3 years ago

I need help on this???

2 answers:

4 0

Use the geometric mean for right triangles here. Like this $\frac{9}{y}= \frac{y}{7}$ . Cross multiply to get $y^{2}=63$ , and $y= \sqrt{63}$ . That simplifies down to $y= \sqrt{9*7}$ . Pull out the 9 as a perfect square of 3 and you're left with $y=3 \sqrt{7}$ , first choice above.

Hitman42 [59]3 years ago

3 0

Another way to do this is to apply pythagoras 3 times, since there are 3 right triangles. Assuming the sides not named here are a and b, you get:

a^2 + b^2 = 16^2
9^2 + y^2 = a^2
y^2 + 7^2 = b^2

You can solve these to y=3√7

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Find the length of the tangent segment AB to the circles centered at O and O' whose radii are a and b respectively when the circ

bixtya [17]

Answer:

$AB=2\sqrt{ab}$

Step-by-step explanation:

From figure,

$OA=a, \quad \quad O'B=b\\\Rightarrow OO'= (a+b) \quad \quad \text{and}\quad OD=(a-b)$

In triangle $OO'D$

$(OO')^2=(OD)^2+(O' D)^2$

$\Rightarrow (a+b)^2=(a-b)^2+(O' D)^2\\\Rightarrow a^2+b^2+2ab-a^2-b^2+2ab=(O' D)^2\\\Rightarrow 4ab=(O' D)^2\\\Rightarrow O'D=2\sqrt{ab} \\\Rightarrow O' D=2\sqrt{ab}=AB \quad \quad [\because O' DAB\;\; \text{is a rectangle.}]$