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Kuis 100÷2 Points dari Stingray:[Kalo lagi pusing mending gk usah

Berikut ini adalah pertanyaan dari Stingray pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

Kuis 100÷2 Points dari Stingray:[Kalo lagi pusing mending gk usah jawab]Jika π = 22/7 maka....



Buktikan luas yg Diarsir Adalah -465½ + 350√2 cm²


Ngasal = Report​
Kuis 100÷2 Points dari Stingray:[Kalo lagi pusing mending gk usah jawab]Jika π = 22/7 maka....Buktikan luas yg Diarsir Adalah -465½ + 350√2 cm²Ngasal = Report​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawab:

Luas yang diarsir [terbukti] =

350√2 - 465½ cm² ✅

Penjelasan dengan langkah-langkah:

Lihat gambar untuk nama-nama garis:

Luas (i) / merah =

Luas persegi luar - lingkaran luar =

(ab)² - (ab)²π/4

7² - π7²/4 =

49 - 49π/4 cm²

Luas (ii) / kuning =

Cari eh

oe = ½ab = ½·7 = 7/2

maka

eh² = 2(oe)² = 2·(7/2)² = 49/2

eh = √49/√2; eh = (7√2)/2 cm

maka

ij = (ab-eh)/2

ij = (7-(7√2)/2)/2, ij = (14-7√2)/4 cm

jk = (14-7√2)/4 · √2 = [7(2-√2)]/(2√2) = 14√2-14/4 = 7(√2-1)/2 cm

luas (ii) = 2·((lingkaran-kotak) + (tembereng))

= 2·((π(ij)² - (jk)²) + (½π(jk)² - (jk)²)

= 2·((π((14-7√2)/4)² - (7(√2-1)/2)²) + (½π(7(√2-1)/2)² - (7(√2-1)/2)²))

= 2·((π(294-196√2/16) - (49(3-2√2)/4) + (¼π(49(3-2√2)/2) - (49(3-2√2)/4)))

= 2·((π(294-196√2/16) - (49(3-2√2)/4) + (¼π(49(3-2√2)/2) - (49(3-2√2)/4))

= 2·((49π(3-2√2)/8) - (49(3-2√2)/4) + (49π(3-2√2)/8) - (49(3-2√2)/4))

= 2·((49π(3-2√2)/4) - (49(3-2√2)/2))

= (49π(3-2√2)/2) - 49(3-2√2)

= (49π(3-2√2)/2) -147+98√2 cm²

Luas (iii) / hijau =

(eh)² - [(½π(eh)² - (eh)²] =

((7√2)/2)² - [(½π((7√2)/2)² - ((7√2)/2)²] =

49/2 - (49π/4 - 49/2) =

49 - 49π/4 cm²

Luas (iv) / biru =

oL = jk = 7(√2-1)/2 cm

mn = 7(√2-1)/2 · √2 = 7(√2-1)/√2 = 7√2(√2-1)/2 =

mn = (14-7√2)/2 cm

maka

Luas (iv) = π(oL)² - (mn²) = π(7(√2-1)/2)² - ((14-7√2)/2)² =

(49π(3-2√2)/4) - (49(3-2√2)/2) cm²

Luas (v) / ungu =

om = (½mn)√2

om = (½(14-7√2)/2)√2

om = (14-7√2)/4 · √2 = [7(2-√2)]/(2√2) = 14√2-14/4 = 7(√2-1)/2 cm

op = mn - om

op = (14-7√2)/2 - 7(√2-1)/2 = 21-14√2/2 cm

Luas (v) = ½π(mn)² - (mn)² - π(op)²

Luas (v) = ½π((14-7√2)/2)² - ((14-7√2)/2)² - π(21-14√2/2)²

Luas (v) = ½π(49(3-2√2)/2) - (49(3-2√2)/2) - π(833-588√2/4)

Luas (v) = (49π(3-2√2)/4)-(49(3-2√2)/2)-(833-588√2/4)

Luas (v) = 49π(5√2-7)/2 - 49(3-2√2)/2 cm²

Luas total =

(i) + (ii) + (iii) + (iv) + (v) =

49-49\cdot \frac{\pi }{4}+49\pi \frac{3-2\sqrt{2}}{2}-147+98\sqrt{2}+49-49\cdot \frac{\pi }{4}+49\pi \frac{3-2\sqrt{2}}{4}-49\cdot \frac{3-2\sqrt{2}}{2}+49\pi \frac{5\sqrt{2}-7}{2}-49\cdot \frac{3-2\sqrt{2}}{2} =\\\\49\pi \frac{3-2\sqrt{2}}{2}-49\cdot \frac{3-2\sqrt{2}}{2}-49\cdot \frac{3-2\sqrt{2}}{2}-49\cdot \frac{\pi }{4}-49\cdot \frac{\pi }{4}+49\pi \frac{3-2\sqrt{2}}{4}+49\pi \frac{5\sqrt{2}-7}{2}+49-147+98\sqrt{2}+49 =

