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If pq=3x, sq=5x-8, and os=x+1, find a.pq b.sq c.os d.oq * i will draw the shape
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|dw:1334613812899:dw|
PA and SQ are tangents to the circle?
PQ = SQ so you can solve for x: 3x= 5x-8
@dpaInc presumably. I giving him the benefit of the doubt.
***PQ
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The only "tricky" part is knowing that the tangent and radius form a right angle, so you can use pythagoras to find QO, then subtract the radius OP to get QR
pq = sq tangents from an external point are equal then pq = sq 3x = 5x - 8 8 = 2x x = 4 pq = 12, sq = 12
os = 4 + 1 = 5
and to find oq use pythagoras' theorem the tangent and radius intersect at right angles... so \[oq^2 = 12^2 + 5^2\] oq = 13
thank you guys.
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