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સમરૂપતા અને પાયથાગોરસ પ્રમેય

∆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

∴ BM² = AM•CM.

AB² = AM•AC અને

BC² = CM•AC

BM² = AM•CM

∴ y² = x•CM

∴ CM = $bold y to the power of bold 2 over bold x$

AC = AM + CM

bold equals bold space bold x bold space bold plus bold space bold y to the power of bold 2 over bold x bold AC bold space bold equals bold space fraction numerator bold x to the power of bold 2 bold space bold plus bold space bold y to the power of bold 2 over denominator bold x end fraction bold AB to the power of bold 2 bold space bold equals bold space bold AM bold times bold AC bold equals bold space bold x bold times bold space fraction numerator bold x to the power of bold 2 bold space bold plus bold space bold y to the power of bold 2 over denominator bold x end fraction bold AB bold space bold equals bold space square root of bold x to the power of bold 2 bold space bold plus bold space bold y to the power of bold 2 bold space end root bold space bold BC to the power of bold 2 bold space bold equals bold space bold CM bold times bold AC bold space bold equals bold y to the power of bold 2 over bold x to the power of bold 2 bold times fraction numerator bold x to the power of bold 2 bold space bold plus bold space bold y to the power of bold 2 over denominator bold x end fraction bold equals bold space bold y to the power of bold 2 over bold x to the power of bold 2 bold times bold left parenthesis bold x to the power of bold 2 bold space end exponent bold plus bold space bold y to the power of bold 2 bold right parenthesis bold space bold equals bold space bold BC bold space bold equals bold space bold y over bold x square root of bold x to the power of bold 2 bold space bold plus bold space bold y to the power of bold 2 bold space end root

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

∆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).

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સમરૂપતા અને પાયથાગોરસ પ્રમેય

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