Q. 6 D5.0( 1 Vote )

# m∠<s

Answer : AB = AM = 2.5

Applying Pythagoras theorem in ΔABM

AB2 = AM2 + BM2

BM2 = 10 – 6.25 = 3.75

Let CM = x

AC = x + 2.5

Applying Pythagoras theorem in ΔCBM

BC2 = BM2 + CM2

BC2 = x2 + 3.75 …Equation (i)

Applying Pythagoras theorem in ΔABC

AC2 = AB2 + BC2

(x + 2.5)2 = BC2 + 10

x2 + 6.25 + 5x = BC2 + 10

BC2 = x2 – 3.75 + 5x …Equation (ii)

Equating Equation (i) and Equation (ii)

x2 + 3.75 = x2 – 3.75 + 5x

5x = 7.5

x = 1.5

So CM = 1.5

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Match the Columns38 mins  Application of Trigo Important Questions44 mins  Chemical Properties of Metal and Non Metal61 mins  Trigo Marathon Part 146 mins  Heights and Distances - II45 mins  Heights and Distances - I54 mins  Trigo Marathon Part 245 mins  Arithmetic Progression and 'nth' Term of AP55 mins  Extraction of Metal56 mins  Purification of Metals60 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

AB = AC and <spanGujarat Board Mathematics

<spGujarat Board Mathematics

In ΔPQR, <img widGujarat Board Mathematics

m<sGujarat Board Mathematics

In ΔABC, mGujarat Board Mathematics

In ΔABC, mGujarat Board Mathematics

In ΔXYZ, mGujarat Board Mathematics

In ΔABC, mGujarat Board Mathematics

In ΔPQR, mGujarat Board Mathematics

â<spGujarat Board Mathematics