Question

Find hypotenuse & perimeter of Δ ABC, where AB = 4 cm & BC = 5cm.​

Answer 1

Question:-

Find hypotenuse & perimeter of Δ ABC, where AB = 4 cm & BC = 5 cm.

Solution:-

We use Pythagoras theorem

=> ( AC )² = ( AB )² + ( BC )²

=> ( AC )² = ( 4 )² + ( 5 )²

=> ( AC )² = 16 + 25

=> ( AC)² = 41

=> AC = √41 cm

=>Perimeter of Δ ABC = AB+BC+AC

=>Perimeter of Δ ABC = 4+5+√41

=>Perimeter of Δ ABC = 9+√41cm

Hence Hypotenuse & Perimeter are

√41 cm & 9+√41cm respectively.

i hope it helps you.

Answer 2

Question:-

Find hypotenuse & perimeter of A ABC, where AB = 5 cm & BC = 6 cm.

Solution:-

We use Pythagoras theorem => ( AC )² = ( AB )² + ( BC )²

=> ( AC )² = ( 5 )² + ( 6 )²

=> ( AC )² = 25 + 36

=> ( AC)? = 61

=> AC = $\sqrt{61 \: cm}$

=>Perimeter of A ABC = AB+BC+AC

=>Perimeter of A ABC = 5+6+v61

=>Perimeter of A ABC = 11+$\sqrt{61 \: cm}$

Hence Hypotenuse & Perimeter are v61 cm & 11+v61cm respectively. i hope it helps you.