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# Lots of questions

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1. A triangle has side lengths of  $$10, 24,$$ and $$26.$$ Let $$a$$ be the area of the circumcircle. Let $$b$$ be the area of the incircle. Compute $$a-b.$$

2. In triangle $$ABC, AB =BC=17$$ and $$AC =16.$$ Find the inradius of triangle $$ABC.$$

3. In triangle $$ABC,M$$ is the midpoint of $$\overline{AB}$$. Let $$D$$ be the point on $$\overline{BC}$$ such that $$\overline{AD}$$ bisects $$\angle{BAC},$$ and let the perpendicular bisector of $$\overline{AB}$$ intersect $$\overline{AD}$$ at $$E.$$ If $$AB=44$$ and $$ME=12,$$ then find the distance from $$E$$ to line $$AC.$$

Feb 22, 2020

#1
+21586
+1

I'll try the first one.

Since the triangle has side lengths of 10, 24, and 26, this triangle is a right triangle (102 + 242 = 262).

The center of the circumcenter of a right triangle is the midpoint of the hypotenuse.

Since the hypotenuse has a length of 26, the distance from this point to any of the vertices is 13.

Therefore, the area of the circumcircle is:  pi·132  =  169·pi.

The center of the incircle is the point of intersection of the angle bisectors.

Call the vertex between the 10 and 24 sides = C (it is a right angle).

Call the vertex between the 24 and 26 sides = A.

Call the center of the incircle X.

Draw the perpendicular from X to side AC; call this point Y; this is a radius of the incircle;

call its value 'h'.

Angle C = 90o; therefore, angle(XCY) = 45o.

I will need to find the size of angle(XAY).

I will use the sin(A) in triangle ABC) and find one-half of that value.

sin(A) = 10/26   --->   A  =  sin-1(10/26)  =  22.61986o   --->   angle(XAY)  =  11.31o.

Now, to find the value of h:

Call the distance from C to Y 'x'; therefore, the distance from A to Y is '24 - x'.

In triangle(XCY), tan(45o)  =  h/x   --->   h  =  x·tan(45o).

In triangle (XAY), tan(11.31o)  =  h/(24 - x)   --->  h  =  (24 - x)·tan(11.31o).

Setting these two equation equal to each other:  x·tan(45o)  =  (24 - x)·tan(11.31o)

--->   x·tan(45o)  = 24·tan(11.31o) - x·tan(11.31o)

--->   x·tan(45o) + x·tan(11.31o)  =  24·tan(11.31o)

--->   x·( tan(45o) + tan(11.31o) )  =  24·tan(11.31o)

--->   x  =  24·tan(11.31o) / [ tan(45o) + tan(11.31o) ]

--->   x  =  4

Therefore, the area of the incircle is:  pi·42  =  16·pi.

169·pi  -  16·pi  =  153·pi

Feb 22, 2020
#2
+21586
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#2:  Let M be the midpoint of side AC. Draw BM.

Since triangle ABC is an isosceles triangle BM is the angle bisector of angle(B) and also the altitude from B to side AC.

This makes angle(AMB) a right angle.

Since AC = 16, AM = 8.

Using the Pythagorean Theorem on triangle(AMB), we can find that BM = 15.

In right triangle(AMB), cos(A)  =  8/17   --->   angle(A)  =  cos-1(8/17)  =  61.9275o.

Draw the bisector of angle(A). Call the point where this angle bisector intersects BM point 'X'.

Angle(MAX) = ½·angle(A)  =  ½·61.9275o  =  30.96o.

To find XM of triangle(MAX)  --->   tan(30.96o)  =  XM/8   ---XM  =  8·tan(30.96o)  =  4.8.

The radius of the incircle is  4.8.

Feb 22, 2020
#3
-1

3. By similar triangles, the distance from E to line AC is 12^2/44 = 36/11.

Feb 23, 2020