Question

The area of a right angled triangle is 30 cm2 and
the length of its hypotenuse is 13 cm. The length of
the shorter leg is

Answer 1

Step-by-step explanation:

area = 30cm^2

1/2 * b * h = 30

bh = 60 --(1)

hypotenuse = 13cm

√(b^2 + h^2) = 13

on squaring both sides ,we get

b^2 + h^2 = 169

b^2 + h^2 +2bh -2bh = 169

(b - h)^2 + 2(60) = 169

(b - h)^2 = 49

b - h = 7 => h = b - 7 --(2)

from(1) b (b - 7) = 60

b^2 - 7b - 60 = 0

b^2 - 12b + 5b - 60 = 0

b(b - 12) + 5( b - 12) = 0

b = - 5, 12

neglect -5 so b = 12 cm

from (2) h = 12 -7 = 5cm

therefore, shortest leg of triangle is of 5cm Answer

Answer 2

Answer:

Shortest side is 5 cm

Step-by-step explanation:

Given :

Area = 30 cm²

Hypotenuse = 13 cm

To find :

The length of the shorter leg

Solution :

$\boxed{\sf{Area = \frac{1}{2} \times Base \times Height}}$

⇒ 30 = 1/2 × Base × Height

⇒ 60 = Base × Height ------ (I)

$\rule{300}{1.5}$

Then, the height of the triangle -

⇒ Height = 60/Base

⇒ h = 60/Base

$\rule{300}{1.5}$

Multiply both sides of Equation I by 2,

⇒ 60 = b × h

⇒ 120 cm² = 2bh

$\rule{300}{1.5}$

$\boxed{\sf{(Hypotenuse)^{2} = (Base)^{2} + (Height)^{2}}}$

⇒ 13² = b² + h²

⇒ 169 = b²+h² ------ (II)

$\rule{300}{1.5}$

By adding Equation II and III

⇒ 289 = b²+h²+2bh

⇒ 289 = (b + h)²

⇒ $\sf{\sqrt{289}}$ = (b + h)

⇒ 17 = b + h ------ (III)

$\rule{300}{1.5}$

Place value of h in this equation ,

⇒ 17 cm = b + 60/b

⇒ 17 = (b² + 60)/b

⇒ 17b = b² + 60

⇒ b² + 60 - 17b = 0

⇒ b² - 17b + 60 = 0

$\rule{300}{1.5}$

Split the middle term,

⇒ b² - 5b - 12b + 60 = 0

⇒ b(b - 5) - 12(b - 5) = 0

⇒ (b - 5) (b - 12) = 0

⇒ b = 5 or b = 12

$\rule{300}{1.5}$

If b is 5 cm

Height = 12 cm

Shortest leg = 5 cm (Base)

If b is 12 cm

Height = 5 cm

Shortest leg = 5 cm (Height)