Question

PQ = −2y + 15 and PS = 3y + 5. Find the radius of the inscribed circle of △LMN.

Answer 1

Answer:

11

Step-by-step explanation:

I can assume that as △ LMN has an inscribed circle, PQ & PS are perpendicular bisectors? If so then PQ & PS are also the radii of this inscribed circle since it is tangent to the points on △LMN...

PQ = PS,

-2y + 15 = 3y + 5,

10 = 5y,

y = 2

If y = 2, radius length = -2(2) + 15 = -4 + 15 = 11 units

Answer 2

The radius of the inscribed circle of △LMN is; 22 units

The image of the triangle △LMN is missing and so i have attached it.

Now, from the attached triangle, we can say that PQ = PS = PR because the distance to the centroid from the midpoint of any side of a triangle are always equal.

Thus;

-2y + 15 = 3y + 5

15 - 5 = 3y + 2y

10 = 5y

y = 10/5

y = 2

Thus; PQ = -2(2) + 15

PQ = 11

Since PQ = PS = PR, then;

PR = 11

Now, the radius of the circle will be;

PL = PM = PN.

We know that from centroid theorem, the distance of the centroid to each vertex is 2/3 the length of each median.

Thus;

2PR = PL

Thus; PL = 2 × 11

PL = 22 Units

In conclusion, radius is 22 units.

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