Question

The perimeter of a right triangle is 60cm. Its hypotenuse is 26cm, find the other two sides and the area of the
triangle?

Answer 1

Answer

120 cm²

Explanation

Let the other two sides be a and b

Then,

a + b + 26 = 60

(since 60 is the perimeter of the triangle, sum of sides should equal 60)

⇒ a + b = 60 - 26

⇒ a + b = 34

Now, by Pythagoras theorem we have,

a² + b² = 26²

⇒ a² + b² = 676

Now,

a + b = 34

Squaring both sides, we get

(a + b)² = (34)²

⇒ a² + b² + 2ab = 1156

⇒ 676 + 2ab = 1156

⇒ 2ab = 1156 - 676

⇒ 2ab = 480

⇒ ab = 480/2

⇒ ab = 240

Now, we know that area of triangle = 1/2 × base × height

⇒ area of give triangle = 1/2 × a × b

= ab/2

⇒ 240/2

= 120 cm²

Answer 2

Answer:

⇢ Other two sides = 24 cm & 10 cm

⇢ Area of triangle = 120 cm²

$\rule{200}{2}$

Let the base and altitude be x cm & y cm.

As it is a right triangle,we can provide Pythagoras theorem here.

According to Pythagoras theorem:-

⇒ (Hypotenuse)² = (Base)² + (Altitude)²

⇒ (26)² = x² + y²

⇒ x² + y² = 676 ...................(eq.1)

2nd case:-

As,Perimeter of triangle = Sum of all sides

⇒ 60 = 26 + x + y

⇒ x + y = 60 - 26

⇒ x + y = 34

⇒ x = 34 - y................(ii)

• Putting value of (eq.2) in (eq.1):-

⇒ (34 - y)² + y² = 676

⇒ (34)² + y² - 68y + y² = 676

⇒ 1156 + 2y² - 68y = 676

⇒ 2y² - 68y + 1156 - 676 = 0

⇒ 2y² - 68y + 480 = 0

⇒ 2(y² - 34y + 240) = 0

⇒ y² - 34y + 240 = 0

⇒ y² - 24y - 10y + 240 = 0

⇒ y(y - 24)-10(y - 24) = 0

⇒ (y - 24)(y - 10) = 0

⇒ (y - 24) = 0 or (y - 10) = 0

⇒ y = 24 or y = 10

$\therefore{\underline{\rm{Altitude = 10 \: m \: or, \: 24 \: m}}}$

• Putting value of y in (eq.2)

1st case:-

⇒ Base = 34 - 10 cm

⇒ Base = 24 cm

2nd case:-

⇒ Base = 34 - 24 cm

⇒ Base = 10 cm

$\therefore{\underline{\rm{Other \: two \: sides = 10 \: cm \: \& \: 24 \: cm}}}$

• Now,area of triangle:-

We know,

Area of triangle = ½ × Base × Altitude

⇒ Area of triangle = ½ × 24 × 10

⇒ Area of triangle = 120 cm²

$\therefore{\underline{\rm{Area \: of \: triangle = 120 \: cm^2}}}$