Question-46637

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Question Number 46637 by rahul 19 last updated on 29/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

$${ok}… \\ $$

Commented by rahul 19 last updated on 29/Oct/18

$${Ans}.\:{is}\:{D}….{since}\:{it}\:{is}\:{asking}\:{for} \\ $$$${largest}\:{possible}\:{radius}…. \\ $$

Commented by rahul 19 last updated on 29/Oct/18

for eg: if sides are 12,16,20 then r=4...

$${for}\:{eg}:\:{if}\:{sides}\:{are}\:\mathrm{12},\mathrm{16},\mathrm{20}\:{then}\:{r}=\mathrm{4}… \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

sides are 5,12,13 (1/2)×5×r+(1/2)×12×r+(1/2)×13×r=(1/2)×5×12 30r=60 r=2

$${sides}\:{are}\:\mathrm{5},\mathrm{12},\mathrm{13} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}×{r}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×{r}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{13}×{r}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}×\mathrm{12} \\ $$$$\mathrm{30}{r}=\mathrm{60}\:\:\:{r}=\mathrm{2} \\ $$

Answered by MJS last updated on 31/Oct/18

these are all possible triangles: 9 12 15 r_i =3 12 16 20 r_i =4 5 12 13 r_i =2 12 35 37 r_i =5 ⇒ answer is D

$$\mathrm{these}\:\mathrm{are}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{triangles}: \\ $$$$\mathrm{9}\:\mathrm{12}\:\mathrm{15}\:{r}_{{i}} =\mathrm{3} \\ $$$$\mathrm{12}\:\mathrm{16}\:\mathrm{20}\:{r}_{{i}} =\mathrm{4} \\ $$$$\mathrm{5}\:\mathrm{12}\:\mathrm{13}\:{r}_{{i}} =\mathrm{2} \\ $$$$\mathrm{12}\:\mathrm{35}\:\mathrm{37}\:{r}_{{i}} =\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{D} \\ $$

Commented by rahul 19 last updated on 31/Oct/18

$${Sir},\:{how}\:{you}\:{checked}\:{these}\:{are}\:{only} \\ $$$${possible}\:{triangles}….? \\ $$

Commented by MJS last updated on 31/Oct/18

a^2 +b^2 =c^2 with a, b, c ∈N\{0} we can set c=12 and b=a+n with n∈N a^2 +(a+n)^2 =144 ⇒ a=((√(288−n^2 ))/2)−(n/2)∈N ⇒ no solution we can set b=12 and c=a+n with n∈N a^2 +144=(a+n)^2 ⇒ a=((72)/n)−(n/2)∈N ⇒ n∈{2, 4, 6, 8}

$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\mathrm{with}\:{a},\:{b},\:{c}\:\in\mathbb{N}\backslash\left\{\mathrm{0}\right\} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{set}\:{c}=\mathrm{12}\:\mathrm{and}\:{b}={a}+{n}\:\mathrm{with}\:{n}\in\mathbb{N} \\ $$$${a}^{\mathrm{2}} +\left({a}+{n}\right)^{\mathrm{2}} =\mathrm{144} \\ $$$$\Rightarrow\:{a}=\frac{\sqrt{\mathrm{288}−{n}^{\mathrm{2}} }}{\mathrm{2}}−\frac{{n}}{\mathrm{2}}\in\mathbb{N}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{set}\:{b}=\mathrm{12}\:\mathrm{and}\:{c}={a}+{n}\:\mathrm{with}\:{n}\in\mathbb{N} \\ $$$${a}^{\mathrm{2}} +\mathrm{144}=\left({a}+{n}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{a}=\frac{\mathrm{72}}{{n}}−\frac{{n}}{\mathrm{2}}\in\mathbb{N}\:\Rightarrow\:{n}\in\left\{\mathrm{2},\:\mathrm{4},\:\mathrm{6},\:\mathrm{8}\right\} \\ $$

Commented by rahul 19 last updated on 31/Oct/18

$${perfect}! \\ $$$${Thanks}\:{sir}! \\ $$

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