DG¯¯¯¯¯¯ , EG¯¯¯¯¯ , and FG¯¯¯¯¯ are perpendicular bisectors of the sides of △ABC . DG=5 cm and BD=12 cm.

Mathematics

1 answer:

jonny [76]3 years ago

3 0

Given that DG, EG, and FG are perpendicular bisectors of the sides of △ABC, this means that the point of intersection, G, is the circumcenter of the triangle and hence AG, BG, and CG are equal.

Given that DG = 5 cm and BD = 12 cm, then

$BG= \sqrt{DG^2+BD^2} \\ \\ = \sqrt{5^2+12^2} = \sqrt{25+144} \\ \\ = \sqrt{169} =13$

Since AG = BG = CG, therefore, CG = 13 cm.

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Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

Step-by-step explanation:

(1) (12, 18, 27, ...)

The common ratio is:

$r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$