tinggi segitiga
=√12²+5²
=√144 +45
=√169
=13 cm
Lp = s . s + 4(a.t)/2
=10 . 10 + 4 (10 . 13)/2
=100 + 260
=360 cm²
sisi 1
=> 10/2
=> 5 cm
tinggi∆
= √12² + 5²
= √144 + 25
= √169
= 13cm
Luas permukaan
= luas alas + jumlah luas bidang tegak
= s² + 4(½×a×t)
= 10² + 4(½×10×13)
= 100 + 4(½×130)
= 100 + 4(65)
= 100 + 260
= 360cm²
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tinggi segitiga
=√12²+5²
=√144 +45
=√169
=13 cm
Lp = s . s + 4(a.t)/2
=10 . 10 + 4 (10 . 13)/2
=100 + 260
=360 cm²
Jawaban:
sisi 1
=> 10/2
=> 5 cm
tinggi∆
= √12² + 5²
= √144 + 25
= √169
= 13cm
Luas permukaan
= luas alas + jumlah luas bidang tegak
= s² + 4(½×a×t)
= 10² + 4(½×10×13)
= 100 + 4(½×130)
= 100 + 4(65)
= 100 + 260
= 360cm²
OPSI
C. 360cm²