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# a circle is inscribed in a triangle abc having sides 8cm 10cm and 12cm

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## A circle is inscribed in a triangle ABC, having sides 8cm, 10cm and 12cm. Find AD,BE and CF ( these 3 are altitudes of triangle ABC ) .

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## A circle is inscribed in a triangle ABC, having sides 8cm, 10cm and 12cm. Find AD,BE and CF ( these 3 are altitudes of triangle ABC) .

Medium Open in App Solution Verified by Toppr Given, x+y=8cm   ....(1) y+z=12cm .....(2) x+z=10cm .....(3)

Adding (1), (2) and (3), we get

2(x+y+z)=30 x+y+z=30/2 x+y+z=15  .....(4)

Equation (4)−(2), we get

x+12=15 x=15−12 x=3

Equation (4)−(3),  we get

10+y=15 y=15−10 y=5 (4)−(1), we get 8+z=15 z=15−8 z=7 So, AD=3 cm BE=5 cm CF=7 cm  Answer.

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## A circle is inscribed in a triangle ABC, having sides 8cm, 10cm and 12cm. Find AD, BE and CF (these 3 are altitudes of triangle ABC).

A circle is inscribed in a triangle ABC, having sides 8cm, 10cm and 12cm. Find AD, BE and CF (these 3 are altitudes of triangle ABC).. Ans: Hint: To find this first we will draw a figure of a circle inscribed in a triangle ABC where the sides of tria...

## A circle is inscribed in a triangle ABC, having sides 8cm, 10cm and 12cm. Find AD, BE and CF (these 3 are altitudes of triangle ABC).

Answer Verified 188.7k+ views 4 likes

Hint: To find this first we will draw a figure of a circle inscribed in a triangle ABC where the sides of triangle can touch the edge of the circle by making points D, E, & F. From the figure we can easily understand that AD, BE & CF are tangents of a circle. Then with the help of the given sides of a triangle we will find the length of tangent made by external point of circle by substituting method

From the figure we get to know that the external points of the circle of circle A, B, & C are the tangents and

AB=12cm,BC=8cm,AC=10cm

AB=12cm,BC=8cm,AC=10cm

.

We know that the tangents drawn from the external points of the circle are equal.

Now let us consider –

It is given that

AB=12cm,BC=8cm,AC=10cm

AB=12cm,BC=8cm,AC=10cm

. Where, ⇒AB=AD+BD ⇒AB=AD+BD x+y=12cm x+y=12cm ……………….. (1) ⇒BC=BE+CE ⇒BC=BE+CE y+z=8cm y+z=8cm …………………… (2) ⇒AC=AF+CF ⇒AC=AF+CF x+z=10cm x+z=10cm ………………… (3)

By adding equation (1), (2) and (3), we get –

(x+y)+(y+z)+(x+z)=12cm+8cm+10cm

2x+2y+2z=30cm

(x+y)+(y+z)+(x+z)=12cm+8cm+10cm2x+2y+2z=30cm

By taking 2 as common we get –

2(x+y+z)=30cm 2(x+y+z)=30cm

By dividing both sides by 2, we get –

x+y+z=15cm x+y+z=15cm ………………………….. (4)

Now, we will substitute equation (1) in equation (4). So, we get –

(x+y)+z=15 12+z=15 (x+y)+z=1512+z=15

By subtracting 12 from both the sides, we get –

z=15−12 z=3 z=15−12z=3

Now, we will substitute the value of ‘z’ in equation (3).

x+z=10 x+3=10 x+z=10x+3=10

By subtracting 3 from both sides, we get –

x=10−3 x=7 x=10−3x=7

By substituting the value of ‘z’ in equation (2), we get –

y+z=8 y+3=8 y+z=8y+3=8

By subtracting 3 from both sides, we get –

y=8−3 y=5 y=8−3y=5 Therefore, AD=x=7cm BE=y=5cm CF=z=3cm.

Note: Generally students make mistakes while solving this problem by taking x, y & z as the sides of the triangle then divide it by 2 to get an answer which is completely wrong. Students should know that the external points of the circle which is forming a triangle are the tangents of a circle. The lengths of two tangents from an external point of a circle are equal.

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## A circle is inscribed in ∆ABC having sides 8 cm, 10 cm and 12 cm as shown in the figure, Find AD, BE and CF.

A circle is inscribed in ∆ABC having sides 8 cm, 10 cm and 12 cm as shown in the figure, Find AD, BE and CF.

A circle is inscribed in ∆ABC having sides 8 cm, 10 cm and 12 cm as shown in the figure, Find AD, BE and CF.

### SOLUTION

AD = AF = x   ...(Tangent of the circle)

BD = BE = y   ...(Tangent of the circle)

CE = CF = z   ...(Tangent of the circle)

AB = AD + DB x + y = 12   ...(1) BC = BE + EC y + z = 8   ...(2) AC = AF + FC x + z = 10  ...(3)

Add (1), (2) and (3)

2x + 2y + 2z = 12 + 8 + 10

x + y + z = 302 = 15  ...(4)

By x + y = 12 in (4)

z = 3 y + z = 8 in (4) x = 7 x + z = 10 in (4) y = 5

AD = 7 cm, BE = 5 cm and CF = 3 cm

Concept: Circles and Tangents

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Chapter 4: Geometry - Exercise 4.4 [Page 198]

Q 3 Q 2 Q 4

### APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 10th SSLC Mathematics Answers Guide

Chapter 4 Geometry

Exercise 4.4 | Q 3 | Page 198

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Mohammed 1 month ago

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