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Question Number 114447 by bemath last updated on 19/Sep/20

Commented by bemath last updated on 19/Sep/20

$$gavekudosallmaster$$

Answered by abdullahquwatan last updated on 19/Sep/20

$$MisalkanED=x$$$$Luas\mathrm{\Delta }AEC=\frac{1}{2}\times 9\times (4-x)$$$$CB=\sqrt{{12}^2+4^2}$$$$=4\sqrt{10}$$$$CE=\sqrt{9^2+x^2}$$$$=\sqrt{x^2+81}$$$$EB=\sqrt{{(4-x)}^2+3^2}$$$$=\sqrt{16-8x+x^2+9}$$$$=\sqrt{x^2-8x+25}$$$$$$$$CB=CE+EB$$$$4\sqrt{10}=\sqrt{x^2+81}+\sqrt{x^2-8x+25}$$$$160=x^2+81+2\sqrt{(x^2+81)(x^2-8x+25)}+x^2-8x+25$$$$didapatx=3$$$$$$$$Luas\mathrm{\Delta }AEC$$$$=\frac{1}{2}\times 9\times (4-3)$$$$=\frac{9}{2}satuan$$

Answered by 1549442205PVT last updated on 19/Sep/20

$${\mathrm{S}}_{\mathrm{ABDC}}=\frac{(\mathrm{AB}+\mathrm{CD}).\mathrm{AD}}{2}=\frac{(3+9).4}{2}=24$$$$\mathrm{AB}//\mathrm{CD}\mathrm{hence}\mathrm{\Delta }\mathrm{AEB}\simeq \mathrm{\Delta }\mathrm{DEC}$$$$\Rightarrow \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{AB}}{\mathrm{CD}}=\frac{3}{9}=\frac{1}{3}\left(1\right)$$$$\Rightarrow \frac{{\mathrm{S}}_{\mathrm{AEB}}}{{\mathrm{S}}_{\mathrm{DEC}}}={\left(\frac{1}{3}\right)}^2=\frac{1}{9}\left(2\right)$$$$\mathrm{Since}{\mathrm{S}}_{\mathrm{ACD}}={\mathrm{S}}_{\mathrm{BCD}}=\mathrm{m}(\mathrm{both}\mathrm{triangles}$$$$\mathrm{have}\mathrm{common}\mathrm{base}\mathrm{CD}\mathrm{and}\mathrm{two}$$$$\mathrm{equal}\mathrm{altitude}\mathrm{from}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{to}\mathrm{CD}),$$$${\mathrm{S}}_{\mathrm{AEC}}={\mathrm{S}}_{\mathrm{BED}}=\mathrm{x}(\mathrm{both}\mathrm{equal}\mathrm{to}\mathrm{m}-{\mathrm{S}}_{\mathrm{CDE}})$$$$\mathrm{Put}{\mathrm{S}}_{\mathrm{AEB}}=\mathrm{a},\mathrm{from}\left(2\right)\mathrm{we}{\mathrm{getS}}_{\mathrm{DEC}}=9\mathrm{a}$$$${\mathrm{S}}_{\mathrm{ABCD}}=24=9\mathrm{a}+\mathrm{a}+2\mathrm{x}=10\mathrm{a}+2\mathrm{x}\left(3\right)$$$$\mathrm{We}\mathrm{have}\frac{{\mathrm{S}}_{\mathrm{AEC}}}{{\mathrm{S}}_{\mathrm{AEB}}}=\frac{\mathrm{CE}}{\mathrm{EB}}=3\left(\mathrm{followsfrom}\left(1\right)\right)$$$$\Rightarrow {\mathrm{S}}_{\mathrm{AEC}}=\mathrm{x}=3\mathrm{a}.\mathrm{Replace}\mathrm{into}\left(3\right)\mathrm{we}\mathrm{get}$$$$24=10\mathrm{a}+2.3\mathrm{a}=16\mathrm{a}\Rightarrow \mathrm{a}=24/16=1.5$$$$\Rightarrow \mathrm{x}=3\mathrm{a}=4.5$$$$\mathrm{Thus},{\mathrm{S}}_{\mathrm{AEC}}=4.5$$

Answered by mr W last updated on 19/Sep/20

$$\frac{AE}{AD}=\frac{AB}{AB+CD}=\frac{3}{3+9}=\frac{1}{4}$$$$\Rightarrow AE=\frac{1}{4}\times 4=1$$$${\mathrm{\Delta }}_{AEC}=\frac{AE\times CD}{2}=\frac{1\times 9}{2}=4.5$$

Answered by Aziztisffola last updated on 19/Sep/20

$$\frac{\mathrm{AE}}{\mathrm{AD}-\mathrm{AE}}=\frac{\mathrm{AB}}{\mathrm{CD}}\Rightarrow \mathrm{AE}\times \mathrm{CD}=\mathrm{AB}\times \mathrm{AD}-\mathrm{AB}\times \mathrm{AE}$$$$9\mathrm{AE}=12-3\mathrm{AE}\Rightarrow \mathrm{AE}=\frac{12}{12}=1$$$$\mathrm{ED}=3$$$$△\mathrm{AEC}=△\mathrm{ACD}-△\mathrm{CED}$$$$=\frac{9\times 4}{2}-\frac{9\times 3}{2}$$$$=\frac{9}{2}=4.5$$

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