# 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

Complete step-by-step answer:

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.

Therefore, AD=AF,CF=CE,BE=BD. AD=AF,CF=CE,BE=BD.

Now let us consider –

AD=AF=x BD=BE=y CE=CF=z

AD=AF=xBD=BE=yCE=CF=z

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.

AD=x=7cmBE=y=5cmCF=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.

## 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.

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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.

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### 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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