Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

Solve the question in the picture and write the correct answer​

Solve The Question In The Picture And Write The Correct Answer class=

Sagot :

Answer:

  36.24 cm²

Step-by-step explanation:

The shaded region is the inscribed circle in the isosceles triangle. In order to find its area, we need to know the radius of the circle.

__

One way to find the radius of the circle is to make use of the Angle Bisector theorem. It tells us an angle bisector divides the sides of a triangle proportionately. In the attached figure, that means angle bisector CD divides segment AE in proportion to sides CA and CE.

Side Length

To make use of this theorem, we need to know the length of side CA. That is the hypotenuse of right triangle AEC, so can be found using the Pythagorean theorem. Isosceles triangle ABC is given as having height AE=10 and base BC = 12.

  CA² = CE² +AE²

  CA² = 6² +10² = 136 . . . . . . . CE is half the base length: 12/2 = 6

  CA = √136 = 2√34

__

Radius

Now we can write the proportion using the angle bisector theorem.

  DE/CE = DA/CA

  r/6 = (10 -r)/(2√34)

  r√34 = 3(10 -r) . . . . . . . multiply by 6√34

  r(3 +√34) = 30

  r = 30/(3 +√34)

__

Area

From this, we can find the area of the shaded circle to be ...

  A = πr² = (3.14)(30/(3 +√34))²

  = 3.14·900/(43 +6√34) ≈ 36.2374 . . . cm²

The area of the shaded region is about 36.24 cm².

_____

Additional comment

Using area considerations, the formula for the radius of the inscribed circle in an isosceles triangle with base 'b' and sides 'a' can be found to be ...

  [tex]r=\dfrac{b}{2}\sqrt{\dfrac{2a-b}{2a+b}}[/tex]

For a=√136 and b=12, this becomes ...

  [tex]r=\dfrac{12}{2}\sqrt{\dfrac{2\sqrt{136}-12}{2\sqrt{136}+12}}=6\sqrt{\dfrac{(\sqrt{34}-3)^2}{34-9}}=1.2(\sqrt{34}-3)\approx3.39714[/tex]

This is the same value we found above.

View image sqdancefan