A poster in the shape of a triangle has one side that is five inches more than the length of the shortest side, and another side is three inches less than twice the shortest side. Find the dimensions of the poster if its perimeter is 42 inches.
2007-06-15 18:24:20 · 15 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
well you can deduce that...
the shortest side is = x
one of the side = x + 5
the other one side = 2x-3
since the total perimeter is 42 in
x + (x+5) + (2x-3) = 42
4x +2 = 42
4x = 42-2
4x=40
x=10in
so the dimentions is
x by (x+5) by (2x-3)
which is
10 by 15 by 17
.hope that helps.
2007-06-15 18:30:47 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Name the length of the shortest side x.
You will have the first side u mentioned as x + 5; the second side as 2x - 3.
And form the perimeter by adding three sides up:
x + x + 5 + 2x - 3 = 42
<=> 4x = 40
<=> x = 10 (inches)
Then you can count the other three dimensions:
10 + 5 = 15 (inches) and 2 . 10 - 3 = 17 (inches)
2007-06-16 01:32:46 · answer #2 · answered by The Last Riddle 2 · 0⤊ 0⤋
Assume that the shortest side is x inches.
Then the first side, mentioned in the question, is x + 5 inches.
Now the other side, is 2x - 3 inches, as per the statement in the question.
Addition of all the sides, gives the length of the perimeter, which is also given as 42 inches.
Thus, summing up the three sides, results in the equation:-
x + (x+ 5) + (2x -3) = 42.....(inches)
i.e., 4x + 2 = 42.....(inches)
or, 4x = 40.....(inches)
i.e., x = 10 inches.
x + 5 = 15 inches.
2x - 3 = 17 inches.
Thus, the triangle is a scalene triangle (of unequal sides), i.e.,
10, 15 & 17 inches, respectively.
2007-06-16 01:53:01 · answer #3 · answered by Sam 7 · 0⤊ 0⤋
check this and see if its right, The shortest side would be 10 inches and the one thats five inches more would b 15 and the side doubled the shortest side minus 3 would b 17 .. . perimeter of a triangle is P= a +b+c so 10+15+17=42 . . hope this helps u out
2007-06-16 01:40:45 · answer #4 · answered by fordguy 2 · 0⤊ 0⤋
take the shortest side to be x.
x+5, x, 2x-3
(x+5)+x+(2x-3)=42
x+5+x+2x-3=42
4x+2=42
4x=42-2
4x=40
x=40/4
x=10
replacing x=10 in the legths above,
x+5=10+5=15inches
x=10inches
2x-3=2*10-3=20-3=17inches
therefore the dimensions are 15, 10 and 17 inches.
2007-06-16 01:38:22 · answer #5 · answered by Anonymous · 0⤊ 0⤋
let the length of the shortest side be x.
now the three sides are x, x+5, and 2x-3.
So x + x+5 + 2x-3 = 42
4x + 2 = 42
4x = 40
x = 10
x+5 = 15
2x-3 = 17
2007-06-16 01:30:33 · answer #6 · answered by holdm 7 · 0⤊ 0⤋
Use algebra to set up an equation.
(x+5) + (2x - 3) + (x) = 42
4x + 2 = 42
4x = 40
x = 10
so the dimensions are ***15 : 17 : 10***
for a total of 42
2007-06-16 01:32:21 · answer #7 · answered by Wayne 2 · 0⤊ 0⤋
x = shortest side.
x+5 = 2nd side
2x-3 = 3rd side
x+x+5+2x-3=42
4x =40
x=10 = short side
x+5 = 15 = 2nd side
2x-3 = 17 = 3rd side
2007-06-16 01:30:03 · answer #8 · answered by ironduke8159 7 · 0⤊ 0⤋
10 in, 15 in, 17 in
2007-06-16 01:27:59 · answer #9 · answered by princess 2 · 0⤊ 0⤋
Shortest side = x
2nd side = x+5
3rd side = 2x-3
x+(x+5)+(2x-3) = 42.
4x+2 = 42
4x = 40
x = 10
DIMENSIONS:
Shortest side = 10
2nd = 15
3rd = 17
(10+15+17 = 42)
2007-06-16 01:29:38 · answer #10 · answered by Anonymous · 0⤊ 0⤋