#
Find all sides of a right triangle whose perimeter is equal to 60 cm and its area is equal to 150 square cm.

2
by ChosenBunny

Log in to add a comment

by ChosenBunny

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Let the three sides of the triangle be

base = b

height = h

hypotenuse = H

b² + h² = H²

⇒(b+h)² - 2bh = H²

Area = 150

⇒(1/2)bh = 150

⇒bh = 300

Perimeter = 60

⇒ b + h + H = 60

⇒b + h = 60-H

Thus (b+h)² - 2bh = H²

⇒(60-H)² -2×300 = H²

⇒ 3600 + H² - 2×60×H - 2×300 = H²

⇒3600 - 120H -600 = 0

⇒ -120H + 3000 = 0

⇒120H = 3000

⇒ H = 3000/120 = 25cm

b+h = 60-25 = 35

⇒300/h + h = 35

⇒h² + 300 = 35h

⇒h² -35h + 300 = 0

⇒ h² -15h -20h +300 = 0

⇒(h-15)(h-20) = 0

⇒h = 15 or 20

⇒b = 20 or 15

Thus Three sides are (15,20,25)

base = b

height = h

hypotenuse = H

b² + h² = H²

⇒(b+h)² - 2bh = H²

Area = 150

⇒(1/2)bh = 150

⇒bh = 300

Perimeter = 60

⇒ b + h + H = 60

⇒b + h = 60-H

Thus (b+h)² - 2bh = H²

⇒(60-H)² -2×300 = H²

⇒ 3600 + H² - 2×60×H - 2×300 = H²

⇒3600 - 120H -600 = 0

⇒ -120H + 3000 = 0

⇒120H = 3000

⇒ H = 3000/120 = 25cm

b+h = 60-25 = 35

⇒300/h + h = 35

⇒h² + 300 = 35h

⇒h² -35h + 300 = 0

⇒ h² -15h -20h +300 = 0

⇒(h-15)(h-20) = 0

⇒h = 15 or 20

⇒b = 20 or 15

Thus Three sides are (15,20,25)

in which , perpendicular = p

base = b , and diagonal = h

A/q,

from pythagoras theorem .

p²+b² = h² --------------------------(1)

Area of right tringle = 1/2×b.p

1/2×b.p = 150

or , bp = 300 -------------------(2)

&

perimeter = p+b+h

or, p+b+h = 60

or, p+b = 60-h -------------------(3)

Now squaring both side of the equation (3).,

(p+b)² = (60-h)²

or, p²+b²+2bp = 60²+h²-2×60.h

or, h²+2(300) = 3600+h²-120h

or, 120h = 3600-600 = 3000

or, h = 3000/120 = 25 cm

so from eqn (3) ,p+b = 60-25 = 35 ------------(4)

We know that, (p+b)²-(p-b)² = 4.pb

or, (35)²+(p-b)² = 4 (300) {from eqns (3) & (2)

or, (p-b)² = 1200 - 1225 = 25

or, p-b = (+-) 5 ------(5)

On adding the Eqns (4) & (5)

2p = 35(+-) 5

so, either p = (35+5)/2 = 20 cm ,then b = 35-20 = 15 cm

or, p = (35-5)/2 = 15 cm , then b = 35-15 = 20 cm

Therefore sides of the right triangle are 25 cm, 20 cm and 15 cm .