Question

4. ABC is an isosceles triangle right angled at C. Prove that ABP = 2AC2
​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{ABC is an isosceles triangle right angled at C}$

$\underline{\textsf{To prove:}}$

$\mathsf{AB^2=2\,AC^2}$

$\underline{\textsf{Solution:}}$

$\textsf{Pythagoras theorem:}$

$\boxed{\begin{minipage}{9cm} \\\textsf{In a right angled triangle, square on the hypotenuse is}\\\\\textsf{is equal to sum of the squares on the other two sides}\\ \end{minipage}}$

$\textsf{since  ABC is a right angled isoceles triangle at C, we have}$

$\mathsf{AC= BC}\;\textsf{and AB is hypotenuse}$......(1)

$\textsf{By pythagoras theorem, we get}$

$\mathsf{AB^2=AC^2+BC^2}$

$\mathsf{AB^2=AC^2+AC^2}\;\;\textsf{(Using(1))}$

$\implies\boxed{\mathsf{AB^2=2\,AC^2}}$

$\textsf{Hence proved}$

Find more:

The perimeter of an isosceles right-angled triangle is 6(√2+1) cm . Find its area. Please solve it with explanation.

https://brainly.in/question/2120508

Answer 2

ur answer is attached....