Question

Plzz answer and no spammers.

In the given figure a semicircle is drawn outside the bigger semicircle. diameter BE of smaller semicircle is half of the radius BF of the bigger semicircle. if the radius of bigger semicircle is 43 cm. find the length of the tangent AC from A on a smaller semicircle ​

Answer 1

Given :  a semicircle is drawn outside the bigger semicircle. diameter BE of smaller semicircle is half of the radius BF of the bigger semicircle.  the radius of bigger semicircle is 43 cm.

To find : the length of the tangent AC from A on a smaller semicircle ​

Solution:

radius of bigger semicircle is 43 cm.

=> AF = BF = 43 cm

=> AB = AF + BF = 86 cm

diameter BE of smaller semicircle is half of the radius BF

=> BE = 43/2  cm

=>  Radius of Smaller Semi Circle = OB = OE = BE/2  =  43/4  cm

=> OC = 43/4  cm

AO  = AB - OB  = 86 -  43/4   =   301/4  cm

AC²  =  AO² - OC²

=> AC² =  (301/4)² -  ( 43/4)²  

=>  AC² =  ( 43/4)² (7² - 1)

=> AC² =  ( 43/4)² (49 -1 )

=> AC² =  ( 43/4)² (48)

=>  AC² =  ( 43 )² (3)

=> AC =  43√3

=> AC = 74.5 cm

the length of the tangent AC from A on a smaller semicircle ​ = 43√3 = 74.5 cm

If  radius of bigger semicircle is 4√3 cm.

=> AF = BF = 4√3 cm

=> AB = AF + BF = 8√3 cm

diameter BE of smaller semicircle is half of the radius BF

=> BE =  2√3  cm

=>  Radius of Smaller Semi Circle = OB = OE = BE/2  =  √3  cm

=> OC =√3   cm

AO  = AB - OB  = 8√3 -√3   = 7√3   cm

AC²  = AO² -  OC²

=> AC² =    ( 7√3 )² -  (   √3 )²

=>  AC² =   147 - 3

=> AC² = 144

=> AC = 12 cm

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Answer 2

Step-by-step explanation:

hope it helps!!!!!

the answer above actually helped me understand it but in my question in place of O I took D