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

Answer 1

For a given triangle to be a right-angled, the sum of the squares of the two sides must be equal to the square of the largest side.

(i) Let a = 9 cm, b = 16 cm and c = 18 cm. Then

$(a^2+b^2)=[9^2+(16{)}^2]=(81+256)c{m}^2=337c{m}^{2}And{c}^2=(18{)}^{2}c{m}^2=324c{m}^2$

Therefore,
$(a^2+b^2)\ne c^2$

Hence, the given triangle is not right-angled.

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

$(a^2+b^2)=[7^2+(24{)}^2]=(49+576)c{m}^2=625c{m}^{2}And{c}^2=(25{)}^{2}c{m}^2=625c{m}^2$

Therefore,
$(a^2+b^2)=c^2$

Hence, the given triangle is a right triangle.

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

$(a^2+b^2)=\left[\right(1.4{)}^2+\left(4.8{)}^2\right]=(1.96+23.04)c{m}^2=25c{m}^{2}And{c}^2=(5{)}^{2}c{m}^2=25c{m}^2$

Therefore,
$(a^2+b^2)=c^2$

Hence, the given triangle is a right triangle.

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

$(a^2+b^2)=\left[\right(1.6{)}^2+\left(3.8{)}^2\right]=(2.56+14.44)c{m}^2=17c{m}^{2}And{c}^2=(4{)}^{2}c{m}^2=16c{m}^2$

Therefore,
$(a^2+b^2)\ne c^2$

Hence, the given triangle is not a right triangle.

(v) Let p = (a - 1) cm, q = $2\sqrt{a}$ cm and r = (a + 1) cm

$(p^2+q^2)=\left[\right(a-1{)}^2+\left(2\sqrt{a}{)}^2\right]=(a^2+1-2a+4a)c{m}^2=(a+1{)}^{2}c{m}^{2}And{r}^2=(a+1{)}^{2}cm2$

Therefore,
$(p^2+q^2)=r^2$

Hence, the given triangle is a right triangle.