Question

If the difference between the two sides of a right angled-teiangle is 2 cm and the area of the triangle is 24 cm², find the perimeter of the triangle.​

Answer 1

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Answer:

perimeter of the triangle is 24cm

Step-by-step explanation:

let us consider as a and b as the sides

b = a - 2

area of triangle = a(a-2)/2 = 24

a^2 - 2a = 48

a = 8,-6

c =√8^2+6^2

= √64+36

=√100

=10

∴ perimeter of triangle =10+8+6=24 cm

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Answer 2

➨GIVEN:-

• Diff. between the two sides of a right angled triangle = 2 cm

• The area of triangle = 24 cm²

➨FIND:-

• What is the perimeter of triangle = ?

➨SOLUTION:-

➳In Question it is written that diff. between two sides is 2cm.

So,

Let AB = x

Then, BC = x+2

we, know that

$\tt✮area \: of \: triangle = \frac{1}{2} \times b \times h$

where it is given that,

base b = AB = x

height h = BC = x+2

area of $\triangle$ ABC = 24 cm²

$\tt➤area \: of \: triangle = \frac{1}{2} \times AB \times BC$

substitute these given values in formula

$\tt➠24 = \frac{1}{2} \times x \times (x + 2)$

$\tt➠24 = \frac{1}{2} \times {x}^{2} + 2x$

$\tt➠24 \times 2= {x}^{2} + 2x$

$\tt➠48= {x}^{2} + 2x$

$\tt➠ {x}^{2} + 2x - 48 = 0$

now, apply middle spit term into this eq.

$\tt➠ {x}^{2} + 8x - 6x- 48 = 0$

now, common out the factor.

$\tt➠ x(x + 8) - 6(x + 8) = 0$

$\tt➠ (x- 6)(x + 8) = 0$

now, Either x + 8 = 0

= x = -8

which is not possible

or, x - 6 = 0

then, x = 6

So, x = 6

Now,

=> AB = x = 6 cm

=> BC = x + 2 = 6 + 2 = 8 cm

Now, we apply pythogoras theorem $\triangle$ ABC

we get,

$\tt ➣ AC^2 = AB^2 + BC^2$

where,

• AB = 6 cm
• BC = 8 cm

substitute, these values in formula

$\tt ➣ AC^2 = {(6)}^{2} + {(8)}^{2}$

$\tt ➣ AC^2 = 36 + 64$

$\tt ➣ AC^2 = 100$

$\tt ➣ AC \cancel{^2} = 10 \cancel{^2}$

$\tt ➣ AC = 10 cm$

Now, we know that

Perimeter of $\triangle$ ABC

= AB + BC + CA

where,

• AB = 6 cm
• BC = 8 cm
• AC = 10 cm

substitute, these values in the formula

$\tt Perimeter \: of \triangle ABC = AB + BC + CA \\ \tt = 6 + 8 + 10 \\ \tt = 24cm$

$\tt \tt Hence, the \: perimeter \: of \triangle ABC \underline{\boxed{ \tt24cm}}$