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સમરૂપતા અને પાયથાગોરસ પ્રમેય

∆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, $bold 1 over bold AD to the power of bold 2 equals bold 1 over bold AB to the power of bold 2 plus bold 1 over bold AC to the power of bold 2$

∆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.

∆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).

∆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.

પક્ષ : ∆ABC માં m∠B = 90 અને $top enclose bold BM bold space bold perpendicular bold space top enclose bold AC bold space bold comma bold space bold M bold space bold element of bold space top enclose bold AC$ વળી, AM = 4MC.
સાધ્ય : AB = 2BC

સાબિતી : ∆ABC માં m∠B = 90 અને $top enclose bold BM bold space bold perpendicular bold space top enclose bold AC bold space bold comma bold space bold M bold space bold element of bold space top enclose bold AC$ ,
$bold therefore bold AB to the power of bold 2 over bold BC to the power of bold 2 bold equals fraction numerator bold AM bold times bold AC over denominator bold MC bold times bold AC end fraction bold therefore bold AB to the power of bold 2 over bold BC to the power of bold 2 bold equals bold AM over bold MC bold therefore bold AB to the power of bold 2 over bold BC to the power of bold 2 bold equals fraction numerator bold 4 bold MC over denominator bold MC end fraction bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold because bold space bold AM bold space bold equals bold space bold 4 bold MC bold right parenthesis bold therefore bold AB to the power of bold 2 over bold BC to the power of bold 2 bold space bold equals bold space bold 4 bold therefore bold AB over bold BC bold equals bold space bold 2 bold therefore bold AB bold space bold equals bold 2 bold AB$

∆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)

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સમરૂપતા અને પાયથાગોરસ પ્રમેય

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