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

Question:

The sides of certain triangles are given below. Determine which of them are right triangles.
(i) 9 cm, 16 cm, 18 cm
(ii) 7 cm, 24 cm, 25 cm
(iii) 1.4 cm, 4.8 cm, 5 cm
(iv) 1.6 cm, 3.8 cm, 4 cm

(v) $(a-1) \mathrm{cm}, 2 \sqrt{a} \mathrm{~cm},(a+1) \mathrm{cm}$

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 ab 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.