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1. what is the circumradius on a triangle with side lengths 29, 29, and 40

 

2.In triangle pqr, m is the midpoint of pq. Let x be the point on qr such that px bisects angle qpr and let the perpendicular bisector of  pq intersect px at y If pq=36, pr=22, and my=8  then find the area of triangle pyr

 

3.find  bc and bz

 Mar 5, 2023
 #1
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1. To find the circumradius of a triangle with side lengths 29, 29, and 40, we can use the formula:

R = abc / (4A)

where R is the circumradius, a, b, and c are the side lengths of the triangle, and A is the area of the triangle.

First, we need to find the area of the triangle. We can use Heron's formula:

s = (a + b + c) / 2

A = sqrt(s(s-a)(s-b)(s-c))

where s is the semiperimeter of the triangle.

In this case, a = b = 29 and c = 40. Therefore, s = (29 + 29 + 40) / 2 = 49.

A = sqrt(49(49-29)(49-29)(49-40)) = 420

Now we can use the formula for the circumradius:

R = abc / (4A) = (29)(29)(40) / (4(420)) = 29/6

So the circumradius of the triangle is 29/6.

2. Let's start by drawing a diagram of the triangle PQR:

```
        P
       / \
      /   \
     /     \
    /       \
   /         \
  /           \
 Q-------------R
```

Since M is the midpoint of PQ, we can label PM and MQ as 18 each (half of 36). Let X be the point on QR such that PX bisects angle QPR, as shown:

```
        P
       / \
      /   \
     /     \
    /       \
   /    X    \
  /-----------\
 Q---------Y---R
```

Since PX bisects angle QPR, we have:

angle QPX = angle RPY

Therefore, triangles QPX and RPY are similar, and we can use their side lengths to find the length of PY.

Since M is the midpoint of PQ, we have:

PM = MQ = 18

Since PX bisects angle QPR, we have:

angle QPX = angle RPY

Therefore, triangles QPX and RPY are similar, and we have:

PX / RY = QX / PY

We can solve for PY to get:

PY = (QX * RY) / PX

We can use the Pythagorean theorem to find QX and RY:

QX^2 + PX^2 = PQ^2

RY^2 + PY^2 = PR^2

Substituting for QX and RY, we get:

((PQ - PX) / 2)^2 + PX^2 = PQ^2

PY^2 + ((PR - PY) / 2)^2 = PR^2

Simplifying each equation, we get:

5PX^2 - 2PQ * PX + PQ^2 = 0

5PY^2 - 2PR * PY + PR^2 = 0

Using the quadratic formula to solve for PX and PY, we get:

PX = PQ / 5 = 36 / 5

PY = PR / 5 = 22 / 5

Now we can use the fact that MY = 8 to find the length of RY:

RY^2 + MY^2 = MR^2

RY^2 = MR^2 - MY^2

RY^2 = (PR / 2)^2 - 8^2

RY^2 = 225

RY = 15

Finally, we can use the formula for the area of a triangle:

A = (1/2) * base * height

to find the area of triangle PYR:

A = (1/2) * PY * RY = (1/2) * (22/5) * 15 = 33

So the area of triangle PYR is 33.

3. It is not clear what is meant by "find bc a". Please provide more information or clarification.

 Mar 5, 2023
 #2
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0

3. By the angle bisector theorem,  BC = 22/5 and BZ = 10/3.

 Mar 5, 2023
 #3
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0

Acute isosceles triangle.

Sides: a = 29   b = 29   c = 40

Area: T = 420
Perimeter: p = 98
Semi-perimeter: s = 49

 

Circumradius = [abc] / 4T =33,640 / 1,680 =841 / 42 = 20.0238

 Mar 5, 2023

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