Determine hypotenuse of right triangle if its area is 6 cm^2 and perimeter is 12 cm?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

Let the smaller sides of the triangle have a length a and b. The length of the hypotenuse is sqrt( a^2 + b^2)

The area of the triangle is (1/2)*a*b = 6

=> ab = 12

=> a = 12/b

The perimeter is a + b + sqrt (a^2 + b^2) = 12

a + b + sqrt (a^2 + b^2) = 12

replace 12 with ab

=> a + b + sqrt (a^2 + b^2) = ab

=> ab - (a + b) = sqrt (a^2 + b^2)

square the two sides

=> a^2*b^2 + (a +b)^2 - 2ab*(a + b) = a^2 + b^2

=> a^2*b^2 + a^2 + b^2 + 2ab - 2ab*(a + b) = a^2 + b^2

=> a^2*b^2  + 2ab - 2ab*(a + b) = 0

divide all the terms by ab

=> ab + 2 - 2(a + b) = 0

substitute a = 12/b

=> b*12/b + 2 - 2(12/b + b) = 0

=> 14 = 2*(12/b + b)

=> 7 = 12/b + b

=> 7b = 12 + b^2

=> b^2 - 7b + 12 = 0

=> b^2 - 4b - 3b + 12 = 0

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

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

b = 3 or b = 4

a = 4 or a = 3

The hypotenuse is sqrt (a^2 + b^2) = sqrt ( 3^2 + 4^2) = sqrt 25 = 5

The required hypotenuse = 5.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll put the length of the legs of the right triangle as x and y.

A = x*y/2

6 = x*y/2 => x*y = 12 (1)

Since the perimeter is the sum of the lengths of the legs and hypotenuse of right angle triangle, we'll get

P = x + y + sqrt(x^2 + y^2)

Hypothenuse = sqrt(x^2 + y^2) (Pythagorean identity)

12 = x + y + sqrt(x^2 + y^2) (2)

We'll equate (1) and (2):

xy = x + y + sqrt(x^2 + y^2)

We'll subtract both sides  x + y:

xy - (x+y) = sqrt(x^2 + y^2)

We'll raise to square both sides:

[xy - (x+y)]^2 = [sqrt(x^2 + y^2)]^2

(xy)^2 + x^2 + 2xy + y^2 - 2xy(x+y) = x^2 + y^2

We'll eliminate x^2 + y^2:

(xy)^2 + 2xy - 2xy(x+y) = 0

We'll factorize by xy:

xy(xy + 2 - 2x - 2y) = 0

We'll put xy = 0

Since xy = 0 => 2 - 2x - 2y = 0

-2(x+y) = -2

x + y = 1

We'll create the quadratic equation whose sum is 1 and product is 0:

x^2 - x = 12

x^2 - x - 12 = 0

x1 = [1+sqrt(1 + 48)]/2

x1 = (1+7)/2

x1 = 4

x2 = -3

Since a length of a side cannot be negative, we'll reject x = -3.

We'll put x = 4

6 = 4*y/2

12 = 4*y

y = 3

The lengths of the legs of the triangle are x = 4 and y = 3.

The hypothenuse is:

H = sqrt(3^2 + 4^2)

H = sqrt25

H = 5 cm

The length of hypotenuse is H = 5 cm.

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