Question Number 221170 by mr W last updated on 26/May/25

Commented by mr W last updated on 26/May/25

$${the}\:{area}\:{of}\:{the}\:{rectangle}\:{is}\:\mathrm{80}. \\ $$$${find}\:{the}\:{shaded}\:{area}. \\ $$
Answered by Rasheed.Sindhi last updated on 26/May/25

$${let}\:{width}\:{of}\:{rectangle}\:{is}\:\:{x}+\mathrm{5} \\ $$$${and}\:{length}\:{is}\:{y}+\mathrm{6} \\ $$$$\left({x}+\mathrm{5}\right)\left({y}+\mathrm{6}\right)=\mathrm{80} \\ $$$$\begin{array}{|c|}{{xy}+\mathrm{6}{x}+\mathrm{5}{y}=\mathrm{50}}\\\hline\end{array} \\ $$$${Right}\:{triangle}\:\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{2}}{xy} \\ $$$${Right}\:{triangle}\:\mathrm{2}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{6}\right)\left({x}+\mathrm{5}\right) \\ $$$${Right}\:{triangle}\:\mathrm{3}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}\right)\left({y}+\mathrm{6}\right) \\ $$$$?=\mathrm{80}−\frac{\mathrm{1}}{\mathrm{2}}\left({xy}+\mathrm{6}\left({x}+\mathrm{5}\right)+\mathrm{5}\left({y}+\mathrm{6}\right)\right) \\ $$$$\left.\:=\mathrm{80}−\frac{\mathrm{1}}{\mathrm{2}}\left({xy}+\mathrm{6}{x}+\mathrm{5}{y}+\mathrm{60}\right)\right) \\ $$$$\left.\:=\mathrm{80}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{50}+\mathrm{60}\right)\right) \\ $$$$\:\:=\mathrm{80}−\mathrm{55}=\mathrm{25} \\ $$
Answered by mehdee7396 last updated on 26/May/25
![s_(sh) =ab−(1/2)[5a+6b+(a−6)(b−5)]=80−(1/2)[ab+30] ⇒ans=25](https://www.tinkutara.com/question/Q221173.png)
$${s}_{{sh}} ={ab}−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{5}{a}+\mathrm{6}{b}+\left({a}−\mathrm{6}\right)\left({b}−\mathrm{5}\right)\right]=\mathrm{80}−\frac{\mathrm{1}}{\mathrm{2}}\left[{ab}+\mathrm{30}\right] \\ $$$$\Rightarrow{ans}=\mathrm{25}\: \\ $$
Commented by mehdee7396 last updated on 26/May/25

Commented by mr W last updated on 26/May/25

$${thanks}! \\ $$
Answered by mr W last updated on 26/May/25

Commented by mr W last updated on 26/May/25

$${shaded}\:{area}\:+\:{hatched}\:{area}\:=\frac{{rectangle}}{\mathrm{2}} \\ $$$${shaded}\:{area}=\frac{{rectangle}}{\mathrm{2}}−{hatched}\:{area} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{80}}{\mathrm{2}}−\frac{\mathrm{6}×\mathrm{5}}{\mathrm{2}}=\mathrm{25}\:\checkmark \\ $$
Commented by Rasheed.Sindhi last updated on 26/May/25

$$\boldsymbol{{Sir}},\:{how}\:{can}\:{we}\:{prove}\:{that}: \\ $$$${shaded}\:{area}\:+\:{hatched}\:{area}\:=\frac{{rectangle}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 26/May/25

Commented by mr W last updated on 26/May/25

$${rectangle}\:={b}\left({h}_{\mathrm{1}} +{h}_{\mathrm{2}} \right) \\ $$$$\left({A}\right)=\frac{{bh}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\left({B}\right)=\frac{{bh}_{\mathrm{2}} }{\mathrm{2}} \\ $$$${shaded}\:+\:{hatched}\:=\left({A}\right)+\left({B}\right) \\ $$$$\:\:\:\:\:=\frac{{bh}_{\mathrm{1}} }{\mathrm{2}}+\frac{{bh}_{\mathrm{2}} }{\mathrm{2}}=\frac{{b}\left({h}_{\mathrm{1}} +{h}_{\mathrm{2}} \right)}{\mathrm{2}}=\frac{{rectangle}}{\mathrm{2}} \\ $$
Commented by Rasheed.Sindhi last updated on 26/May/25

$$\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{sir}}! \\ $$