Question

Kuis [medium]
Diketahui bangun ruang terdiri dari
tabung yg diameter alasnya 14 cm dan tingginya 63 cm, serta
kerucut yg diameter alasnya 28 cm dan garis pelukisnya 50 cm.
Tentukan volume bangun ruang gabungan itu.

Answer 1

Bangun Ruang Sisi Lengkung

[Tabung dan Kerucut]

..

[tex]\begin{aligned} Volume &= \pi × ({r_t}².t_t + {r_k}².t_k.⅓) \\&= \frac{22}{7} × (7².63 + 14².48.⅓) \\&= \frac{22}{7} × (49.63 + 196.16) \\&= \frac{22}{7} × (3087 + 3136) \\&= \frac{22}{7} × 6223 \\&= 22 × 889 \\&= \boxed{\bold{\underline{19558~cm³}}} \end{aligned}[/tex]

Digunakan

• [tex]\pi = \frac{22}{7}[/tex]
• Triple Pythagoras [tex]14,48,50[/tex]

[tex]\begin{array}{lr}\texttt{Selamat Hari Raya Idul Fitri 1444 H}\end{array}[/tex]

[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 28 - 04 - 2023}}[/tex]

Answer 2

Jawaban:

volume bangun ruang gabungan

= 19.558 cm³

Penjelasan dengan langkah-langkah:

tinggi kerucut:

= √garis pelukis² - (diameter/2)²

= √50² - (28/2)²

= √2.500 - 14²

= √2.500 - 196

= √2.304

= 48 cm

Volume kerucut:

[tex] = \dfrac{1}{3} \pi {r}^{2} t[/tex]

[tex] = \dfrac{1}{3}\pi {14}^{2}.48[/tex]

[tex] = 196.(48 \div 3)\pi[/tex]

[tex] = 196.16\pi[/tex]

[tex] = 3136\pi \: {cm}^{3} [/tex]

volume tabung:

[tex] = \pi {r}^{2} t[/tex]

[tex] = \pi { \frac{d}{2} }^{2} t[/tex]

[tex] = \pi { \frac{14}{2} }^{2} .63[/tex]

[tex] ={7}^{2}.63\pi[/tex]

[tex] = 49.63\pi[/tex]

[tex] = 3087\pi \: {cm}^{3} [/tex]

Volume gabungan:

= volume kerucut + volume tabung

[tex] = 3136\pi + 3087\pi[/tex]

[tex] = 6223\pi = 6223 \times \frac{22}{7}[/tex]

[tex] = (6223 \div 7) \times 22[/tex]

[tex] = 889 \times 22 = 19558 \: {cm}^{3} [/tex]