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Area Unit Test Part 1


Area Unit Test Part 1

jawaban competence test 3 kelas 9 semester 1 part 2​

1. jawaban competence test 3 kelas 9 semester 1 part 2​


Jawaban:

11. a. 225

12. b. minor pains

13. c. reading the information on the carton

14. b. what happened?

15. d. how did it happen?

Penjelasan:

semoga membantu :))


2. Area of shaded part = cm2


Jawaban:

itu sebenarnya luas dari setengah lingkaran dgn jari² 14

Penjelasan dengan langkah-langkah:

jawabannya ½*πr² = ½*22/7*14*14

=½*22*2*14

=11*2*14

=22*14

=308Cm²


3. find the area of the shaded partarea of figure​


[tex] \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 + 2 \times \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 \\ \\ 154 + 154 \\ \\ 308 \: {cm}^{2} [/tex]


4. Area of square = . . . . cm2 Area of 14-circle = . . . . cm2 - Area of shaded part = . . . . cm2


Aos= 14 x 14 = 196
Aoc= phi x r x r x 1/4
Aoc= 22/7 x 14 x 14 x 1/4
Aoc = 22 x 7
Aoc= 154
Aosp= 196 - 154 = 42
Hope this helpful
Sorry if i'm wrong

5. Find the area of the shaded part​


Jawaban:

32cm²

Penjelasan dengan langkah-langkah:

Find The Area Of the Clear part :

8 x 4 : 2 =16

8 x 4 : 2 =16

16 + 16 = 32

Find The Total Area on the square :

8 x 8 = 64

Find The Area Of the Shaded part :

64 - 32 = 32cm²


6. Find the area of the shaded part of the figure.


Jawaban:

81

Penjelasan dengan langkah-langkah:

15*15=225

12*12=144

225-144=81


7. teks mp3 test 7 part a dialogs 3.1​


Jawaban:

where is the file or where is the question please give it to me so I can help u thankyou


8. Find the area of the shaded part in the figure below


4. Area = 49.2075 m²
5. Area = 63.25 m²

Penjelasan/Explanation

Number 4

The area of that figure equals to the area of the square minus the area of the semicircle.

The length of each side: s = 9 mRadius of semicircle: r = ½s

A = s² –  ½πr²
⇔ A = s² – ½π(½s)²
⇔ A = s² – ½(¼πs²)
⇔ A = s² – (1/8)πs²
⇔ A = s²[1 – (1/8)π]
⇔ A = 9²[1 – (1/8)×3.14]
⇔ A = 81(1 – 0.3925)
⇔ A = 81(0.6075)
A = 49.2075 m²

Number 5

The area of that figure equals to the sum of the area of the right-angled triangle and the area of the semicircle.

Base of triangle: b = 8 mHeight of triangle: h = 6 mRadius of semicircle: r = (½×10) m = 5 m

A = ½bh + ½πr²
⇔ A = ½(bh + πr²)
⇔ A = ½(8×6 + 3.14×5²)
⇔ A = ½(48 + 3.14×25)
⇔ A = ½(48 + 78.5)
⇔ A = ½(126.5)
A = 63.25 m²

KESIMPULAN/CONLUSION

4. Area = 49.2075 m²
5. Area = 63.25 m²


9. calculate the area of the shaded part of the shape​


Jawab:

5,375 [tex]cm^2[/tex]

Penjelasan dengan langkah-langkah:

The area of square is

5×5 = 25

The area of circle is

[tex]\pi[/tex]x2,5x2,5

3,14x2,5x2,5

19,625

The area of shaded part is 25-19,625 = 5,375 [tex]cm^2[/tex]


10. What is the answer from test toelf part a listening


Jawaban:

toelf part listening? add media pls :)

Penjelasan:


11. Find the area of the shaded part!​


[tex]luas \: segitiga = \frac{alas \times tinggi}{2}[/tex]

[tex]luas \: persegi = panjang \: \times \: lebar[/tex]

Jumlah bangun,

2 Segitiga (Alas dan Tinggi = 6 cm)

1 Persegi Panjang (Panjang = 20 cm dan Lebar = 6 cm)

Maka,

[tex]2 \times luas \: segitiga + luas \: persegi[/tex]

[tex] = 2 \times \frac{6 \times 6}{2} + 20 \times 6[/tex]

[tex] = 36 + 120[/tex]

[tex]luas = 156 \: {cm}^{2} [/tex]

Tolong jadikan jawaban terbaik ya! :)


12. find the area of the shaded part of the rug


(3,14 × 50cm ×50cm) - (2×¼×3,14×50cm×50cm)
= 7.850cm² - 3.925cm²
= 3.925cm²

i'm sorry if i made a mistake

13. find the area of the shaded parttolong dengan cara​


Cara:

Dicari dulu luas kedua segitiga.

