Question

A circle is touching the side BC of AABC at L and touching AB and AC produced at M and N respectively. If semi-perimeter of AABC is 60 cm, then find the length of AM.​

Answer 1

Answer:

Chapter wise Important Questions for CBSE Class 10 Maths. Chapter 1 - Real Numbers. Chapter 2 - Polynomials. Chapter 3 - Pair of Linear Equations in Two Variables. Chapter 4 - Quadratic Equations. Chapter 5 - Arithmetic Progressions. Chapter 6 - Triangles. Chapter 7 - Coordinate Geometry.

Real Numbers

Answer 2

Question :-

• A circle is touching the side BC of AABC at L and touching AB and AC produced at M and N respectively. If semi-perimeter of AABC is 60 cm, then find the length of AM.

Answer

Given :-

• A circle is touching the side BC of AABC at L and touching AB and AC produced at M and N respectively.
• The semi-perimeter of AABC is 60 cm.

To Find :-

• The length of AM.

Concept used :-

• The length of tangents drawn from external point to a circle are equal.
• Perimeter of triangle is sum of three sides.

Solution :-

Since, BL & BM are tangents from external point B.

$\bf\implies \:BL = BM........(1)$

Also, CL & CN are tangents from external point C.

$\bf\implies \:CL = CN........(2)$

Again, AM & AN are tangents from external point A.

$\bf\implies \:AM = AN........(3)$

Now, it is given Perimeter of triangle ABC = 60 cm

$\bf\implies \:AB + BC + CA = 60$

$\bf\implies \:AB + (BL + LC) + CA = 60$

Using (1) and (2) equation, we get

$\bf\implies \:AB + BM + CN + CA = 60$

$\bf\implies \:AM + AN = 60$

Using equation (3), we get

$\bf\implies \:AM + AM = 60$

$\bf\implies \:2AM = 60$

$\bf\implies \:AM = {\cancel\dfrac{60}{2} \: 30}$

$\bf\implies \:Length \: of \: AM = 30 \: cm$

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