Question

The perimeter of the right angled triangle is 12 cm and its hypotenuse is of length 5 cm. Find the other two sides and calculate its area. Verify the result using heron' formula.

Answer 1

Hint: To get this problem solved we need to find the height and base of the triangle using the information given and then we have to calculate the area and use heron’s formula to verify it. Area of the right angled triangle can be calculated by two ways that id with the help of half of height multiplied by base formula and we can also calculate the value of the area using heron's formula. In heron's formula, Area =$\sqrt{s(s-a)(s-b)(s-c)}$ where s is the sum of all the sides of a triangle divided by 2.

Complete step-by-step answer:

Here in this triangle a is the length of height and b is the length of base and hypotenuse is of 5 cm.
We know the perimeter that is sum of all the sides is 12 cm and one of it is
5 cm so we can do,
a + b + 5 = 12

a + b = 7

We got the value of a as a = 7 − b...(1)

On applying the Pythagoras theorem we get the new equation as:
$$\begin{gathered} {\text{a}}^{\text{2}}\text{ + }{\text{b}}^{\text{2}}\text{ = }{\text{5}}^{\text{2}} \\ {\text{a}}^{\text{2}}\text{ + }{\text{b}}^{\text{2}}\text{ = 25}.............\text{(2)} \end{gathered}$$
Putting the value of a = 7 – b here we get the equation as:
${\text{(7 - b)}}^{\text{2}}\text{ + }{\text{b}}^{\text{2}}\text{ = 25}$
On solving the above quadratic,
$$\begin{gathered} \text{49 + }{\text{b}}^{\text{2}}\text{ - 14b + }{\text{b}}^{\text{2}}\text{ = 25} \\ \text{2}{\text{b}}^{\text{2}}\text{ - 14b + 24 = 0} \\ {\text{b}}^{\text{2}}\text{ - 7b + 12 = 0} \end{gathered}$$
(b−4)(b−3)=0
Then we got the value of b as 4 and 3.
b=4, 3

a=7−b
a=3,4
For the two different values of b we will get two different values of a.

Then we can calculate the Area as area = ${\displaystyle \frac{\text{1}}{\text{2}}}\times height\times base={\displaystyle \frac{1}{2}}(4)(3)$=6 sqcm.
So, the area is 6 sqcm.

Now we have to verify this area using heron’s formula.
In heron's formula,
Area =$\sqrt{s(s-a)(s-b)(s-c)}$
Where $\text{s = }{\displaystyle \frac{\text{a + b + c}}{\text{2}}}$
On putting the values of a, b, c we get the value of s as:
$\text{s = }{\displaystyle \frac{\text{3 + 4 + 5}}{\text{2}}}={\displaystyle \frac{12}{2}}$
s = 6.
Then we get Area according to heron’s formula as:
$$\begin{gathered} \sqrt{\text{6(6 - 4)(6 - 3)(6 - 5)}} \\ \sqrt{\text{6(2)(3)(1)}}\text{ = }\sqrt{\text{36}}\text{ = 6sqcm} \end{gathered}$$
Hence, the calculated area and the area we got from heron's formula are the same so the calculated area is correct.

Note: To get this problem solved we need to recall the formula of area of right angle triangle using height and base after calculating the length of height and base and then we have to check the area calculated with the help of heron’s formula, it is the formula derived by Heron that formula is Area =$\sqrt{s(s-a)(s-b)(s-c)}$, where $\text{s = }{\displaystyle \frac{\text{a + b + c}}{\text{2}}}$. Doing this will solve your problem and will give you the right answer.