Mathematics
Answer
By formula,
By pythagoras theorem,
⇒ (Hypotenuse)2 = (Perpendicular)2 + (Base)2
In right-angled △ ABD,
⇒ AB2 = AD2 + BD2
⇒ c2 = h2 + (a - x)2 ………….(1)
In right-angled △ ACD,
⇒ AC2 = AD2 + CD2
⇒ b2 = h2 + x2
⇒ h2 = b2 - x2 ……….(2)
Substituting value of h2 from equation (2) in (1), we get :
⇒ c2 = b2 - x2 + (a - x)2
⇒ c2 = b2 - x2 + a2 + x2 - 2ax
⇒ c2 = a2 + b2 - 2ax.
Hence, proved that c2 = a2 + b2 - 2ax.
Related Questions
In the given figure, the value of AB × CD is :

AC × BC
AC × CD
AC × AB
AC2 + BC2
ABC is an isosceles triangle right-angled at C. Then 2AC2 is equal to :
BC2
AC2
AC2 - BC2
AB2
In equilateral △ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.
ABC is a triangle, right-angled at B. M is a point on BC. Prove that :
AM2 + BC2 = AC2 + BM2.
