Question

71. A circle is inscribed in a triangle ABC such that it
touches the sides BC, CA and AB at D, E and F
respectively as shown in the figure. If BC = 5 cm
and AB + AC = 13 cm, then AE equals
А
and
F
E
5),
В
С
D
(1) 2 cm
(3) 4 dm
(2) 3 cm
(4) 6 cm​

Answer 1

Answer:

Let there is a circle having centre O touches the sides AB and AC of the triangle at point E and F respectively.

Let the length of the line segment AE is x.

Now in ΔABC,

CF = CD = 6 (tangents on the circle from point C)

BE = BD = 6 (tangents on the circle from point B)

AE = AF = x (tangents on the circle from point A)

Now AB = AE + EB

=> AB = x + 8

BC = BD + DC

=> BC = 8+6 = 14

CA = CF + FA

=> CA = 6 + x

Now

s = (AB + BC + CA )/2

=> s = (x + 8 + 14 + 6 +x)/2

=> s = (2x + 28)/2

=> s = x + 14

Area of the ΔABC = √{s*(s-a)*(s-b)*(s-c)}

= √[(14+x)*{(14+x)-14}*{(14+x)-(6+x)}*{(14+x)-(8+x)}]

= √[(14+x)*x*8*6]

= √[(14+x)*x*2*4*2*3]

=> Area of the ΔABC = 4√[3x(14+x)] .............................1

Now area of ΔOBC = (1/2)*OD*BC

= (1/2)*4*14

= 56/2

= 28

Area of ΔOBC = (1/2)*OF*AC

= (1/2)*4*(6+x)

= 2(6+x)

= 12 + 2x

Area of ΔOAB = (1/2)*OE*AB

= (1/2)*4*(8+x)

= 2(8+x)

= 16 + 2x

Now Area of the ΔABC = Area of ΔOBC + Area of ΔOBC + Area of ΔOAB

=> 4√[3x(14+x)] = 28 + 12 + 2x + 16 + 2x

=> 4√[3x(14+x)] = 56 + 4x

=> 4√[3x(14+x)] = 4(14 + x)

=> √[3x(14+x)] = 14 + x

On squaring bothe side, we get

3x(14 + x) = (14 + x)2

=> 3x = 14 + x (14 + x = 0 => x = -14 is not possible)

=> 3x - x = 14

=> 2x = 14

=> x = 14/2

=> x = 7

Hense

AB = x + 8

=> AB = 7+8

=> AB = 15

AC = 6 + x

=> AC = 6 + 7

=> AC = 13

So value of AB is 15 cm and value of AC is 13 cm

1.5