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The center of the circle is drawn in an isosceles triangle with a height ratio of 12: 5, the side is equal to 30 cm. Find the AC

The center of the circle is drawn in an isosceles triangle with a height ratio of-example-1
User Mr Mixin
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1 Answer

17 votes
17 votes

Answer:

25cm

Explanation:

**see attached diagram**

Let BH = h (height)

OH = radius r = (5/17)h ⇒ h = (17/5)r

Area ∆ABC = 1/2 × base × height = 1/2 x AC x h

substituting h = (17/5)r: = 1/2 x AC x (17/5)r

= 17/10 × AC × r

We can split the isosceles triangle into three separate triangles indicated by the red lines on the diagram attached. Because the radius always meets a tangent (points E, D and H) at a right angle, the area of each triangle will be the length of the side multiplied by the radius of the circle:

Area ∆BOC = Area ∆BOA = 1/2 × 30 × r = 15r

Area ∆AOC = 1/2 × AC × r = 1/2 × AC × r

Therefore, area ∆ABC = ∆AOC + ∆BOC + ∆BOA

=1/2 (AC)r + 15r + 15r

=1/2 (AC)r + 30r

Now we have 2 different equations for the area of the isosceles triangle ABC. Equate both equations and solve to find AC:

Area ∆ABC = Area ∆ABC

17/10 (AC)r = 1/2 (AC)r + 30r

Divide both sides by r: 17/10 (AC) = 1/2 (AC) + 30

Collect & combine like terms: 6/5 (AC) = 30

Divide by 6/5: AC = 25 cm

The center of the circle is drawn in an isosceles triangle with a height ratio of-example-1
User Rebca
by
3.0k points