In the figure below, what value of x makes G the incenter of JKL?

x=8.

Step-by-step explanation:

For triangle GTL

h²=p²+b²

or, 13²=p²+12²

or, 169=p²+144

or, p²= 25

or, p=5

so, GT = 5

now

for triangles GTL and GTJ

<GTL=<GTJ(both are 90° because GT is perpendicular to JL)

<JGT=<LGT=90° (GT is perpendicular to JL so it bisects G)

GT=GT(common)

so

triangles GTL and GTJ is congruent by A.S.A axiom

JT=12(correspong side of congruent ttriangle are equal)

now

JG= 13(corresponding side of congruent triangles are equal)

now

for triangle JGT, it is a right angled triangle as <JGT = 90°)

Let <JGT be G.

tanG=p/b

or, tanG=12/5

G= 67.4°

<JGT=67.4°

now

<JGR=<JGT=67.4°(JG is at midpoint bisecting <RGT)

in JGR,

<JRG=90°(RG is perpendicular to JK as <GRK is 90°)

now

JGR is a right angled triangle so

cosG=b/h

or, cos67.4=(x-3)/13

or, cos67.4(13)=x-3

or, 4.99=x-3

or, 5 = x-3

so, x = 5+3

so, x = 8

circle area = pi * radius^2

if diameter = 10 then radius = 5

circle area = pi * 5^2

circle area = pi * 25

circle area = 78.5398163397 square units

step-by-step explanation: