Find the altitude of an isosceles triangle with sides of 13 inches and a base of 10 inches.
2007-08-10 16:20:25 · 9 answers · asked by G 1 in Science & Mathematics ➔ Physics
it's 12 inches
2007-08-10 16:24:17 · answer #1 · answered by stanley l 5 · 0⤊ 0⤋
since its isoceles, the altitude passes through the center of the base.
if you imagine the altitude line dissecting the triangle two right triangles are created.
for each base = 5 {half the base of the whole triangle}
hypotenuse = 13 {the sides }
altitude = a { the line dissecting the triangle}
13 x 13 = (5 x 5) + (a x a)
169 = 25 +axa
subtract 25 from both sides
144 = a x a
square root of both sides
12 = altitude
2007-08-10 23:27:34 · answer #2 · answered by who knows 2 · 0⤊ 0⤋
Pythagoras theorem
altitude is the perpendicular bisector of the base
so
5^2+x^2=13^2
169=25+x^2
x=12 inches
2007-08-10 23:27:46 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Hi. Divide the base by two which gives you two right triangles with a base of 5 and a hypotenuse of 13. Then just trig it out.
2007-08-10 23:40:03 · answer #4 · answered by Cirric 7 · 0⤊ 0⤋
h^2 = 13^2 - 5^2. h^2 =169-25 =144. h =12. QED
2007-08-10 23:33:26 · answer #5 · answered by Renaissance Man 5 · 0⤊ 0⤋
i think its altitude=(13^2 - (10/2)^2) ^1/2 = 12inches
2007-08-10 23:28:06 · answer #6 · answered by astrid 1 · 0⤊ 0⤋
This is two 5,12,13 triangles joined. Thus, the height is 12
2007-08-13 16:38:44 · answer #7 · answered by galyamike 5 · 0⤊ 0⤋
13^2+10+^2= x^2
x= 16.4
2007-08-10 23:24:31 · answer #8 · answered by Anonymous · 0⤊ 1⤋
16.4 inches.
2007-08-14 11:13:43 · answer #9 · answered by johnandeileen2000 7 · 0⤊ 0⤋