Two equal circles pass through each other’s centre. If the radius of each circle is 5 cm, whtat is length of the common chord?

Dear Student


Given that ,
2 equal circles with centre A & radii AP, AB, AQ = 5 cm And circle with centre B, & radii BP, BA & BQ = 5 cm. PQ is a common chord to both the circles.

To find :length of chord PQ= ?

Since APBQ is a rhombus, as all its sides = 5 cm & one diagonal AB = 5 cm

So, triangle PAB & triangle QAB are equilateral triangles ( as all 3 sides are equal in each triangle)
Area of equilateral triangle = (√3/4) side²

=> area of triangle PAB = (√3/4)* 25 cm²

=> area of both triangles = 2* ( √3/4) *25

=> (25√3 /2 ) cm² = area of rhombus PAQB …(1)

NOW, area of rhombus = 1/2 * diagonal AB * diagonal PQ

=> 1/2 * 5 * PQ = 25√3 /2…. ( by (1) )

PQ = (25√3/2) * (2/5)

=> PQ = 5√3 cm Ans

Thanks