Mohon bantuan no 29 dan 30 dengan penjelasan
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NO 29
Diketahui:
Panjang jari-jari tabung = 7 cm
Tinggi tabung = 10 cm
Tinggi kerucut = 24 cm
Ditanya:
Luas permukaan = ?
Jawab:
Luas permukaan tabung tanpa tutup
= π x r x (r + 2t)
= 22/7 x 7 x (7 + 2.10)
= 22/7 x 7 x (7 + 20)
= 22 x 27
= 594 cm²
Luas permukaan kerucut
= π x r x (r + s)
= 22/7 x 7 x √(7² +24²))
= 22/7 x 7 x √(49 + 576))
= 22/7 x 7 x (7 + √625))
= 22 x (7 + 25)
= 22 x 32
= 704 cm²
Luas permukaan = 594 cm² + 704 cm²
= 1.298 cm²
JAWABAN: A
NO 30
Diketahui:
Panjang AB = 32 cm
Panjang BC = 30 cm
Panjang CT = 25 cm
Ditanyakan:
Volume limas = ?
Jawab:
Panjang AC
Panjang AC = √32² + 30²
Panjang AC = √1.024 + 900
Panjang AC = √1.924
Panjang AC = 43,86 cm
Panjang AO = ½ x AC
= 21,93 cm
Panjang TO
TO² = TC² - AO²
TO² = 25² – 21,93²
TO² = 625 - 481
TO² = 144
TO = √144
TO = 12 cm
Volume limas
= 1/3 x p x l x t
= 1/3 x 32 x 30 x 12
= 3.840 cm³ → JAWABAN: A
Mapel: Matematika
Kelas: 9
Materi: Bangun Ruang Sisi Lengkung