Solve the following questions :
( 1 ) Chords AB and CD of a circle intersect each other in point M . The centre of the circle is P . The radius of the circle is 13 cm and PM = 5 cm . Find the product CM x DM .
( 2 ) Two poles of height a metres and b metres are P metres apart . Prove that the height h of the point of intersection N of the lines joining the top of each pole to the foot of the opposite pole is ab/a+b metres .
Answers
solution 1.
Answer:
CM * DM = 144
Step-by-step explanation:
Given :
The chords AB and CD intersect at M.
Center of the circle is P.
Radius of the circle = 13 cm
PM = 5 cm
To find :
CM * DM
Solution:
Intersecting chords theorem states that when two chords intersect each other inside a circle, the product of their segments are equal.
Also, diameter is the largest chord of the circle.
Now, draw a diameter of the circle passing through points P and M. It is the line XY as shown in the attachment.
Length XY = 2 * radius = 2 * 13 =26 cm
MY = PY - PM = 13 - 5 =8 cm
XM = XP + PM = 13 + 5 = 18 cm
By the intersecting chords theorem we have,
CM * DM = XM * YM =18 * 8 = 144
∴ CM * DM = 144
solution 2.
Let AB & CD be two poles of heights 'a' m and 'b' m .Let the point of intersection of lines AD & BC be E and the its height be 'h' m. Let the perpendicular drawn from E be EF. Let DF = 'x' m and BF = 'y' m.
BD = 'p' m [ GIVEN]
Then x + y= p
In ∆ DEF and ∆DAB.
∠DBA = ∠DFE [each equal to 90°]
∠D = ∠D [common]
∆DEF ~ ∆DAB [by AA Similarity ]
DF/DB = EF /AB
x/p = h/a
x = ph/a………………………. (1)
In ∆ BFE and ∆BDC.
∠BFE = ∠BDC [each equal to 90°]
∠B = ∠B [common]
∆ BFE ~ ∆BDC [by AA Similarity ]
BF/BD = EF /CD
y/p = h/b
y= ph/b……………………... (2)
On Adding eq (1) and (2)
x + y = ph/a + ph/b
p = ph(1/a +1/b) [x +y = p]
p/p = h(1/a +1/b)
1 = h[(a+b)/ab]
ab = h(a+b)
h = ab/(a+b) m
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