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Question Number 214228 by mr W last updated on 02/Dec/24

Commented by mr W last updated on 02/Dec/24

$$findareaofthesquare$$

Answered by aleks041103 last updated on 02/Dec/24

$$letthepointhaveanameG$$$$andletitbeat(y,z)$$$$letBbeat(0,0)$$$$letCbeat(x,0)$$$$letAbeat(0,x)$$$$then:$$$$y^2+{(x-z)}^2=7^2$$$$y^2+z^2=4^2$$$${(x-y)}^2+z^2=9^2$$$$$$$$x^2+y^2+z^2-2xz=49$$$$x^2+y^2+z^2-2xy=81$$$$y^2+z^2=16$$$$$$$$x^2-33=2xz$$$$x^2-65=2xy$$$$y^2+z^2=16$$$$$$$$\Rightarrow x(z-y)=16,4x^{2}yz=(x^2-33)(x^2-65)$$$$\Rightarrow x^2{(z-y)}^2={16}^2=$$$$=x^2(y^2+z^2)-2x^{2}yz=$$$$=16x^2-\frac{1}{2}(x^2-33)(x^2-65)$$$$\Rightarrow S=x^{2}and$$$$2.{16}^2=2.16S-(S-33)(S-65)$$$$(S-33)(S-65)-32S+512=0$$$$S^2-130S+33.65+512=0$$$$S^2-130S+2657=0$$$$\Rightarrow S=65\pm \sqrt{{65}^2-2657}=65\pm \sqrt{1568}$$$$$$$$remainstobecheckedwhetherthese$$$$arevalidsolutions.$$$$(alsoimayhavemadeamistakesomewhere;))$$

Commented by mr W last updated on 02/Dec/24

$$S=65+\sqrt{1568}=65+28\sqrt{2}$$$$\Rightarrow correct!$$

Answered by mr W last updated on 02/Dec/24

Commented by mr W last updated on 02/Dec/24

$$\mathrm{\Delta }BEA\equiv \mathrm{\Delta }BFG$$$$\cos \beta =\frac{{\left(4\sqrt{2}\right)}^2+7^2-9^2}{2\times 4\sqrt{2}\times 7}=0\Rightarrow \beta =90°$$$${AB}^2=4^2+7^2-2\times 4\times 7\cos 135°$$$$=65+28\sqrt{2}=areaofsquare$$

Commented by MATHEMATICSAM last updated on 06/Dec/24

$$\mathrm{How}\mathrm{did}\mathrm{you}\mathrm{get}\mathrm{that}9\mathrm{unit}\mathrm{side}\mathrm{of}\mathrm{the}$$$$\mathrm{triangle}\mathrm{constructed}\mathrm{outside}\mathrm{of}\mathrm{AB}?$$

Commented by MATHEMATICSAM last updated on 06/Dec/24

$$\mathrm{you}\mathrm{actually}\mathrm{wanted}4\mathrm{units},9\mathrm{units},\mathrm{A}$$$$\mathrm{limited}\mathrm{AB}\mathrm{and}\mathrm{you}\mathrm{also}\mathrm{defined}\mathrm{the}\mathrm{angle}$$$$\mathrm{of}4\mathrm{units}\mathrm{side}\mathrm{with}\mathrm{AB}.\mathrm{how}\mathrm{is}\mathrm{it}$$$$\mathrm{possible}\mathrm{to}\mathrm{draw}\mathrm{a}\mathrm{triangle}\mathrm{with}$$$$\mathrm{all}\mathrm{four}\mathrm{conditions}?$$

Commented by mr W last updated on 06/Dec/24

$$sinceAB=BC,ifyousetAE=9and$$$$BE=4,then\mathrm{\Delta }ABE\equiv \mathrm{\Delta }CBF.$$

Commented by mr W last updated on 06/Dec/24

$${it}^{\prime }saveryclearthing,soi{don}^{\prime }t$$$$reallyknow{what}^{\prime }syourproblemis.$$

Answered by TonyCWX08 last updated on 02/Dec/24

Answered by TonyCWX08 last updated on 02/Dec/24

$$BF=2\sqrt{2}$$$${BF}^2+{FC}^2={BC}^2$$$${\left(2\sqrt{2}\right)}^2+{(2\sqrt{2}+7)}^2={BC}^2$$$${BC}^2=8+8+28\sqrt{2}+49=65+28\sqrt{2}=Area$$

Commented by TonyCWX08 last updated on 02/Dec/24

$$PE=\sqrt{4^2+4^2}=4\sqrt{2}$$$${PE}^2+{EC}^2$$$$={\left(4\sqrt{2}\right)}^2+{\left(7\right)}^2$$$$=81$$$$={PC}^2$$$${Triangle}_{PEC}isarightangletriangle.$$$$ECiscollinearwithFC.$$

Commented by mr W last updated on 02/Dec/24

$$youmustproveatfirstthatFEand$$$$ECarecollinear!otherwiseyou$$$$cannottakeFC=FE+EC=2\sqrt{2}+7.$$

Commented by TonyCWX08 last updated on 02/Dec/24

$$Mydiagramisabitmisleading.$$

Commented by mr W last updated on 02/Dec/24

$$yes!goodapproach!$$

Commented by TonyCWX08 last updated on 02/Dec/24

$$Fixed.$$

Answered by ajfour last updated on 03/Dec/24

$$p^2+q^2=16$$$$s=q+\sqrt{49-p^2}=p+\sqrt{81-q^2}$$$$\Rightarrow s^2-2qs=33$$$$\mathrm{\&}{s}^2-2ps=65$$$$\Rightarrow {(s^2-33)}^2+{(s^2-65)}^2=64s^2$$$$2s^2-260s^2+1089+4225=0$$$$s=65\pm \sqrt{4225-\frac{1089}{2}-\frac{4225}{2}}$$$$s=65\pm \sqrt{1568}$$$$s=65\pm 28\sqrt{2}$$

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