Question

Sum of the area of the two squares
in 500 metre. If the difference of their
Perimetens is 40 metre. Find the side of
the two squares​

Answer 1

Step-by-step explanation:

Let the length of side be a in first triangle and b in second

Area = (side)²

So, a²+b² = 500. EQUATION 1

Also, perimeter = 4(side)

So, 4a-4b = 40. (Assuming a>b)

a-b= 10 (dividing by 4 on both sides). EQUATION 2

From equation 2,

a=b+10

Putting a=b+10 in equation 1 ,

(b+10)² + b² = 500

b² + 20b + 100 + b² =500

2b²+20b+100=500

2b²+20b-400=0

b²+10b-200=0 (dividing by 2 on both sides)

b²+50b-40b-200=0 (mid-term factorisation)

b(b+50)-40(b+50)=0

(b-40)(b+50)=0

Therefore, b= 40 OR -50

But length cannot be negative so b=40 ONLY

Putting b = 40 in equation 2,

a-40=10

a=50

Therefore, a=50. & b = 40

Answer 2

Answer:

a1 = 20 and a2 = 10

Step-by-step explanation:

Let the sides of the square A1 and A2 be a1 and a2 respectively and P1 and P2 be the perimeter of the same

A1 + A2 = 500m²

So, a1² + a2²= 500....(1)

P1 + P2 = 40m

So, 4a1 - 4a2 = 40

a1 = 10 + a2

put the value of a1 in (1)

(10 + a2)² + a2² = 500

100 + a2² + 20a2 + a2² = 500

a2² + 10a2 - 200 = 0

solve quaditic eq..

a1 = 20m and a2 = 10m