m∠B=90 in ΔABC. BM is altitude to AC.If BM=15, AC=34, find AB
Answers
Answered by
7
SOLUTION IS IN THE ATTACHMENT.
This corollary is used in this question.
If an altitude is drawn to hypotenuse of a right angled triangle ,then the length of altitude is the Geometric mean of length of the segment of hypotenuse formed by the altitude.
Length of each side other then the hypotenuse is the geometric mean of length of the hypotenuse and segment of hypotenuse adjacent to the side.
BM² = AM × CM
AB² = AM × AC
** Geometric mean : geometric mean of two positive numbers a and b is √ab.
HOPE THIS ANSWER WILL HELP YOU….
Attachments:
Answered by
2
In ∆ABC , <B = 90°
BM perpendicular to AC.
BM = 15 , AC = 34 , AB = ?
In ∆ABC and ∆AMB ,
<A = <A ( common angle )
<ABC = <AMB = 90°
∆ABC ~ ∆AMB ( A.A similarity )---( 1 )
Similarly ,
∆ABC ~ ∆BMC ----( 2 )
FROM ( 1 ) and ( 2 ) we get ,
∆AMB ~ ∆BMC
Therefore ,
MA/BM = BM/CM
BM² = CM × MA
15² = ( 34 - x )x
225 = 34x - x²
x² - 34x + 225 = 0
x² - 25x - 9x + 225 = 0
x( x - 25 ) - 9( x - 25 ) = 0
( x - 25 ) ( x - 9 ) = 0
Therefore ,
x - 25 = 0 or x - 9 = 0
x = 25 or x = 9
i ) if AM = x = 25
In ∆AMB , <M = 90
AB² = BM² + AM²
= 15² + 25²
= 225 + 625
= 850
AB = √ 850
AB = 5√34
ii ) if AM = x = 9
AB² = 15² + 9²
= 225 + 81
= 306
AB = √306
AB = 3√34
I hope this helps you.
: )
BM perpendicular to AC.
BM = 15 , AC = 34 , AB = ?
In ∆ABC and ∆AMB ,
<A = <A ( common angle )
<ABC = <AMB = 90°
∆ABC ~ ∆AMB ( A.A similarity )---( 1 )
Similarly ,
∆ABC ~ ∆BMC ----( 2 )
FROM ( 1 ) and ( 2 ) we get ,
∆AMB ~ ∆BMC
Therefore ,
MA/BM = BM/CM
BM² = CM × MA
15² = ( 34 - x )x
225 = 34x - x²
x² - 34x + 225 = 0
x² - 25x - 9x + 225 = 0
x( x - 25 ) - 9( x - 25 ) = 0
( x - 25 ) ( x - 9 ) = 0
Therefore ,
x - 25 = 0 or x - 9 = 0
x = 25 or x = 9
i ) if AM = x = 25
In ∆AMB , <M = 90
AB² = BM² + AM²
= 15² + 25²
= 225 + 625
= 850
AB = √ 850
AB = 5√34
ii ) if AM = x = 9
AB² = 15² + 9²
= 225 + 81
= 306
AB = √306
AB = 3√34
I hope this helps you.
: )
Attachments:
Similar questions