import{w as e,W as h,m,aL as q}from"./index-XCg2QAX4.js";function d(t="A",$="B",i="C",g=1,b=3,a=4,o=5,f="cm",n="#f15929"){let r,s;return r=`The triangle $${t+$+i}$ is right-angled in $${t}$ so according to the Pythagorean theorem, we have:`,r+=`<br> $${$+i}^2=${t+$}^2+${t+i}^2$`,g===1?(r+=`<br> $${$+i}^2=${e(b,2)}^2+${e(a,2)}^2$`,r+=`<br> $${$+i}^2=${e(b**2+a**2,2)}$`,r+=`<br> $${$+i}=\\sqrt{${e(b**2+a**2,2)}}$`,h(Math.sqrt(b**2+a**2),1)===h(Math.sqrt(b**2+a**2),5)?s="=":s="\\approx",r+=`<br> So $${$+i} ${s} ${m(e(o,2),n)}$ ${q(f,n)}.`):g===2&&(r+=`<br> Hence $${t+$}^2=${$+i}^2-${t+i}^2$.`,r+=`<br> $${t+$}^2=${e(o,2)}^2-${e(a,2)}^2$`,r+=`<br> $${t+$}^2=${e(o**2-a**2,2)}$`,r+=`<br> $${t+$}=\\sqrt{${e(o**2-a**2,2)}}$`,h(Math.sqrt(o**2-a**2),1)===h(Math.sqrt(o**2-a**2),5)?s="=":s="\\approx",r+=`<br> So $${t+$} ${s} ${m(e(b,2),n)}$ ${q(f,n)}.`),[r,s]}export{d as RedactionPythagore};
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