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Anuj Kumar Pal

· started a discussion

· 1 Months ago

Q-54 (Question: 201621.0 )

Firstly, In a right triangle base or altitude can not be larger than the hypotenuse.
Second, how can you assume a=2b. Nowhere this condition is given in the question.

Question:
In the ABC, BD bisects \(\angle \)B and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in the terms of a and b, find the value of a and b.


Options:
A) \(\cfrac{18}{5}, \cfrac{9}{5}\)
B) \(\cfrac{5}{18}, \cfrac{5}{9}\)
C) \(\cfrac{5}{18}, \cfrac{34}{57}\)
D) \(\cfrac{27}{41}, \cfrac{62}{47}\)
Solution:
Ans: (a)

  

Knowledge Expert

· commented

· 1 Months ago

Dear Student,

In This question, Triangle ABD and Triangle BCD

angle ABD and angle CBD is equal due to BD bisect angle B.
BD=BD is common sides, in both triangle, BD perpendicular to Ac thus angle ADB and CDB, are 90 degree
Thus by the angle angle and side both Triangle ABD and Triangle BCD are congruent.
if triangles are congruent, we can write AB/AD = BC/CD
please see the solution for further steps,
AD =BC , AD bisect the BC thus put the value
7a=14b
a=2b.

Keep learning with us!!
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Team Toprankers

Knowledge Expert

· commented

· 1 Months ago

Yet to be approved!
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