Question

Chords AB and CD of a circle meet inside the circle at P. If PA \[ = 4cm\] , AB \[ = 7cm\] and PD \[ = 6cm\] , then length of CD is

Answer 1

Hint: A chord of a circle is nothing but a line whose endpoints lie on the circle. From the properties of a chord, we have that whenever any two chords intersect at some point inside the circle then the product of the parts of the one chord will be equal to the product of the parts of the other.
Using this statement, we can solve the given problem.

Complete step by step answer:
It is given that there are two chords in a circle AB and CD. These two chords meet inside the circle at a point P. It is also given that PA \[ = 4cm\] , AB \[ = 7cm\] , and PD \[ = 6cm\] . We aim to find the length of the CD from the given diagram.
From the diagram, we can see that the chords have two parts. The chord AB has two parts they are AP and PB where P is its point of intersection with the other chord. And the chord CD has two parts they are CP and PD where P is the point of intersection with the other chord.
Then by the properties of chords, we have that the product of parts of the chords intersecting at a point inside the circle will be the same. Using this statement, we get that \[AP \times PB = CP \times PD\].
We already know the values of AB and PA from this we can find the value of PB.
From the diagram, we can see that \[AB = AP + PB\].
Therefore \[7 = 4 + PB\] from this, we get that \[PB = 3cm\].

Now let us substitute the values we have in the expression \[AP \times PB = CP \times PD\].
On substituting the values, we get
\[ \Rightarrow 4 \times 3 = 6 \times CP\]
Let’s simplify these to get the value of CP. Taking \[6\] to the other side we get
\[ \Rightarrow \dfrac{{4 \times 3}}{6} = CP\]
On further simplification we get
\[ \Rightarrow CP = 2\]
Thus, we got that the length of CP is \[2cm\] .

Note: A diameter is also a line with its endpoints on the circle thus, the largest chord of any circle is its diameter. In the given question the two chords meet at a point inside the circle so we have used the statement (given above) to solve the problem. This statement also applies to the chords that meet at a point outside the circle.