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→Hints for difficult problems
=8-3<br>=5<br>
=8-3<br>=5<br>
'''to back to concept page'''
'''to back to concept page'''
[[http://karnatakaeducation.org.in/KOER/en/index.php/Quadratic_Equations#Hints_for_difficult_problems| quadratic equation problems]]<br>
2.The altitude of a triangle is 6cm greter than its base. If its area is 108cmsq .Find its base.<br>
[[solution]]<br>
Statement: Solving problem based on quadratic equations.<br>
[[solution]]
*Interpretation of the problem:<br> * Converting data in to eqn.<br> *Knowledge about area of a triangle.<br>*knowledge of the formula of area of triangle.<br>*Methods of finding the roots of the eqn.<br> *Methods of finding the roots of the
*Different approches to solve the problem: <br>*Factorisation
*Using formula
*using graph
*Concept used:Forming the eqn. 216=x(x+6)
216=x2+6x<br>
x2 +6x -216=0<br>
Substitution: x 2 +18x-12x -216=0<br>
Simplification: x(x+18)-12(x+18)=0<br>
(x+18)( x-12)=0<br>
(x+18)=0 (x-12)=0<br>
x=-18, x=12<br>.
# Base=12cm, <br> Altitude=x+6
=12+6=18cm.<br>
'''Prior Knowledge''' -<br>
*Methods of solving the Eqn<br>
*Factorisation<br>
*Using Formula<br>
*Using Graph<br>
[[http://karnatakaeducation.org.in/KOER/en/index.php/Quadratic_Equations#Hints_for_difficult_problems| quadratic equation problems]]
[[http://karnatakaeducation.org.in/KOER/en/index.php/Quadratic_Equations#Hints_for_difficult_problems| quadratic equation problems]]