Question

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

please ans me fast with explaintion ​

Answer 1

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GIVEN: 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: 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 …

Answer 2

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$\large \underline\red{ᴀɴsᴡᴇʀ:- }$

ɢɪᴠᴇɴ :- 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.

ᴛᴏ ꜰɪɴᴅ : ᴘQ= ?

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

ꜱᴏ , triangle PAB & triangle QAB are equilateral triangles ( as all 3 sides are equal in each triangle)

ᴀʀᴇᴀ ᴏꜰ ᴇQᴜᴀɪʟᴀᴛᴇʀᴀʟ ᴛʀɪᴀɴɢʟᴇ = (√3/4) side²

⇨ ᴀʀᴇᴀ ᴏꜰ ᴛʀɪᴀɴɢʟᴇ ᴘᴀʙ = (√3/4) × 25 cm²

⇨ ᴀʀᴇᴀ ᴏꜰ ʙᴏᴛʜ ᴛʀᴀɪɴɢʟᴇ = 2 ×( √3/4) × 25

⇨(25√3 /2 ) cm² = ᴀʀᴇᴀ ᴏꜰ ʀʜᴏᴍʙᴜꜱ ᴘᴀQʙ …(1)

ɴᴏᴡ , ᴀʀᴇᴀ ᴏꜰ ʀhombus = 1/2 × ᴅiagonal AB ×

ᴅiagonal PQ

⇨ 1/2 × 5 × PQ = 25√3 /2…. ( by (1) )

PQ = (25√3/2) × (2/5)

⇨PQ = 5√3 cm

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