What is the area of this triangle ?

The area of the triangle is: (3 + 2√2) units^2

from a point outside the circle, the tangent drawn to the circle are congruent

from C to the circle measures 1 unit

from A to the circle measures x units

from B to the circle measures x units

using Pythagorean theorem, solve for x...

(1 + x)^2 + (1 + x^2) = (2x)^2

expand and solve...

2(1 + 2x + x^2) = 4x^2

2x^2 - 4x - 2 = x^2 - 2x - 1 = 0

x = 1 ± √2 [the negative solution is extraneous]

length of the legs: 2 + √2

area of triangle = (1/2)(2 + √2)^2 = (1/2)(4 + 4√2 + 2) = (3 + 2√2) units^2

qed

5 explain my answer below

Imagine a square whose vertices are, O, C, A point on CB below O (which touches the circle), and A point on AC to the left of O (which touches the circle). The length of each side of that square is the same as the radius of the circle: 1. So, by Pythagoras, the distance from O to C is sqrt(1 + 1) = sqrt(2)

Let D = a point on AB which touches the circle (in fact, it must be the midpoint of AB). So the length of OD is the radius of the circle, which is 1. So the length of CD is 1 + sqrt(2).

Triangle CD breaks up triangle ABC into two identical halves; CD is perpendicular to AB. So BDC and ADC are right-angled triangles, with AD = BD = CD = 1 + sqrt(2). So the area of BDC and ADC are each (1/2)*(1 + sqrt(2))*(1 + sqrt(2)) = 1.5 + sqrt(2)

So the area of ABC is the sum of the areas of BDC and ADC, i.e. 1.5 + sqrt(2) + 1.5 + sqrt(2) = 3 * 2*sqrt(2) =~ 5.8284271247461900976033774484194

Answer (E) is correct.