1=5$ Our final answer is B, 5. [Note: don't get tricked into picking answer choice C! Even though each line in the figure is worth 1 and there are 6 lines, line $\ov{CE}$ is NOT part of the perimeter. This is an answer choice designed to bait you, so be careful to always answer only what the question asks.) The Area of a Figure: We are told that the length of the rug is 8 feet and that the length is also 2 feet more than the width. This means that the width must be: $8−2=6$ Now we also know that we find the area of a rectangle by multiplying width and length. So: $8*6=48$ The area of the rug is 48 square feet. Our final answer is B, 48. And now time for some practical how-to's, from tying a bow to solving your polygon questions. How to Solve a Polygon Question Now that we’ve seen the typical kinds of questions you’ll be asked on the SAT and gone through the process of finding our answers, we can see that each solving method has a few techniques in common. In order to solve your polygon problems most accurately and efficiently, take note of these strategies: #1: Break up figures into smaller shapes Don’t be afraid to write all over your diagrams. Polygons are complicated figures, so always break them into small pieces when you can. Break them apart into triangles, squares, or rectangles and you’ll be able to solve questions that would be impossible to figure out otherwise. Alternatively, you may need to expand your figures by providing extra lines and creating new shapes in which to break your figure. Just always remember to disregard these false lines when you’re finished with the problem. Because this is an awkward shape, let us create a new line and break the figure into two triangles. Next, let us replace our given information. From our definitions, we know that every triangle will have interior angles that add up to 180 degrees. We also know that the two angles we created will be equal. We can use this information to find the missing, equal, angle measures by subtracting our givens from 180 degrees. $180−30−20−20$ $110$ Now, we can divide that number in half to find the measurement of each of the two equal angles. $110/2$ $55$ Now, we can look at the smaller triangle as its own independent triangle in order to find the measure of angle z. Again, the interior angles will measure out to 180 degrees, so: $180−55−55$ $70$ Angle $z$ is 70 degrees. Our final answer is B, 70. #2: Use your shortcuts If you don’t feel comfortable memorizing formulas or if you are worried about getting them wrong on test day, don’t worry about it! Just understand your shortcuts (for example, remember that all polygons can be broken into triangles) and you’ll do just fine. #3: When possible, use PIA or PIN Because polygons involve a lot of data, it can be very easy to confuse your numbers or lose track of the path you need to go down to solve the problem. For this reason, it can often help you to use either the plugging in answer strategy (PIA) or the plugging in numbers strategy (PIN), even though it can sometimes take longer (for more on this, check out our guides to PIA and PIN). #4: Keep your work organized There is a lot of information to keep track of when working with polygons (especially once you break the figure into smaller shapes). It can be all too easy to lose your place or to mix-up your numbers, so be extra vigilant about your organization and don’t let yourself lose a well-earned point due to careless error. Ready? Test Your Knowledge Now it's time to test your knowledge with real SAT math problems. 1. 2. 3. Answers: D, B, 6.5 Answer Explanations 1. Again, when dealing with polygons, it's useful to break them into smaller pieces. For this trapezoid, let us break the figure into a rectangle and a triangle by dropping down a height at a 90 degree angle. This will give us a rectangle, which means that we will be able to fill in the missing lengths. Now, we can also find the final missing length for the leg of the triangle. Since this is a right triangle, we can use the Pythagorean theorem. $a^2+b^2=c^2$ $x^2+15^2=17^2$ $x^2+225=289$ $x^2=64$ $x=8$ Finally, let us add up all the lines that make up the perimeter of the trapezoid. $17+20+15+20+8$ $80$ Our final answer is D, 80. 2. We are told that the larger polygon has equal sides and equal angles. We can also see that the shaded figure has 4 sides and angles, which means it is a quadrilateral. We know that a quadrilateral has 360 degrees, so let us subtract our givens from 360. $x+y=80$ $360−80=280$ Again, we know that the polygon has all equal angles, so we can find the individual degree measures by dividing this found number in half. $280/2=140$ Each interior angle of the polygon will have 140 degrees. Now, we can find the number of sides by either reversing our polygon side formula or by plugging in answers. Let's look at both methods. Method 1: Formula $${(n−2)180}/n$$ We know that this formula gives us the measure of each interior angle, so let us use the knowledge of our individual interior angle (our found 140 degrees) and plug it in to find n, the number of sides. $140={(n−2)180}/n$ $140n=(n−2)180$ $140n=180n−360$ $−40n=−360$ $n=9$ Our polygon has 9 sides. Our answer is B, 9. Method 2: Plugging in answers We can also use our method of plugging in answers to find the number of sides in our polygon. As always, let us select answer option C. Answer choice C gives us 8 sides. We know that a polygon with eight sides will be broken into 6 triangles. So it will have: $180*6$ $1080$ degrees total Now, if we divide this total by the number of sides, we get: $1080/8$ $135$ Each interior angle will be 135 degrees. This answer is close, but not quite what we want. We also know that the more sides a regular polygon has, the larger each interior angle measure will be (an equilateral triangle's angles are each 60 degrees, a rectangle's angles are each 90 degrees, and so on), so we need to pick a polygon with more than 8 sides. Let us then try answer choice B, 9 sides. We know that a 9-sided polygon will be made from 7 triangles. This means that the total interior degree measure will be: $180*7$ $1260$ And we know that each angle measure will be equal, so: $1260/9$ $140$ We have found our correct answer- a 9-sided polygon will have individual angle measures of 140 degrees. Our final answer is B, nine. 3. Let us begin by breaking up our figure into smaller, more manageable polygons. We know that the larger rectangle will have an area of: $2*1$ $2$ The smaller rectangle will have an area of: $1*x$ $x$ (Note: we are using $x$ in place of one of the smaller sides of the small rectangles, since we do not yet know its length) We are told that the total area is $9/4$, so: $2+x=9/4$ $x=9/4−2$ $x=9/4−8/4$ $x=1/4$ Now that we know the length of x, we can find the perimeter of the whole figure. Let us add all of the lengths of our exposed sides to find our perimeter. $1+2+1+0.25+1+0.25+1$ $6.5$ Our perimeter is $6.5.$ Our final answer is 6.5. I think you deserve a present for pushing through on polygons, don't you? The Take Aways Though polygon questions may seem complicated, all polygons follow just a handful of rules. You may come across irregular polygons and ones with many sides, but the basic strategies and formulas will apply regardless. So long as you follow your solve steps, keep your work well organized, and remember your key definitions, you will be able to take on and solve polygon questions that once seemed utterly obscure. What’s Next? Phew! You knocked out polygons and now it's time to make sure the rest of your math know-how is in top shape. First, make sure you have working knowledge of all the math topics on the SAT so that you can get a sense of your strengths and weaknesses. Next, find more topic-specific SAT math guides like this one so that you can turn those weak areas into strengths. Need to brush up on your probability questions? Fractions and ratios? Lines and angles? No matter what topic you need, we've got you covered. Running out of time on the SAT math? Look to our guide on how to best boost your time (and your score!). Worried about test day? Take a look at how you should prepare for the actual day in question. Want to get a perfect score? 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