49\pi \frac{3-2\sqrt{2}}{2}-98\cdot \frac{3-2\sqrt{2}}{2}-98\cdot \frac{\pi }{4}+49\pi \frac{3-2\sqrt{2}}{4}+49\pi \frac{5\sqrt{2}-7}{2}+49-147+98\sqrt{2}+49

maka

\frac{49\pi \left(3-2\sqrt{2}\right)}{2}-49\left(3-2\sqrt{2}\right)-\frac{49\pi }{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{4}+\frac{49\pi \left(5\sqrt{2}-7\right)}{2}+49-147+98\sqrt{2}+49 =\\\\-\frac{49\pi }{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{4}+\frac{49\pi \left(5\sqrt{2}-7\right)}{2}-49\left(3-2\sqrt{2}\right)+49-147+98\sqrt{2}+49

sederhanakan!

\frac{49\pi \left(3-2\sqrt{2}\right)+49\pi \left(5\sqrt{2}-7\right)-49\pi }{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{4}-49\left(3-2\sqrt{2}\right)+49-147+98\sqrt{2}+49 =\\\\\frac{147\sqrt{2}\pi -245\pi }{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{4}-49\left(3-2\sqrt{2}\right)+49-147+98\sqrt{2}+49 =\\\\-49\left(3-2\sqrt{2}\right)+98\sqrt{2}+\frac{196\sqrt{2}\pi -343\pi }{4}+49-147+49 =\\\\-147+98\sqrt{2}+98\sqrt{2}+\frac{-343\pi +196\sqrt{2}\pi }{4}+49-147+49

hitung semuanya

\frac{196\sqrt{2}\pi -343\pi }{4}+196\sqrt{2}-196 =\\\\\frac{196\sqrt{2}\pi -343\pi +196\sqrt{2}\cdot \:4-196\cdot \:4}{4} =

