the lengths of the sides of a triangle are consecutive odd integers.if the perimeter of the triangle is 27 what is are the lengt
2 answers:
Bearing in mind that, if you multiply any integer whatsoever, by 2, you get an even number, 3*2 or 17*2 or 4*2 and so on.
to get an ODD integer, simply hop once forward of backwards from any even integer, 2-1, or 2+1 or 8+1 or 8-1.
to get a consecutive ODD integer, simply hop twice from any odd integer, say 3+2, or 7+2 and so on.
now, let's pick a reference integer, say hmmm "a".
we know that a * 2, or 2a is an even integer, therefore (2a + 1) is an ODD integer, and (2a +1 ) + 2 is a consecutive odd integer. So, there you have it, our first two odd integers, and the next one, is, you guessed it, (2a + 1) + 2 + 2
so, our three odd consecutive integers are (2a+1), (2a+3) and (2a+5)
so, the triangle has a perimeter of 27, so, let's add them up then
so, our first odd integer is
2a + 1
2(3) + 1 --> 7
and surely you know what the other two are.
to get an ODD integer, simply hop once forward of backwards from any even integer, 2-1, or 2+1 or 8+1 or 8-1.
to get a consecutive ODD integer, simply hop twice from any odd integer, say 3+2, or 7+2 and so on.
now, let's pick a reference integer, say hmmm "a".
we know that a * 2, or 2a is an even integer, therefore (2a + 1) is an ODD integer, and (2a +1 ) + 2 is a consecutive odd integer. So, there you have it, our first two odd integers, and the next one, is, you guessed it, (2a + 1) + 2 + 2
so, our three odd consecutive integers are (2a+1), (2a+3) and (2a+5)
so, the triangle has a perimeter of 27, so, let's add them up then
so, our first odd integer is
2a + 1
2(3) + 1 --> 7
and surely you know what the other two are.
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Answer:
which agrees with option"B" of the possible answers listed
Step-by-step explanation:
Notice that in order to solve this problem (find angle JLF) , we need to find the value of the angle defined by JLG and subtract it from , since they are supplementary angles. So we focus on such, and start by drawing the radii that connects the center of the circle (point "O") to points G and H, in order to observe the central angles that are given to us as
and
. (see attached image)
We put our efforts into solving the right angle triangle denoted with green borders.
Notice as well, that the triangle JOH that is formed with the two radii and the segment that joins point J to point G, is an isosceles triangle, and therefore the two angles opposite to these equal radius sides, must be equal. We see that angle JOH can be calculated by :
Therefore, the two equal acute angles in the triangle JOH should add to:
resulting then in each small acute angle of measure
.
Now referring to the green sided right angle triangle we can find find angle JLG, using:
Finally, the requested measure of angle JLF is obtained via:
