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2 years ago

The perimeter of a right angled triangle is 30cm and its area is 30cm square then find the side of triangle

Mathematics

2 answers:

wariber [46]2 years ago

7 0

Perimeter = a + b + c = 30

Area = 1/2 x a x b = 30

Multiples of 30: 2, 3, 5, 6, 10, 12, 15

For perimeter c = 30- (a+b)

C= sqrt( a^2 + b^2)

Using the possible combinations of the above:

5 and 12:

C = sqrt(5^2 + 12^2) = 13

5 + 12 + 13 = 30 for the perimeter

Area = 1/2 x 5 x 12 = 30

The sides are 5, 12 and 13 cm

maria [59]2 years ago

5 0

Answer:

12 cm, 5 cm, 13 cm

Step-by-step explanation:

Sides of triangle:

a, b, c

Perimeter:

P= a+b+c = 30

Area:

1/2ab = 30 ⇒ ab = 60

As per Pythagorean theorem, side c= √(a²+b²):

a+b+c = 30
a+b+ √(a²+b²) = 30
a+b - 30= - √(a²+b²)
(a+b - 30)²= (- √(a²+b²))²
a²+b²+900 + 2ab - 60a - 60b = a²+b²

Considering ab = 60

900+120=60a+60b ⇒ a+b = 17

Again using ab= 60 to find out sides:

ab=60
a(17-a)= 60
17a - a² -60 = 0
a² -17 a +60= 0

Solving the quadratic equation, its positive root is:

a= 12

Then we get the rest:

b = 17 - 12= 5
c= 30 -17 = 13

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Answer:

Check attachment for complete question

Step-by-step explanation:

Given that,

y=Coskt

We are looking for value of k, that satisfies 4y''=-25y

Let find y' and y''

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y'=-kSinkt

y''=-k²Coskt

Then, applying this 4y'"=-25y

4(-k²Coskt)=-25Coskt

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k²=-25/-4

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Then, k=√(25/4)

k= ± 5/2

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Since k= ± 5/2

y=A•Sin(±5/2t)+ B •Cos(±5/2t)

y'=±5/2ACos(±5/2t)-±5/2BSin(±5/2t)

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Then, inserting this to our equation given to check if it a solution to y=ASinkt+BCoskt

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-25A•Sin(±5/2t) - 25B •Cos(±5/2t)

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