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Question-41103

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Question Number 41103 by ajfour last updated on 02/Aug/18
Commented by ajfour last updated on 02/Aug/18
Q.41062 solution
$${Q}.\mathrm{41062}\:\:{solution} \\ $$
Commented by MJS last updated on 02/Aug/18
∣AF∣=∣BF∣ ⇒ ∣AE∣≠∣DE∣ ⇒ area(ACE)≠area(CDE)
$$\mid{AF}\mid=\mid{BF}\mid\:\Rightarrow\:\mid{AE}\mid\neq\mid{DE}\mid\:\Rightarrow\:\mathrm{area}\left({ACE}\right)\neq\mathrm{area}\left({CDE}\right) \\ $$
Commented by ajfour last updated on 02/Aug/18
no sir it is given in question AE=DE and the shaded area is asked not the coloured area. I have coloured the asked shaded area.
$${no}\:{sir}\:{it}\:{is}\:{given}\:{in}\:{question} \\ $$$${AE}={DE}\:{and}\:{the}\:{shaded}\:{area}\:{is} \\ $$$${asked}\:{not}\:{the}\:{coloured}\:{area}. \\ $$$${I}\:{have}\:{coloured}\:{the}\:{asked}\:{shaded} \\ $$$${area}. \\ $$
Commented by MJS last updated on 02/Aug/18
but my solution should work for any triangle if ∣AF∣=∣BF∣ then it should be easy to see that ∣AE∣=∣DE∣ is not possible. just consider the intercept theorems... or look at the values of both lines in my answer. if the other areas are meant, I get 14−((11)/2)=((17)/2)
$$\mathrm{but}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{should}\:\mathrm{work}\:\mathrm{for}\:\mathrm{any}\:\mathrm{triangle} \\ $$$$\mathrm{if}\:\mid{AF}\mid=\mid{BF}\mid\:\mathrm{then}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{that} \\ $$$$\mid{AE}\mid=\mid{DE}\mid\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}.\:\mathrm{just}\:\mathrm{consider}\:\mathrm{the} \\ $$$$\mathrm{intercept}\:\mathrm{theorems}…\:\mathrm{or}\:\mathrm{look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{values} \\ $$$$\mathrm{of}\:\mathrm{both}\:\mathrm{lines}\:\mathrm{in}\:\mathrm{my}\:\mathrm{answer}. \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{other}\:\mathrm{areas}\:\mathrm{are}\:\mathrm{meant},\:\mathrm{I}\:\mathrm{get}\:\mathrm{14}−\frac{\mathrm{11}}{\mathrm{2}}=\frac{\mathrm{17}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 02/Aug/18
where did you read AF=BF , Sir ?
$${where}\:{did}\:{you}\:{read}\:{AF}={BF}\:,\:{Sir}\:? \\ $$
Commented by MJS last updated on 02/Aug/18
yes, where did I read this? sorry I looked at the picture only once and then obviously confused something while thinking it over...
$$\mathrm{yes},\:\mathrm{where}\:\mathrm{did}\:\mathrm{I}\:\mathrm{read}\:\mathrm{this}? \\ $$$$\mathrm{sorry}\:\mathrm{I}\:\mathrm{looked}\:\mathrm{at}\:\mathrm{the}\:\mathrm{picture}\:\mathrm{only}\:\mathrm{once}\:\mathrm{and}\:\mathrm{then} \\ $$$$\mathrm{obviously}\:\mathrm{confused}\:\mathrm{something}\:\mathrm{while}\:\mathrm{thinking} \\ $$$$\mathrm{it}\:\mathrm{over}… \\ $$
Answered by ajfour last updated on 02/Aug/18
2p=(3/4)(Area △ABC) ⇒ p= (3/8)×14 = ((21)/4) cm^2 r = ((p+q)/3) ⇒ 3r−q = ((21)/4) ...(i) Area(△ABD)= 2q+r = ((Area(△ABC))/4) ⇒ 2q+r = ((14)/4) =(7/2) ...(ii) 2(i)+(ii) gives 7r = ((21)/2)+(7/2) = 14 ⇒ r = 2 cm^2 then q = (1/2)((7/2)−r) ⇒ q = (1/2)((7/2)−2) = (3/4)cm^2 Total Area of coloured regions = p+q+r = ((21)/4)+(3/4)+2 = ((32)/4) ⇒ the shaded area in question is 8cm^2 .
$$\mathrm{2}{p}=\frac{\mathrm{3}}{\mathrm{4}}\left({Area}\:\bigtriangleup{ABC}\right) \\ $$$$\Rightarrow\:\:\:{p}=\:\frac{\mathrm{3}}{\mathrm{8}}×\mathrm{14}\:=\:\frac{\mathrm{21}}{\mathrm{4}}\:{cm}^{\mathrm{2}} \\ $$$${r}\:=\:\frac{{p}+{q}}{\mathrm{3}}\:\:\Rightarrow\:\mathrm{3}{r}−{q}\:=\:\frac{\mathrm{21}}{\mathrm{4}}\:\:\:…\left({i}\right) \\ $$$${Area}\left(\bigtriangleup{ABD}\right)=\:\mathrm{2}{q}+{r}\:=\:\frac{{Area}\left(\bigtriangleup{ABC}\right)}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{2}{q}+{r}\:=\:\frac{\mathrm{14}}{\mathrm{4}}\:=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$$\:\:\mathrm{2}\left({i}\right)+\left({ii}\right)\:{gives} \\ $$$$\:\:\:\mathrm{7}{r}\:=\:\frac{\mathrm{21}}{\mathrm{2}}+\frac{\mathrm{7}}{\mathrm{2}}\:=\:\mathrm{14} \\ $$$$\Rightarrow\:\:\:{r}\:=\:\mathrm{2}\:{cm}^{\mathrm{2}} \:\:\:{then}\:{q}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{2}}−{r}\right) \\ $$$$\Rightarrow\:\:{q}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{2}}−\mathrm{2}\right)\:=\:\frac{\mathrm{3}}{\mathrm{4}}{cm}^{\mathrm{2}} \\ $$$${Total}\:{Area}\:{of}\:{coloured}\:{regions} \\ $$$$\:\:\:=\:{p}+{q}+{r}\:=\:\frac{\mathrm{21}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}+\mathrm{2}\:=\:\frac{\mathrm{32}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:{the}\:{shaded}\:{area}\:{in}\:{question} \\ $$$$\:\:\:\:\:\:\:{is}\:\mathrm{8}{cm}^{\mathrm{2}} \:. \\ $$

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