Challenge for Brainly Users.
∆ABC is an isoceles triangle
AC = BC = 4cm
AB = 6 cm
EC : (AE + AD)
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Answers
∆ABC is an isoceles triangle
AC = BC = 4cm
AB = 6 cm
AB is a tangent to the circle touching at D.
The vertex C lies on the circle.
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There is no data about placement of circle with respect to the triangle. If there was any data about length of AE or EC or CF or FB or BD or AD or the radius of circle, it placement would have been fixed.
Since there is no data about it, i am placing the circle symmetrically in the triangle WITHOUT CHANGING ANY 'Given data'
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Since circle is symmetrically places, altitude CD will also be the diameter of the circle.
Since it is a isosceles triangle, the altitude bisects the base. So AD = DB = AB/2 = 3cm
See the attachment for answer.
Answer: 1/3
Answer:
Step-by-step explanation:
With due respect to @TPS , I am only proving that AD = BD first before entering into the problem .
We do not need to go to complex processes and in fact the problem was very easy .
To prove : AD = DB
Join C and D .
Let O be the centre .
So CO and DO are the radii
CD is the diameter of the circle .
In quadrilateral ODAC and quadrilateral ODBC ,
CO = CO [ Common ]
∠CDA = ∠CDB [ 90° each ]
AC = BC [ 4 cm each ]
DO = DO [ Common ]
Hence ODAC ≅ ODBC [ S.A.S.S ]
I think when you prove 3 sides and any 1 angle equal , then the quadrilaterals are congruent .
But when 3 angles are equal , they are similar .
AD = BD [ by congruency ]
∴ AD + BD = 6 cm
= > AD + AD = 6 cm
= > 2 AD = 6 cm
= > AD = 6 cm / 2
= > AD = 3 cm
Now there is a theorem :
When a chord and a tangent externally meet at a point , then the product of the chord and the distance of the chord from the external point is equal to the square of the tangent .
This means we have :
AC × AE = AD²
AC = 4 .
AD = 3
So 4 AE = 3²
= > 4 AE = 9
= > AE = 9/4
EC = AC - EC
= > EC = 4 - 9/4
= > EC = ( 16 - 9 ) / 4
= > EC = 7/4
Calculate the ratio :
EC : ( AE + AD )
⇒ 7/4 : ( 9/4 + 3 )
⇒ 7/4 : ( 9 + 12 ) / 4
⇒ 7/4 : 21/4
⇒ 7 : 21
⇒ 1 : 3
The ratio is 1 : 3 .
