Menu Close

The-perimeter-of-an-isosceles-right-angled-triangle-is-2p-Find-out-the-area-of-the-triangle-

Tinku Tara
Pin




Question Number 99 by sagarwal last updated on 25/Jan/15
The perimeter of an isosceles right−angled triangle is 2p. Find out the area of the triangle.
$$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{right}−\mathrm{angled} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{2}{p}.\:\mathrm{Find}\:\mathrm{out}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}. \\ $$
Answered by 123456 last updated on 14/Dec/14
lets ΔABC be a triangle with lenght a,b,c by it isosceles mean that two or more angle and lenght are equal since its right angle and isosceles, we have a angle 90°, then the other will be 45° then we gets seting a,b catetes a=b a^2 +b^2 =2a^2 =c^2 the perimet is a+b+c=2a+c=2p then the area is A=((ab)/2)=(a^2 /2) solving 2a^2 −c^2 =0 2a+c=2p c=2(p−a) c^2 =4p^2 −8ap+4a^2 −2a^2 +8ap−4p^2 =0 2a^2 −8ap+4p^2 =0 Δ=(−8p)^2 −4(2)(4p^2 ) =64p^2 −16p^2 =48p^2 a=(((8±2(√(12)))p)/4)=(2±(√(12)))p c=2p−2a we gets a=(2+(√(12)))p c=2((√(12))−1)p A=(((2+(√(12)))^2 p^2 )/2)=(((16+4(√(12)))p^2 )/2)=(8+2(√(12)))p^2
$$\mathrm{lets}\:\Delta\mathrm{A}{BC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{lenght}\:{a},{b},{c} \\ $$$$\mathrm{by}\:\mathrm{it}\:\mathrm{isosceles}\:\mathrm{mean}\:\mathrm{that}\:\mathrm{two}\:\mathrm{or}\:\mathrm{more}\:\mathrm{angle}\:\mathrm{and}\:\mathrm{lenght}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\mathrm{since}\:\mathrm{its}\:\mathrm{right}\:\mathrm{angle}\:\mathrm{and}\:\mathrm{isosceles},\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{angle}\:\mathrm{90}°,\:\mathrm{then}\:\mathrm{the}\:\mathrm{other}\:\mathrm{will}\:\mathrm{be}\:\mathrm{45}° \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{gets}\:\mathrm{seting}\:{a},{b}\:\mathrm{catetes} \\ $$$${a}={b} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{perimet}\:\mathrm{is} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{2a}+\mathrm{c}=\mathrm{2p} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{area}\:\mathrm{is} \\ $$$$\mathbb{A}=\frac{{ab}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{solving} \\ $$$$\mathrm{2a}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2a}+\mathrm{c}=\mathrm{2p} \\ $$$$\mathrm{c}=\mathrm{2}\left(\mathrm{p}−\mathrm{a}\right) \\ $$$$\mathrm{c}^{\mathrm{2}} =\mathrm{4p}^{\mathrm{2}} −\mathrm{8ap}+\mathrm{4a}^{\mathrm{2}} \\ $$$$−\mathrm{2a}^{\mathrm{2}} +\mathrm{8ap}−\mathrm{4p}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2a}^{\mathrm{2}} −\mathrm{8ap}+\mathrm{4p}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(−\mathrm{8}{p}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}\right)\left(\mathrm{4}{p}^{\mathrm{2}} \right) \\ $$$$=\mathrm{64}{p}^{\mathrm{2}} −\mathrm{16}{p}^{\mathrm{2}} =\mathrm{48}{p}^{\mathrm{2}} \\ $$$${a}=\frac{\left(\mathrm{8}\pm\mathrm{2}\sqrt{\mathrm{12}}\right){p}}{\mathrm{4}}=\left(\mathrm{2}\pm\sqrt{\mathrm{12}}\right){p} \\ $$$${c}=\mathrm{2}{p}−\mathrm{2}{a} \\ $$$$\mathrm{we}\:\mathrm{gets} \\ $$$$\mathrm{a}=\left(\mathrm{2}+\sqrt{\mathrm{12}}\right){p} \\ $$$$\mathrm{c}=\mathrm{2}\left(\sqrt{\mathrm{12}}−\mathrm{1}\right){p} \\ $$$$\mathbb{A}=\frac{\left(\mathrm{2}+\sqrt{\mathrm{12}}\right)^{\mathrm{2}} {p}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\mathrm{16}+\mathrm{4}\sqrt{\mathrm{12}}\right){p}^{\mathrm{2}} }{\mathrm{2}}=\left(\mathrm{8}+\mathrm{2}\sqrt{\mathrm{12}}\right){p}^{\mathrm{2}} \\ $$

Pin