ABCD is a square of side 8, M is the centre of the circle taking AD as diameter, E is a point on the side AB such that CE is tangent to the circle. Find the area of the triangle CBE.
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Answers
Concept
Square is an object having four sides equal
Given
BD=30cm,DC=7cm,∠BAC=90
AB, BC and AC are tangents to the circle at E, D and F.
Find
We need to find the area of triangle CBE
Solution
From the theorem stated,
BE=BD=30cm
Also, FC=DC=7cm
Let, AE=AF=x …. (1)
Then AB=BE+AE=(30+x)
AC=AF+FC=(7+x)
BC=BD+DC=30+7=37cm
Consider right triangle ABC, by Pythagoras theorem we have
BC2=AB2+AC
2(37)2=(30+x)
2+(7+x)21369=900+60x+x2+49+14x+x22x2+74x+949–1369=0
2x2+74x–420=0
x2+37x–210=0
x2+42x–5x–210=0
x(x+42)–5(x+42)=0
(x–5)(x+42)=0
(x–5)=0or(x+42)=0
x=5orx=–42
x=5[Since x cannot be negative]
∴AF=5cm[From (1)]
ThereforeAB=30+x=30+5=35cm
(ii)
AC=7+x=7+5=12cm
Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.
Join point O, F; points O, D and points O, E.
From the figure,
21×AC×AB=21×AB×OE+21×BC×OD+ 21×AC×OC
AC×AB=AB×OE+BC×OD+AC×OC
12×35=35×r+37×r+12×r
420=84r
∴r=5
Thus the radius of the circle is 5 cm.
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