Question

The sides of certain triangles are given below. Determine which of them are right angled triangles.
(i) a = 7 cm, b = 24 cm and c = 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm
(iii) a = 1.6 cm, b = 3.8 cm and c = 4 cm
(iv) a = 8 cm, b = 10 cm and c = 6 cm

Answer 1

$\left(i\right)Given,$ $a=7,cm,b=24cm$ and $c=25cm$

$c^2=a^2+b^2$ [ By using Pythagoras theorem ]

$c^2=7^2+{24}^2$

$c^2=49+576$

$c^2=625$

$\therefore$ $c=25cm$

Hence, the given triangle is right angled triangle.

$\left(ii\right)$ given, $a=9cm,b=16cm$ and $c=18$

$c^2=a^2+b^2$ [ By using Pythagoras theorem ]

$c^2=9^2+{16}^2$

$c^2=81+256$

$c^2=337$

$\therefore$ $c=18.35cm$

Hence, the given triangle is not right angled triangle.

$\left(iii\right)$ given, $a=1.6,cm,b=3.8cm$ and $c=4cm$

$c^2=a^2+b^2$ [ By using Pythagoras theorem ]

$c^2=(1.6{)}^2+(3.8{)}^2$

$c^2=2.56+14.44$

$c^2=17$

$\therefore$ $c=4.12cm$

Hence, the given triangle is not right angled triangle.

$\left(iv\right)$ given, $a=8,cm,b=10cm$ and $c=6cm$

$b^2=a^2+c^2$ [ By using Pythagoras theorem ]

$b^2=(8{)}^2+(6{)}^2$

$b^2=64+36$

$b^2=100$

$\therefore$ $b=10cm$

Hence, the given triangle is right angled triangle.