Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

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 .

$$\mathrm{2}{p}=\frac{\mathrm{3}}{\mathrm{4}}\left({Area}\:\bigtriangleup{ABC}\right) \\ $$$$\Rightarrow\:\:\:{p}=\:\frac{\mathrm{3}}{\mathrm{8}}×\mathrm{14}\:=\:\frac{\mathrm{21}}{\mathrm{4}}\:{cm}^{\mathrm{2}} \\ $$$${r}\:=\:\frac{{p}+{q}}{\mathrm{3}}\:\:\Rightarrow\:\mathrm{3}{r}−{q}\:=\:\frac{\mathrm{21}}{\mathrm{4}}\:\:\:...\left({i}\right) \\ $$$${Area}\left(\bigtriangleup{ABD}\right)=\:\mathrm{2}{q}+{r}\:=\:\frac{{Area}\left(\bigtriangleup{ABC}\right)}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{2}{q}+{r}\:=\:\frac{\mathrm{14}}{\mathrm{4}}\:=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:...\left({ii}\right) \\ $$$$\:\:\mathrm{2}\left({i}\right)+\left({ii}\right)\:{gives} \\ $$$$\:\:\:\mathrm{7}{r}\:=\:\frac{\mathrm{21}}{\mathrm{2}}+\frac{\mathrm{7}}{\mathrm{2}}\:=\:\mathrm{14} \\ $$$$\Rightarrow\:\:\:{r}\:=\:\mathrm{2}\:{cm}^{\mathrm{2}} \:\:\:{then}\:{q}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{2}}−{r}\right) \\ $$$$\Rightarrow\:\:{q}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{2}}−\mathrm{2}\right)\:=\:\frac{\mathrm{3}}{\mathrm{4}}{cm}^{\mathrm{2}} \\ $$$${Total}\:{Area}\:{of}\:{coloured}\:{regions} \\ $$$$\:\:\:=\:{p}+{q}+{r}\:=\:\frac{\mathrm{21}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}+\mathrm{2}\:=\:\frac{\mathrm{32}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:{the}\:{shaded}\:{area}\:{in}\:{question} \\ $$$$\:\:\:\:\:\:\:{is}\:\mathrm{8}{cm}^{\mathrm{2}} \:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com