Solve for b. a = 1/2bcsquared
The length of one of the equal legs of an isosceles triangle is 8 cm less than 4 times the length of the base. If the perimeter is 29 cm, find the length of one of the equal legs.
Solve. 8x – 11 = 2x + 1
2007-01-21 10:44:07 · 3 answers · asked by robpurpleblazekamp 2 in Science & Mathematics ➔ Mathematics
1. a=(1/2)bc^2
Divide each side by c^2
a/(c^2)=(1/2)b
Multiply each side by 2.
2a/(c^2)=b
2. Draw an isosceles triangle with a base (x) and two equal sides (y). We know that 2y+x=29 because that is the perimeter. Solve the equation for x.
x=29-2y
We also know that 4x-8=y because of the given statements. Plug x into the equation (x=29-2y).
4(29-2y)-8=y
116-8y-8=y
108=9y
y=12
One of the equal legs is equal to 12 cm.
I hope that this helps.
2007-01-21 10:54:30 · answer #1 · answered by Anonymous · 0⤊ 0⤋
8x-11 = 2x + 1
8x = 2x + 12
6x = 12
x = 2
2.
The isosceles triangle will have perimeter 2S + B (2 times side, plus base) = 29.
S = 4B -8
2(4B -8) + B = 29
8B -16 + B = 29
9B - 16 = 29
9B = 45
B = 5
29 - 5 = 2S
S = 12
2007-01-21 18:57:44 · answer #2 · answered by John T 6 · 0⤊ 0⤋
1) b=2a/csquared
2) call a the base, and b each of the two eual legs. b=4a-8: 29=a+2b; 29=a+2(4a-8); 29=a+8a-16; 45=9a; a=45/9; a=5
3) 8x-11=2x+1; 6x=12; x=12/6; x=2
2007-01-21 19:00:01 · answer #3 · answered by Anonymous · 0⤊ 0⤋