in triangle ABC shown below, L is the midpoint of BC, M is the midpoint of AB, and N is the midpoint of AC. Find the perimeter of trapazoid BMNC.

35

since m and n are the midpoints of ab and ac resistively

then ,bc=2mn=16

since m and l are the midpoints of ba and bc respectively

then, ac =2mn =10

then,nc =1/2 × 10 =5

similary, mb=6

then ,perimeter=nc+mb+mn+bc=5+6+8+16=35

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm