Question Number 213626 by Tawa11 last updated on 10/Nov/24
Commented by Tawa11 last updated on 10/Nov/24
The area of ABCD is 40.5 square units .
Find the area of the unshaded region
Find the area of the unshaded region
Commented by Frix last updated on 10/Nov/24
$$\mathrm{In}\:\mathrm{any}\:\mathrm{trapezoid}\:\left({AB}\parallel{CD}\right)\:\mathrm{with}\:{AB}={a}, \\ $$$${CD}={c},\:\mathrm{height}\:{h} \\ $$$$\mathrm{the}\:\mathrm{areas}\:\mathrm{are} \\ $$$$\Box_{{ABCD}} =\frac{\left({a}+{c}\right){h}}{\mathrm{2}} \\ $$$$\bigtriangleup_{{ABM}} =\frac{{a}^{\mathrm{2}} {h}}{\mathrm{2}\left({a}+{c}\right)} \\ $$$$\bigtriangleup_{{ADM}} =\bigtriangleup_{{BDM}} =\frac{{ach}}{\mathrm{2}\left({a}+{c}\right)} \\ $$$$\bigtriangleup_{{CDM}} =\frac{{c}^{\mathrm{2}} {h}}{\mathrm{2}\left({a}+{c}\right)} \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{given}\:\mathrm{case}\:\mathrm{we}\:\mathrm{have} \\ $$$${a}={x}=\mathrm{5} \\ $$$${b}={y}=\mathrm{4} \\ $$$$\Box_{{ABCD}} =\frac{\mathrm{81}}{\mathrm{2}} \\ $$$$\Rightarrow\:{h}=\mathrm{9} \\ $$$$\Rightarrow \\ $$$$\bigtriangleup_{{ABM}} =\frac{\mathrm{25}}{\mathrm{2}} \\ $$$$\bigtriangleup_{{ADM}} =\bigtriangleup_{{BDM}} =\mathrm{10} \\ $$$$\bigtriangleup_{{CDM}} =\mathrm{8} \\ $$$$ \\ $$$$\mathrm{shaded}=\mathrm{10} \\ $$$$\mathrm{unshaded}=\frac{\mathrm{61}}{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 11/Nov/24
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Answered by A5T last updated on 10/Nov/24
Commented by A5T last updated on 10/Nov/24
![Assuming a trapezium(AB∥DC) ((h(4+5))/2)=40.5⇒h=9 ⇒[ABD]=[ABC]=((9×5)/2)=22.5; [BCD]=((4×9)/2)=18 [ADE]=x=[BCE]⇒[ABE]=22.5−x;[DCE]=18−x (x/(22.5−x))=((18−x)/x)⇒[ADE]=x=10](https://www.tinkutara.com/question/Q213630.png)
$${Assuming}\:{a}\:{trapezium}\left({AB}\parallel{DC}\right) \\ $$$$\frac{{h}\left(\mathrm{4}+\mathrm{5}\right)}{\mathrm{2}}=\mathrm{40}.\mathrm{5}\Rightarrow{h}=\mathrm{9} \\ $$$$\Rightarrow\left[{ABD}\right]=\left[{ABC}\right]=\frac{\mathrm{9}×\mathrm{5}}{\mathrm{2}}=\mathrm{22}.\mathrm{5};\:\left[{BCD}\right]=\frac{\mathrm{4}×\mathrm{9}}{\mathrm{2}}=\mathrm{18} \\ $$$$\left[{ADE}\right]={x}=\left[{BCE}\right]\Rightarrow\left[{ABE}\right]=\mathrm{22}.\mathrm{5}−{x};\left[{DCE}\right]=\mathrm{18}−{x} \\ $$$$\frac{{x}}{\mathrm{22}.\mathrm{5}−{x}}=\frac{\mathrm{18}−{x}}{{x}}\Rightarrow\left[{ADE}\right]={x}=\mathrm{10} \\ $$
Commented by Tawa11 last updated on 11/Nov/24
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$