J's answer to Yos's Secondary 4 E Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Recall Volume/capacity of regular shaped container = base area x height or cross section area x length.
So cross section area here
= capacity/volume ÷ length of container
= 2 000 000 cm³ ÷ 150 cm
= 13 333 ⅓ cm²
Now area of cross section = area of sector AOB
13 333 ⅓ cm² = πr²θ/360
13 333 ⅓ cm² = π(135/360)r²
r² = (13 333 ⅓ ÷ (135π/360) ) cm²
r = √(13 333 ⅓ ÷ (135π/360) ) cm
r ≈ 106.384608 cm
r = 106 cm (3s.f)
So cross section area here
= capacity/volume ÷ length of container
= 2 000 000 cm³ ÷ 150 cm
= 13 333 ⅓ cm²
Now area of cross section = area of sector AOB
13 333 ⅓ cm² = πr²θ/360
13 333 ⅓ cm² = π(135/360)r²
r² = (13 333 ⅓ ÷ (135π/360) ) cm²
r = √(13 333 ⅓ ÷ (135π/360) ) cm
r ≈ 106.384608 cm
r = 106 cm (3s.f)
Date Posted:
4 years ago
360 Tutoring Program