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સમરૂપતા અને પાયથાગોરસ પ્રમેય

∆PQR માં m∠Q = 90 અને $top enclose bold QM$ એક વેધ છે અને M $bold not an element of bold space top enclose bold PR$ . જો QM = 12, PR = 26, તો PM અને RM શોધો. જો PM < RM, તો Pq અને QR શોધો.

∆ABCમાં m∠B = 90, $top enclose bold BM bold space bold perpendicular bold space top enclose bold AC bold comma bold space bold M bold space bold element of bold space top enclose bold AC bold.$ જો AM = x, BM = y તો AB, BC અને CM ને x અને y ના સ્વરૂપમાં મેળવો. (x > 0, y > o)

∆ABC માં m∠B = 90 અને $top enclose bold BM bold space bold perpendicular bold space top enclose bold AC bold space bold comma bold space bold M bold space bold element of bold space top enclose bold AC$ જો AM = 4MC, તો સાબિત કરો કે, AB = 2BC.

∆ABC માં m∠A= 90 અને $top enclose bold AD bold space bold perpendicular bold space top enclose bold BC bold comma bold space bold D bold space bold element of bold space top enclose bold BC$ તો સાબિત કરો કે, $bold 1 over bold AD to the power of bold 2 equals bold 1 over bold AB to the power of bold 2 plus bold 1 over bold AC to the power of bold 2$

∆ABC માં $top enclose bold AD bold comma bold space top enclose bold BE bold space bold અન ે bold space top enclose bold CF$ મધ્યગાઓ છે. સાબિત કરો કે, 4(AD² + BE² + CF²) = 3(AB² + BC² + AC²).

પક્ષ : ∆ABC માં $top enclose bold AD bold comma bold space top enclose bold BE bold space bold અન ે bold space top enclose bold CF$ મધ્યગાઓ છે.
સાધ્ય : 4(AD² + BE² + CF²) = 3(AB² + BC² + AC²)

$bold સ ા બ િ ત ી bold space bold colon bold space bold increment bold ABCમ ાં bold space top enclose bold AD bold comma end enclose bold space top enclose bold BE bold space bold અન ે bold space top enclose bold CF bold space bold મધ ્ યગ ા ઓ bold space bold છ ે bold. bold therefore bold space bold BD bold space bold equals bold space bold 1 over bold 2 bold BC bold comma bold space bold AF bold space bold equals bold space bold 1 over bold 2 bold AB bold space bold space bold અન ે bold space bold CE bold equals bold 1 over bold 2 bold AC bold therefore bold space bold BD to the power of bold 2 bold space bold equals bold space bold 1 over bold 4 bold BC to the power of bold 2 bold comma bold space bold AF to the power of bold 2 bold space end exponent bold equals bold space bold 1 over bold 2 bold AB to the power of bold 2 bold space end exponent bold space bold અન ે bold CE to the power of bold 2 bold space bold equals bold space bold 1 over bold 4 bold AC to the power of bold 2 bold space end exponent bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold. bold. bold. bold. bold left parenthesis bold 1 bold right parenthesis$

∆abc માં મધ્યગા છે.

∴ AB² + AC = 2(AD² + BD²) ........(2)

∆ABC માં મધ્યગા છે.

∴ AC² + BC² = 2(BE² + CE²) .......(3)

∆ABC માં મધ્યગા છે.

∴ AC² + BC² = 2(CF² + AF²) .........(4)

પરિણામ (2), (3) અને (4) નો સરવાળો લેતાં,

2(AB² + BC² + AC²) = 2(AD² + BD² + BE² + CE² + CF² + AF²)

∴AB² + BC² + AC² = (AD² + BE² + CF²) + (BD² + CE² + AF²)

∴(AB² + BC² + AC² )= (AD² + BE² + CF²) + $open parentheses bold 1 over bold 4 bold BC bold 2 bold space bold plus bold space bold 1 over bold 4 bold AC bold 2 bold space bold plus bold space bold 1 over bold 4 bold AB bold 2 close parentheses$ (પરિણામ (1) પરથી)

∴ $bold 3 over bold 4$ (AB² + BC² + AC²) = (AD² + BE² + CF²)

∴3(AB² + BC² + CF²) = 3(AB² + BE² + CF²)

∴4(AD² + BE² +CF²) = 3(AB² + BC² + AC²)

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સમરૂપતા અને પાયથાગોરસ પ્રમેય

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