Question
The sides of certain triangles are given below. Determine them are right triangles:
$(\text{a} - 1)\text{cm},2\sqrt{\text{a}}\text{ cm},(\text{a} + 1)\text{cm}.$
$(\text{a} - 1)\text{cm},2\sqrt{\text{a}}\text{ cm},(\text{a} + 1)\text{cm}.$
✓
Answer
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.
Let $\text{p}=(\text{a} - 1)\text{cm},\text{q}=2\sqrt{\text{a}}\text{ cm}$ and $\text{r}=(\text{a}+1)\text{cm}^2$
$\text{p}^2+\text{q}^2=\Big[(\text{a}-1)^2+\big(2\sqrt{\text{a}}\big)^2\Big]\text{cm}^2$
$=\big(\text{a}^2+1-2\text{a}+4\text{a}\big)\text{cm}^2$
$=\big(\text{a}^2+1+2\text{a}\big)\text{cm}^2=(\text{a}+1)^2\text{cm}^2$
$\text{r}^2=(\text{a}+1)^2\text{cm}^2$
$\therefore\text{p}^2+\text{q}^2=\text{r}^2$
Hence, the given triangle is a right triangle.
Let $\text{p}=(\text{a} - 1)\text{cm},\text{q}=2\sqrt{\text{a}}\text{ cm}$ and $\text{r}=(\text{a}+1)\text{cm}^2$
$\text{p}^2+\text{q}^2=\Big[(\text{a}-1)^2+\big(2\sqrt{\text{a}}\big)^2\Big]\text{cm}^2$
$=\big(\text{a}^2+1-2\text{a}+4\text{a}\big)\text{cm}^2$
$=\big(\text{a}^2+1+2\text{a}\big)\text{cm}^2=(\text{a}+1)^2\text{cm}^2$
$\text{r}^2=(\text{a}+1)^2\text{cm}^2$
$\therefore\text{p}^2+\text{q}^2=\text{r}^2$
Hence, the given triangle is a right triangle.
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