In the figure above, line segments AB and AC are tangent to the circle O. If the lenght of OB=1 and the lenght of OA = √10. What is the area of quadrilateral ABOC? (Figure not drawn to scale)
A. 3/2
B. √3
C. 2√2
D. 3
E. 2√3
The OA is D.
I don't have clear this PS question, I appreciate if any expert explain it for me. Thank you so much.
Hi AAPL,
Let's take a look at your question.
We know that the tangents are perpendicular to the radius at the point where the tangent is intersecting the circle. Therefore, triangles ABO and ACO are right triangles.
In Triangle ABO,
$$OB=1$$
$$OA=\sqrt{10}$$
We can find AB using Pythagoras theorem,
$$\left(OA\right)^2=\left(AB\right)^2+\left(OB\right)^2$$
$$\left(\sqrt{10}\right)^2=\left(AB\right)^2+\left(1\right)^2$$
$$10=\left(AB\right)^2+1$$
$$\left(AB\right)^2=10-1$$
$$\left(AB\right)^2=9$$
$$AB=3$$
Now, we can find the area of triangle ABO,
$$\text{Area of Triangle ABO}=\frac{1}{2}\left(Base\right)\left(Height\right)$$
$$\text{Area of Triangle ABO}=\frac{1}{2}\left(OB\right)\left(AB\right)$$
$$\text{Area of Triangle ABO}=\frac{1}{2}\left(1\right)\left(3\right)=\frac{3}{2}$$
Since Triangles ABO and ACO are congruent, therefore they have the same area. Hence,
$$\text{Area of Triangle ACO}=\frac{3}{2}$$
Area of quadrilateral ABOC = Area of triangle ABO + Area of triangle ACO
$$\text{Area of quadrilateral ABOC}=\frac{3}{2}+\frac{3}{2}=3$$
Therefore, Option
D is correct.
Hope it helps.
I am available if you'd like any follow up.
