I need help with this geometry question
1 answer:
Answer:
radius ≈ 15.5
Step-by-step explanation:
the radius is RS
the angle between a tangent and the radius at the point of contact is 90°
then Δ RST is a right triangle
using Pythagoras' identity in the right triangle.
the square on the hypotenuse is equal to the sum of the squares on the other 2 sides , then
RS² + ST² = RT² ( substitute values )
RS² + 7² = 17²
RS² + 49 = 289 ( subtract 49 from both sides )
RS² = 240 ( take square root of both sides )
RS = ≈ 15.5 ( to 1 dec. place )
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Set up the given triangle on x-y coordinates with right angle at (0,0). So the two vertices are at (5,0) and (0,2![sqrt{x} n]{3}](https://tex.z-dn.net/?f=sqrt%7Bx%7D%20n%5D%7B3%7D%20)
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let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>
for a pt (x,y) on line sx+ty=1, the minimum of
equals to
smallest value happens at
so area is
hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145
let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>
for a pt (x,y) on line sx+ty=1, the minimum of
equals to
smallest value happens at
so area is
hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145