¼(196√2π-343π+784√2-784) cm²

jika π = 22/7 maka

¼ · (196√2·(22/7)-(343·(22/7))+784√2-784) =

¼ · (616√2-1078+784√2-784) =

(1400√2-1862)/4 = 14(100√2-133)/4 =

7(100√2-133)/2 cm²

7×(100√2)/2 = 700√2/2 = 350√2 [terbukti] ✅

-7×133/2 = -931/2 = -465½ [terbukti] ✅

_____________

#Jenius - kexcvi

Jawab:Luas yang diarsir [terbukti] =350√2 - 465½ cm² ✅Penjelasan dengan langkah-langkah:Lihat gambar untuk nama-nama garis:Luas (i) / merah =Luas persegi luar - lingkaran luar =(ab)² - (ab)²π/47² - π7²/4 =49 - 49π/4 cm²Luas (ii) / kuning =Cari ehoe = ½ab = ½·7 = 7/2makaeh² = 2(oe)² = 2·(7/2)² = 49/2eh = √49/√2; eh = (7√2)/2 cmmakaij = (ab-eh)/2ij = (7-(7√2)/2)/2, ij = (14-7√2)/4 cmjk = (14-7√2)/4 · √2 = [7(2-√2)]/(2√2) = 14√2-14/4 = 7(√2-1)/2 cmluas (ii) = 2·((lingkaran-kotak) + (tembereng))= 2·((π(ij)² - (jk)²) + (½π(jk)² - (jk)²)= 2·((π((14-7√2)/4)² - (7(√2-1)/2)²) + (½π(7(√2-1)/2)² - (7(√2-1)/2)²))= 2·((π(294-196√2/16) - (49(3-2√2)/4) + (¼π(49(3-2√2)/2) - (49(3-2√2)/4)))= 2·((π(294-196√2/16) - (49(3-2√2)/4) + (¼π(49(3-2√2)/2) - (49(3-2√2)/4))= 2·((49π(3-2√2)/8) - (49(3-2√2)/4) + (49π(3-2√2)/8) - (49(3-2√2)/4))= 2·((49π(3-2√2)/4) - (49(3-2√2)/2))= (49π(3-2√2)/2) - 49(3-2√2)= (49π(3-2√2)/2) -147+98√2 cm²Luas (iii) / hijau =(eh)² - [(½π(eh)² - (eh)²] =((7√2)/2)² - [(½π((7√2)/2)² - ((7√2)/2)²] =49/2 - (49π/4 - 49/2) =49 - 49π/4 cm²Luas (iv) / biru =oL = jk = 7(√2-1)/2 cmmn = 7(√2-1)/2 · √2 = 7(√2-1)/√2 = 7√2(√2-1)/2 =mn = (14-7√2)/2 cmmakaLuas (iv) = π(oL)² - (mn²) = π(7(√2-1)/2)² - ((14-7√2)/2)² =(49π(3-2√2)/4) - (49(3-2√2)/2) cm²Luas (v) / ungu =om = (½mn)√2om = (½(14-7√2)/2)√2om = (14-7√2)/4 · √2 = [7(2-√2)]/(2√2) = 14√2-14/4 = 7(√2-1)/2 cmop = mn - omop = (14-7√2)/2 - 7(√2-1)/2 = 21-14√2/2 cmLuas (v) = ½π(mn)² - (mn)² - π(op)²Luas (v) = ½π((14-7√2)/2)² - ((14-7√2)/2)² - π(21-14√2/2)²Luas (v) = ½π(49(3-2√2)/2) - (49(3-2√2)/2) - π(833-588√2/4)Luas (v) = (49π(3-2√2)/4)-(49(3-2√2)/2)-(833-588√2/4)Luas (v) = 49π(5√2-7)/2 - 49(3-2√2)/2 cm²Luas total = (i) + (ii) + (iii) + (iv) + (v) =[tex]49-49\cdot \frac{\pi }{4}+49\pi \frac{3-2\sqrt{2}}{2}-147+98\sqrt{2}+49-49\cdot \frac{\pi }{4}+49\pi \frac{3-2\sqrt{2}}{4}-49\cdot \frac{3-2\sqrt{2}}{2}+49\pi \frac{5\sqrt{2}-7}{2}-49\cdot \frac{3-2\sqrt{2}}{2} =\\\\49\pi \frac{3-2\sqrt{2}}{2}-49\cdot \frac{3-2\sqrt{2}}{2}-49\cdot \frac{3-2\sqrt{2}}{2}-49\cdot \frac{\pi }{4}-49\cdot \frac{\pi }{4}+49\pi \frac{3-2\sqrt{2}}{4}+49\pi \frac{5\sqrt{2}-7}{2}+49-147+98\sqrt{2}+49 =[/tex][tex]49\pi \frac{3-2\sqrt{2}}{2}-98\cdot \frac{3-2\sqrt{2}}{2}-98\cdot \frac{\pi }{4}+49\pi \frac{3-2\sqrt{2}}{4}+49\pi \frac{5\sqrt{2}-7}{2}+49-147+98\sqrt{2}+49[/tex]maka[tex]\frac{49\pi \left(3-2\sqrt{2}\right)}{2}-49\left(3-2\sqrt{2}\right)-\frac{49\pi }{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{4}+\frac{49\pi \left(5\sqrt{2}-7\right)}{2}+49-147+98\sqrt{2}+49 =\\\\-\frac{49\pi }{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{4}+\frac{49\pi \left(5\sqrt{2}-7\right)}{2}-49\left(3-2\sqrt{2}\right)+49-147+98\sqrt{2}+49[/tex]sederhanakan![tex]\frac{49\pi \left(3-2\sqrt{2}\right)+49\pi \left(5\sqrt{2}-7\right)-49\pi }{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{4}-49\left(3-2\sqrt{2}\right)+49-147+98\sqrt{2}+49 =\\\\\frac{147\sqrt{2}\pi -245\pi }{2}+\frac{49\pi \left(3-2\sqrt{2}\right)}{4}-49\left(3-2\sqrt{2}\right)+49-147+98\sqrt{2}+49 =\\\\-49\left(3-2\sqrt{2}\right)+98\sqrt{2}+\frac{196\sqrt{2}\pi -343\pi }{4}+49-147+49 =\\\\-147+98\sqrt{2}+98\sqrt{2}+\frac{-343\pi +196\sqrt{2}\pi }{4}+49-147+49[/tex]hitung semuanya[tex]\frac{196\sqrt{2}\pi -343\pi }{4}+196\sqrt{2}-196 =\\\\\frac{196\sqrt{2}\pi -343\pi +196\sqrt{2}\cdot \:4-196\cdot \:4}{4} =[/tex]¼(196√2π-343π+784√2-784) cm²jika π = 22/7 maka¼ · (196√2·(22/7)-(343·(22/7))+784√2-784) =¼ · (616√2-1078+784√2-784) =(1400√2-1862)/4 = 14(100√2-133)/4 =7(100√2-133)/2 cm²7×(100√2)/2 = 700√2/2 = 350√2 [terbukti] ✅-7×133/2 = -931/2 = -465½ [terbukti] ✅_____________#Jenius - kexcvi

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Last Update: Thu, 08 Jul 21
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