Question

If the sides of a triangle are in ap the perimeter of the triangle is 30 cm. The difference between the longer and shorter side is4cm.then find all sides of triangle.

Answer 1

Answer:

8 cm , 10 cm , 12 cm.

Step-by-step explanation:

Given,    

     sides of the triangle are in AP. They must have a common difference.

Let the common difference between them be d.

  And, let the smallest angle be a - d.

So,  second angle should be a - d + d = a

      3rd( largest ) angle should be a - d + 2d = a + d

Here,

  Perimeter of Δ = 30 cm

⇒ Sum of all side = 30

⇒ ( a - d ) + a + ( a + d ) = 30

⇒ a - d + a + a + d = 30

⇒ 3a = 30

⇒ a = 30 / 3

⇒ a = 10

     Also,

Difference b/w longer and shortest side is 4 cm.

This means : longest side - shortest side = 4

  ⇒ ( a + d ) - ( a - d ) = 4

  ⇒ a + d - a + d = 4

  ⇒ 2d = 4

  ⇒ d = 2

Therefore,

Sides are = a - d = 10 - 2

                     ⇒ 8 cm

 2nd = a = 10 cm

 3rd = a + d = 10 + 2

               ⇒ 12 cm

Answer 2

Solution :

It is given that the sides of a triangle are in AP  the perimeter of the triangle is 30 cm. Also, difference = 4 cm.

So, keeping this in mind,

Let ' a ' be the number and ' d ' be the difference.

sides of given triangle will be :-

• a - d
• a
• a + d

According to question :

$\sf (a+d) - ( a - d ) = 4$

$\sf a + d - a + d = 4$

$\sf 2d = 4$

$\sf d =\dfrac{4}{2}$

$\sf\therefore d = 2$

Now,

we know that, perimeter of triangle is equal to sum of all sides.

So,

$\sf a-d + a + a + d = 30$

$\sf 3a = 30$

$\sf a =\dfrac{30}{3}$

$\sf\therefore a = 10$

Finally,

Sides of given triangle :-

• a - d = 10 - 2 = 8 cm
• a = 10 cm
• a + d = 10 + 2 = 12 cm

$\rule{200}2$