Simple Math Puzzle on triangles.
ABCD is a square. Green circle's radius = 2 cm. Blue circle has a radius = 4 cm.
Find the length "a" of the edge of the square.
Use the formulas related to triangles.
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kvnmurty:
incircle formula ?
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2
Heya User,
Clearly.
--> PF = KO = KF = 4cm
--> BP = BT ; CK = CT --> [ Tangents to a circle from common point ]
=> Perimeter [ BFC ] = BP + PF + KF + KC + BC
--> Perimeter [ BFC ] = ( BT + CT ) + PF + KF + BC
--> Perimeter [ BFC ] = a + a + 4 + 4
--> Perimeter [ BFC ] = 2a + 8; --------> (i)
Also,
==> AREA [ BFC ] =inradius[ Semi-Perimeter [ BFC ] ]--------> (ii)
Now, Area [ EFGH ] = 4² = 16; ---------> (iii)
Observe :-> ΔBFC, AEB, AHD, DGC are same !!
---> .'. 4(ii) + (iii) ---> gives :->
----> 16[ 2a + 8 ]/2 + 16 = a²
==> 16a + 64 + 16 = a²
==> a² - 16a - 80 = 0
==> a = 20, -4
Hence, a = 20 is the soln.
Clearly.
--> PF = KO = KF = 4cm
--> BP = BT ; CK = CT --> [ Tangents to a circle from common point ]
=> Perimeter [ BFC ] = BP + PF + KF + KC + BC
--> Perimeter [ BFC ] = ( BT + CT ) + PF + KF + BC
--> Perimeter [ BFC ] = a + a + 4 + 4
--> Perimeter [ BFC ] = 2a + 8; --------> (i)
Also,
==> AREA [ BFC ] =inradius[ Semi-Perimeter [ BFC ] ]--------> (ii)
Now, Area [ EFGH ] = 4² = 16; ---------> (iii)
Observe :-> ΔBFC, AEB, AHD, DGC are same !!
---> .'. 4(ii) + (iii) ---> gives :->
----> 16[ 2a + 8 ]/2 + 16 = a²
==> 16a + 64 + 16 = a²
==> a² - 16a - 80 = 0
==> a = 20, -4
Hence, a = 20 is the soln.
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16
Nice question.
Answer explained below.
Comment below if any doubt occurs to you.
#OriginalAnswer
#NoCopy
Answer explained below.
Comment below if any doubt occurs to you.
#OriginalAnswer
#NoCopy
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