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

Answer 1

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,
$$\begin{gathered} a^2+b^2=9^2+{16}^2 \\ =81+256 \\ =337 \\ {c}^2={19}^2 \\ =361 \\ {a}^2+b^2\ne c^2 \end{gathered}$$

Thus, the given triangle is not right-angled.

(ii)
a = 7 cm, b = 24 cm and c = 25 cm
Then,
$$\begin{gathered} a^2+b^2=7^2+{24}^2 \\ =49+576 \\ =625 \\ {c}^2={25}^2 \\ =625 \\ {a}^2+b^2=c^2 \end{gathered}$$
Thus, the given triangle is a right-angled.

(iii)
a = 1.4 cm, b = 4.8 cm and c = 5 cm
Then,
$$\begin{gathered} a^2+b^2=(1.4{)}^2+(4.8{)}^2 \\ =1.96+23.04 \\ =25 \\ {c}^2=5^2 \\ =25 \\ {a}^2+b^2=c^2 \end{gathered}$$
Thus, the given triangle is right-angled.

(iv)
a = 1.6 cm, b = 3.8cm and c = 4 cm
Then,
$$\begin{gathered} a^2+b^2=(1.6{)}^2+(3.8{)}^2 \\ =2.56+14.44 \\ =17 \\ {c}^2=4^2 \\ =16 \\ {a}^2+b^2\ne c^2 \end{gathered}$$
Thus, the given triangle is not right-angled.

(v)
p = (a $-$ 1) cm, q = 2$\sqrt{a}$ cm and r = (a + 1) cm
Then,
$$\begin{gathered} p^2+q^2=(a-1{)}^2+(2\sqrt{a}{)}^2 \\ =a^2+1-2a+4a \\ =a^2+1+2a \\ =(a+1{)}^2 \\ {r}^2=(a+1{)}^2 \\ {p}^2+q^2={r^2}^{} \end{gathered}$$
Thus, the given triangle is right-angled.