<p>One side of a triangle has a lngth 6 and a second side has length 7. Which of the following could be the area of this triangle?</p>
<p>I 13
II 21
III 24</p>
<p>A. I oonly
B II only
C III only
D I and II only
E I II and III</p>
<p>the answe is D I and II only. please help me understand how you arrived at this conclusion mathematically....</p>
<p>arhhhh!! i keep getting B only as an answer lol.</p>
<p>For the area to be x the altitude a from the side that is 6 long must be x/3 (since x=(a*6)/2=a/3). Now think of the side that is 7 inches long as attached to one end of the 6-inch side by a hinge. Swing it around, starting with it flush against the six inch side, then going until it is fully extended, so that the two sides form a single straight line. It is clear that the altitude from the 6-inch side can be anything from 0 to 7, but no higher than 7 (i.e., the altitude when the 7- and 6-inch sides are perpendicular). We can eliminate 24, since 24/3=8, but 13 and 21 are possible areas because 13/3 and 21/3 are less than 8. I wish I could draw a picture but to help you visualize it use two pencils and pretend one is 6 inches long and one is 7 inches long.</p>
<p>easier explanation:</p>
<p>the maximum area the triangle can have, the maximum being a right triangle, is .5 X b X h. so that would give u 42/2 or 21. so the total area has to be 21 or less, thus 21 and 13 are the answer, and 24 is above the maximum.</p>
<p>I had trouble w/ this question as well... how do you KNOW that those are the bases and heights... for all we know they can be the lengths of a scalene triangle. This question was bad.</p>
<p>The area of a triangle with sides a, b with included angle c is given by:</p>
<p>Area = (ab*sin(c))/2</p>
<p>Since a and b are known, we know area will be at a max when sin(c) = pi/2</p>
<p>So, yes, the maximum possible area is 21.</p>
<p>How do you know for sure a = that and b = that. Like you i used that formula (is that heron's or is heron's square root of s (s-a) (s-b) (s-c)?). But i still didn't like the wording.</p>
<p>heron's formula!!! god chill out. i don't think u ever have to use that.</p>
<p>they tell u what a = and be =. they r equal to the sides they give u.</p>
<p>oh btw if u know calculus u can find a graph and integrate it.</p>
<p>Another way to look at the problem is to treat the area as 1/2 the area of a parrallelogram with sides 6 and 7. As you vary the angle (choose any corner) from 0 through 180 degrees, you will find the area maximized for a 90 degree corner.</p>
<p>The maximum area is then 1/2 of 6*7 = 21</p>
<p>"how do you KNOW that those are the bases and heights... for all we know they can be the lengths of a scalene triangle. This question was bad."</p>
<p>We know that two of the sides are 6 and 7 inches long. We do NOT know the angle between them, but whatever it is, the area of the triangle is (1/2)6h, where h is the altitude of the triangle when it is turned so that the 6 inch side is the base. So the question is, what values are possible for h? Well, it can be very small indeed; think of the case where the angle between the 6 and 7 inch sides is 1/100 of a degree, or if you prefer, pi/10000. Or it can be 7 inches; picture the case where the angle is 90 degrees. It cannot, however, be more than 7. But it can be anything in between. Draw a few triangles that satisfy the condition and you will see why. Thus, h can be 6 (so that the area is 21) or 4.333... (so that the area is 13), but not 8, so that the area can't be 24. </p>
<p>You don't need Heron's formula, the law of cosines, calculus, or anything as complicated as that. It helps to remember that the maximum altitude of a triangle with two given sides is in the case where the two given sides are perpendicular, but if you think visually you don't even have to remember that.</p>
<p>yea i did this all today in the morning. I was pondering over it and finally understood why AFTER i drew a diagram.</p>
<p>yeah, but can you find out the minimum area of the triangle, or do you just say ok the max is 21 . and it can be NETHING less than that. o wait as i type that out it sounds right.. hmmm</p>
<p>There is no minimum. You can get the area arbitrarily close to zero by making the angle between the 6-inch and 7-inch sides as small (sharp) as you want. Then the altitude from the 6-inch side to the opposite angle is very small, and so the area is very small. (But you can't get all the way to 0 because then you wouldn't have a triangle any more. (Though you can technically refer to that case as a "degenerate" triangle.)) Draw a picture and you'll see. Read post #10 again.</p>
<p>Just remember that the maximum area is the right triangle whose legs are the two given sides:</p>
<p>If h is the altitude on base a, then the area of the triangle is A = 2ah. Now 2ah is clearly a maximum when h is largest, and this occurs when h = b. If you draw a quick sketch, you will see this. So the maximum area should equal (1/2)(a)(b). Doesn't matter what side you plug in for a or b.</p>
<p>Anything less than (1/2)ab is valid.</p>