The sides of certain triangles are given below. Determine which of them are right triangles.

Solution:

For the given triangle to be right-angled, the sum of square of the two sides must be equal to the square of the third side.
Here, let the three sides of the triangle be a, b and c.
(i)
a = 9 cm, b = 16 cm and c = 18 cm
Then,

$a^{2}+b^{2}=9^{2}+16^{2}$

$=81+256$

$=337$

$c^{2}=19^{2}$

$=361$

$a^{2}+b^{2} \neq c^{2}$

Thus, the given triangle is not right-angled.

(ii)
a = 7 cm, b = 24 cm and c = 25 cm
Then,

$a^{2}+b^{2}=7^{2}+24^{2}$

$=49+576$

$=625$

$c^{2}=25^{2}$

$=625$

$a^{2}+b^{2}=c^{2}$

Thus, the given triangle is a right-angled.

(iii)
a = 1.4 cm, b = 4.8 cm and c = 5 cm
Then,

$a^{2}+b^{2}=(1.4)^{2}+(4.8)^{2}$

$=1.96+23.04$

$=25$

$c^{2}=5^{2}$

$=25$

$a^{2}+b^{2}=c^{2}$

Thus, the given triangle is right-angled.

(iv)
a = 1.6 cm, b = 3.8cm and c = 4 cm
Then,

$a^{2}+b^{2}=(1.6)^{2}+(3.8)^{2}$

$=2.56+14.44$

$=17$

$c^{2}=4^{2}$

$=16$

$a^{2}+b^{2} \neq c^{2}$

Thus, the given triangle is not right-angled.

(v)

$p=(\mathrm{a}-1) \mathrm{cm}, \quad q=2 \sqrt{a} \mathrm{~cm}$ and $r=(\mathrm{a}+1) \mathrm{cm}$

Then,

$p^{2}+q^{2}=(a-1)^{2}+(2 \sqrt{a})^{2}$

$=a^{2}+1-2 a+4 a$

$=a^{2}+1+2 a$

$=(a+1)^{2}$

$r^{2}=(a+1)^{2}$

$p^{2}+q^{2}=r^{2}$

Thus, the given triangle is right-angled.