Book Store

Download books and chapters from book store.
Currently only available for
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for
Class 10 Class 12

સમરૂપતા અને પાયથાગોરસ પ્રમેય

∆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.

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

પક્ષ : ∆ABC માં m∠A= 90 અને $top enclose bold AD bold space bold perpendicular bold space top enclose bold BC bold comma bold space bold D bold space bold element of bold space top enclose bold BC$ .

સાધ્ય : $bold 1 over bold AD to the power of bold 2 bold equals bold 1 over bold AB to the power of bold 2 bold plus bold 1 over bold AC to the power of bold 2$ .

સાબિતી : ∆ABC માં m∠A= 90 અને $top enclose bold AD bold space bold perpendicular bold space top enclose bold BC bold comma bold space bold D bold space bold element of bold space top enclose bold BC$ .
∴AB² = BD•BC,

AC² = CD•BC અને AD² = BD•CD

$bold 1 over bold AB to the power of bold 2 bold plus bold 1 over bold AC to the power of bold 2 bold equals fraction numerator bold 1 over denominator bold BD bold times bold BC end fraction fraction numerator bold 1 over denominator bold CD bold times bold BC end fraction bold equals fraction numerator bold CD bold plus bold BD over denominator bold BD bold times bold CD bold times bold BC end fraction bold equals fraction numerator bold BC over denominator bold BD bold times bold CD bold times bold BC end fraction bold space bold space bold left parenthesis bold because bold D bold element of top enclose bold BC bold space bold right parenthesis bold therefore bold 1 over bold AB to the power of bold 2 bold plus bold 1 over bold AC to the power of bold 2 bold plus fraction numerator bold 1 over denominator bold BC bold times bold CD end fraction bold therefore bold space bold 1 over bold AD to the power of bold 2 bold plus bold 1 over bold AB to the power of bold 2 bold equals bold 1 over bold AC to the power of bold 2 bold therefore bold space bold 1 over bold AD to the power of bold 2 bold plus bold 1 over bold AB to the power of bold 2 bold equals bold 1 over bold AC to the power of bold 2$

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

Switch

Chapter Chosen

Book Chosen

Subject Chosen

Book Store

Previous Year Papers

સમરૂપતા અને પાયથાગોરસ પ્રમેય

Switch