Question

The pthe perimeter of a right angled triangle is 12cm and it hypotenuse is of length 5 cm find the other two side and calculate the area verify the result using heron's formula

Answer 1

Answer:

The other two sides of the Δ are 3cm and 4cm

Step-by-step explanation:

Let "a", "b" and "h" be the three sides of the Δ, such that "h" is the hypotenuse (longest side).

We are given that perimeter of Δ = 12cm

=> a + b + h = 12

But h = 5 (given)

=> (a + b) + 5 = 12

=> (a + b) = 7  ........(i)

Now, using Pythagoras theorem:

h² = a² + b²

=> a² + b² = 5²

From (i), we get: b = 7 - a

=> a² + (7 - a)² = 5²

=> a² + 49 - 14a + a² = 25

=> 2a² - 14a + 24 = 0

=> a² - 7a + 24 = 0

=> (a - 4)(a - 3) = 0

=> a = 3 or a = 4

Now, b = 7 - a

=> If a = 3, then b = 4

&  If a = 4 then b = 3

Both these statements only mean that a = 3 or 4 & b = 3 or 4

=> Remaining sides of the Δ are: 3cm and 4cm

Heron's Formula verification:

Let s1, s2, s3 be the respective lengths of the sides of a Δ, and let:

S = (s1 + s2 + s3)/2 ..................(S is the "half perimeter")

Heron's Formula states that:

Area of Δ = $\sqrt{S(S-S1)(S-S2)(S-S3)}$

In our example, we have:

S1 = 3, S2 = 4, S3 = 5

S = (3+4+5)/2 = 12/2 = 6

Area of Δ = $\sqrt{6*(6-3)*(6-4)*(6-5)}$

                = $\sqrt{6*2*3*1}$

                 = $\sqrt{36}$

                = 6

Area of Δ by Heron's formula = 6 cm² ...........Statement 1

Area of Δ by conventional method:

Area = (1/2)*(base)*(height)

In our Δ, base = 3cm and height = 4cm

(Note that base and height can be interchanged. We only need to make sure that we are NOT involving the hypotenuse in calculating the area.)

Area of Δ = (1/2)*(3)*(4)

                = (1/2)*12

                = 6

Area of Δ by conventional (base-height) formula = 6 cm² ...Statement 2

Result verified from Statement 1 and Statement 2