Haven't heard "catheter" used as being a side in a right-angled triangle in a long time ....
I made the diagram, let AD = x, then DC = 9.4-x
(9.4-x)^2 + 6.2^2 = x^2
88.36 - 18.8x + x^2 + 38.44 = x^2
18.8x = 126.8
x = appr 6.74
so DC = ...
b) Let the 2 legs be 2x and 3x
if h is the hypotenuse:
h^2 = 4x^2 + 9x^2
h = √13 x
Look at your sketch, you have similar triangles, so set up ratios of corresponding sides
I see: 2/(2x) = 3x/√13 x = 3/√13
6x = 2√13
x = √13/3
then the hypotenuse = √13(√13/3) = 13/3
Two questions here:
1. In a right-angled triangle ABC, catheter AC is 9.4 cm long and catheter BC is 6.2 cm long. D is a point on the catheter AC so that DA = DB. Determine the length of the CD.
2. In a right-angled triangle, the ratio of catheters is 2: 3. The height towards the hypotenuse is 2.0 cm long. Determine the length of the hypotenuse.
4 answers
catheter?
AD^2 = 6.2^2 + 9.4^2 = 38.44 + 88.36 = 126.8
so
AD = 11.26
let CD = q
let DA = DB =p
q + p = 9.4 so p = 9.4-q
q^2 + 6.2^2 = p^2 = (9.4-q)^2
q^2 + 38.44 = 88.36 - 18.8 q + q^2
18.8 q = 49.92
q = 2.65
AD^2 = 6.2^2 + 9.4^2 = 38.44 + 88.36 = 126.8
so
AD = 11.26
let CD = q
let DA = DB =p
q + p = 9.4 so p = 9.4-q
q^2 + 6.2^2 = p^2 = (9.4-q)^2
q^2 + 38.44 = 88.36 - 18.8 q + q^2
18.8 q = 49.92
q = 2.65
2. In a right-angled triangle, the ratio of catheters is 2: 3. The height towards the hypotenuse is 2.0 cm long. Determine the length of the hypotenuse.
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Get that Catheter out of me !
2 a and 3 a legs so
hypotenuse = a * sqrt 13
sin of one angle in the corner = 2/sqrt 13 = 2/3a
3 a = sqrt 13
a =sqrt 13/3
hyp = a sqrt 13 = 13/3
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Get that Catheter out of me !
2 a and 3 a legs so
hypotenuse = a * sqrt 13
sin of one angle in the corner = 2/sqrt 13 = 2/3a
3 a = sqrt 13
a =sqrt 13/3
hyp = a sqrt 13 = 13/3
plural of cathetus is catheti, not catheter !
2 answers