Question

Pa is a tangent to the circle with center o if BC=3cm and AC=4cm of triangle ACB similar PAO then find OA and OP/AP

Answer 1

Answer:

The answer to this question is 5/4 = 1.25

Step-by-step explanation:

Considering triangle ΔACB, As PA is the tangent to the circle having centre at O then Angle C will be 90 degrees from the point of contact.

Now

$AB^{2}  = AC^{2} + BC^{2}$

AC = 4cm

BC = 3cm

Hence,

$AB^{2}  = 4^{2} + 3^{2}$

$AB^{2}  = 16 + 9</p><p>[tex]AB^{2}  = 25 [tex]cm^{2}$

AB = $\sqrt{25}$

AB = 5 cm

As AB is the diameter of the circle with centre O.

OA = $\frac{1}{2} AB$

As given in the question triangle ACB is similar to PAO

$\frac{AC}{PA} = \frac{BC}{AO} = \frac{AB}{PO}$

$\frac{4}{PA} = \frac{3}{2.5} = \frac{5}{PO}$

$\frac{4}{PA} = \frac{5}{PO}$

Hence proved that,

$\frac{OP}{AP} = \frac{5}{4}$

Answer 2

Answer:

Answer:

The answer to this question is 5/4 = 1.25

Step-by-step explanation:

Considering triangle ΔACB, As PA is the tangent to the circle having centre at O then Angle C will be 90 degrees from the point of contact.

Now

AB^{2} = AC^{2} + BC^{2}AB

2

=AC

2

+BC

2

AC = 4cm

BC = 3cm

Hence,

AB^{2} = 4^{2} + 3^{2}AB

2

=4

2

+3

2

AB^{2} = 16 + 9 < /p > < p > [tex]AB^{2} = 25 [tex]cm^{2}AB

2

=16+9</p><p>[tex]AB

2

=25[tex]cm

2

AB = \sqrt{25}

25

AB = 5 cm

As AB is the diameter of the circle with centre O.

OA = \frac{1}{2} AB

2

1

AB

As given in the question triangle ACB is similar to PAO

\frac{AC}{PA} = \frac{BC}{AO} = \frac{AB}{PO}

PA

AC

=

AO

BC

=

PO

AB

\frac{4}{PA} = \frac{3}{2.5} = \frac{5}{PO}

PA

4

=

2.5

3

=

PO

5

\frac{4}{PA} = \frac{5}{PO}

PA

4

=

PO

5

Hence proved that,

\frac{OP}{AP} = \frac{5}{4}

AP

OP

=

4

5