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Question Number 199940 by cortano12 last updated on 11/Nov/23

Commented by Frix last updated on 11/Nov/23

$$\mid AD\mid =3$$$$\mid BD\mid =\frac{9}{2}$$$$r=\frac{\sqrt{15}}{4}$$

Commented by cortano12 last updated on 11/Nov/23

$$\mathrm{Find}\mathrm{AD}$$

Commented by cortano12 last updated on 11/Nov/23

$$\mathrm{how}\mathrm{get}\mathrm{AD}=3?$$

Answered by Frix last updated on 11/Nov/23

$$\mid AD{\mid }^2=\mid BD{\mid }^2+p\mid BD\mid +q$$$$\mathrm{We}\mathrm{know}$$$$1.\mid BD\mid =0\iff \mid AD\mid =6$$$$2.\mid BD\mid =8\iff \mid AD\mid =4$$$$\Rightarrow$$$$\mid AD\mid =\sqrt{\mid BD{\mid }^2-\frac{21}{2}\mid BD\mid +36}$$$$\mathrm{We}\mathrm{have}\mathrm{now}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{both}\mathrm{triangles}$$$${\mathrm{\Delta }}_1:6,\mid BD\mid ,\mid AD\mid$$$${\mathrm{\Delta }}_2:4,8-\mid BD\mid ,\mid AD\mid$$$$\mathrm{Now}\mathrm{use}\mathrm{the}\mathrm{formula}\mathrm{for}\mathrm{the}\mathrm{incircle}$$$$r_I(a,b,c)=\frac{\sqrt{2(a^2{b}^2+b^2{c}^2+c^2{a}^2)-(a^4+b^4+c^4)}}{2(a+b+c)}$$$$\mathrm{to}\mathrm{find}\mid BD\mid$$

Answered by mr W last updated on 11/Nov/23

$$AD=x$$$$BD=y,DC=8-y$$$$\frac{6+x+y}{y}=\frac{4+x+8-y}{8-y}$$$$\Rightarrow y=\frac{4(x+6)}{5+x}$$$$\cos B=\frac{6^2+y^2-x^2}{2\times 6y}=\frac{6^2+8^2-4^2}{2\times 6\times 8}$$$$\frac{6^2+y^2-x^2}{y}=\frac{21}{2}$$$$x^4+10x^3+15x^2-90x-216=0$$$$(x-3)(x+3)(x+4)(x+6)=0$$$$\Rightarrow x=3=AD$$

Commented by cortano12 last updated on 11/Nov/23

$$\mathrm{very}\mathrm{simple}$$

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