The Two Lines are Tangent To The Circle. GMAT Problem Solving

Question: The two lines are tangent to the circle. If AC = 10 and AB = 10√3, what is the area of the circle?

1. 100\(\pi\)
2. 150\(\pi\)
3. 200\(\pi\)
4. 250\(\pi\)
5. 300\(\pi\)

“The two lines are tangent to the circle.”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

Given:

• The two lines are tangent to the circle.
• AC = 10 and AB = 10√3

Find Out:

• The area of the circle

Approach:

• The problem has been solved using geometric diagrams for better understanding.

Now, if If AC = 10, then BC is also equals to 10.
Hence, we get the below structure.

Now, Since ABC is an isosceles triangle, the following gray line will create two right triangles.

Now, if we focus on the following blue triangle. The measurements have a lot in common with the BASE 30-60-90 special triangle.

In fact, if we take the BASE 30-60-90 special triangle and multiply all sides by 5 we see that the sides are the same as the sides of the blue triangle.

So, we can now add in the 30-degree and 60-degree angles

After we get the angles, we will now add a point for the circle's center and draw a line to the point of tangency. The two lines will create a right triangle (circle property)

We can see that the missing angle is 60 degrees. So, we will create the following right triangle (in Green)

We already know that one side has length 5√3. Since we have a 30-60-90 special triangle, we know that the hypotenuse is twice as long as the side opposite the 30-degree angle.

So, the hypotenuse must have length 10√3
In other words, the radius has length 10√3
What is the area of the circle?
Area = \(\pi\)r²
= \(\pi\)(10√3)²
= \(\pi\)(10√3)(10√3)
= 300\(\pi\)

Correct Answer: E

Approach Solution 2:

Given:

• The two lines are tangent to the circle.
• AC = 10 and AB = 10√3

Find Out:

• The area of the circle

Approach:

• The problem has been solved using geometric diagrams for better understanding.

Let us look at triangle ACD.

It is 30:60:90 triangle as sides are _:5√3:10
Now join OA, where O is the centre..
angle OAD = OAC-DAC=90-30=60... OAC is 90 as it is tangent..
So OAD becomes 30:60:90.
Radius is hypotenuse =2*AD=2*5√3=10√3
Area =\(\pi\)*(10√3)^2=300\(\pi\)

Correct Answer: E

Approach Solution 3:

In above sketch, OA=OB=OA=OB= radius
∠OAC=∠OBC=∠ADC=∠BDC=90∘∠OAC=∠OBC=∠ADC=∠BDC=90∘
AD=DB=103√2=53√AD=DB=1032=53

AD=53√AD=53 and∠ADC=90∘∠ADC=90∘ , along with AC=10AC=10,
We can say that △ADC△ADC is 30∘−60∘−90∘30∘−60∘−90∘ Triangle with ∠ACD=60∘∠ACD=60∘

Now in △△ OACOAC ,∠AOC=30∘∠AOC=30∘

∠AOB=2∗∠AOC=60∘∠AOB=2∗∠AOC=60∘

This imply that △AOB△AOB is equilateral △△ , with all sides equal to 103√103.
OA=OB=OA=OB= radius =103√=103

Area of circle=π∗103√∗103√=300π

Correct Answer: E

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