The three sides of an isosceles triangle have lengths of 34, 34, and 32. Find the length of the altitude to the shortest side of this isosceles triangle.
2006-12-16 11:21:55 · 7 answers · asked by Quagmire77 1 in Science & Mathematics ➔ Mathematics
a² = b² + c²
34² = (32/2)² + c²
1156 - 256 = c²
c² = 900
c = \/900
c = 30
Answer: The length of the altitude to the shortest side of this isosceles triangle is 30.
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2006-12-16 12:42:54 · answer #1 · answered by aeiou 7 · 0⤊ 0⤋
Use the Pythagorean Theorem.
One leg of the triangle is 32/2 = 16. The other is h.
So the formula is:
h^2 + 16^2 = 34^2
h^2 = 34^2 - 16^2 = 1156 - 256 = 900
h = 30
2006-12-16 11:55:14 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
The altitude is perpendicular to, and bisects, the 32 side, so take 1/2 of that and use 34 as the hypotenuse to find the altitude
2006-12-16 11:27:24 · answer #3 · answered by arbiter007 6 · 0⤊ 0⤋
the altitude divides the base in half
a^2+b^2=c^2
a^2+16^2=34^2
a^2+256=1156
a^2=900
a=30
2006-12-16 11:27:32 · answer #4 · answered by 7 · 0⤊ 0⤋
X=length of shortest part X+4=length of part A 2X=length of part B X+(X+4)+2X=40 4 ==> 4X+4=40 4 ==> 4X=40 ==> X=10 short part = 10 meters part A = X+4 = 14 meters part B = 2X = 20 meters
2016-11-26 23:12:04 · answer #5 · answered by Anonymous · 0⤊ 0⤋
altitude=sq.rt. of (32^2-17^2)
=sq.rt.of(32+17)(32-17)
=sq.rt.of 49*15
=7rt15
=27.11
area=/2*34*27.11
this mustalso be equal to 1/2*32*h
equating
1/2*34*27.11=1/2*32*h
h=(34/32)*27.11
=28.80
2006-12-16 11:29:42 · answer #6 · answered by raj 7 · 0⤊ 0⤋
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2006-12-16 11:33:47 · answer #7 · answered by lori b 2 · 0⤊ 0⤋