[tex]l = \frac{axt}{2} [/tex]

L = Luas a = alas t = tinggi

[tex]l 1 = \frac{12x12}{2} = 72cm {}^{2} [/tex]

[tex]l2 = \frac{12 \times 5}{2} = 30cm {}^{2} [/tex]

angka 5 dari mana? dari 12 - 7 = 5 cm, yang merupakan tinggi segitiga kecil.

[tex]ls = 72 - 30 = 42cm {}^{2} [/tex]


14. The area of shaded part is . . . . cm2


Jawaban:

400

Penjelasan dengan langkah-langkah:

20 × 20 = 400 cm²

semoga membantu

jadikan saya tercerdas dong


15. 12. Banyak mobil baris depan padaarea pabrik mobil 21 unit dan banyakmobil pada baris di belakangnyaselalu lebih 5 unit. Jika dalam areaparbrik mobil tersebut terdapat 16baris, maka jumlah mobil yangterdapat pada area parkir pabrikadalah .....​


Jawaban:

10 mobil

Penjelasan dengan langkah-langkah:

karena 16 mobil didalam ,sedangkan ada 26 mobil.jadi 26 - 16 =10 mobil parkir diarea pabrik


16. cara mendapatkan unit area dan unit volume


Untuk mendapatkan unit area dan volume sangat bergantuk pada bentuk bangun yang akan dicari.

Pembahasan  

Pada soal ini dapat diselesaikan dengan konsep besaran, satuan dan dimensi

Jawaban:

Untuk mencari satuan area atau luas dan satuan volume bergantung pada dimensi benda, misalkan pada persegi panjang maka besar luasnya adalah

A=p*l

dan pada balok besar volumenya adalah

V=p*l*t

Pada bangun lanin, tentunya rumus yang digunakan akan berbeda

Materi lebih lanjut terkait dengan besaran satuan dan dimensi dapat dipelajari pada link berikut ini

Pelajari lebih lanjut  

1.Materi tentang besaran https://brainly.co.id/tugas/13775864

2.Materi tentang Konversi satuan https://brainly.co.id/tugas/3618501

3.Materi tentang dimensi https://brainly.co.id/tugas/11667054

4.Materi tentang besaran https://brainly.co.id/tugas/20337168

5.Materi tentang besaran dan turunan https://brainly.co.id/tugas/20619770

Detail jawaban  

Kelas: 10

Mapel: Fisika

Bab: Bab 1 - Besaran dan Satuan

Kode: 10.6.1

Kata Kunci: Besaran, satuan, besaran pokok, besaran turunan, dimensi


17. Super Minds Unit 6 Part 1. tekan pada gambar jika tidak kelihatan.


2. Thats $40 + $55 = $95
3. Thats $310
4. Thats $35 + $160 = $195
5. Thats $12 + $13 = $15

Maaf kalo salah harga.. fotonya ngeblur jadi harga harganya ga keliatan jelas

18. I did ..... Of all on part IV of the test.(badly)


i did the worst of all (dlm sepengetahuan sy, tdk ada kt worstly)

19. Happy New Year 2017What is the white part area ?( in square unit )a. 57.5b. 63.5c. 69.5d. 81.5e. 97.5B./R.,@ Wello1☺☺☺..


mungkin cara klasik ini bisa menjawab

20. Find the area of the shaded part.​


We have width 8 cm. Each of the width that has been bisect is 4 cm. Now, we need to find the smallest white triangle (with the base 4 cm) and a triangle with the base 8 cm.

[tex] \displaystyle \underset{\text{White}}{\text{Area}} = \frac{1}{2} \times 4 \times 14 = 28 [/tex]

[tex] \displaystyle \underset{\text{White}}{\text{Area}} = \frac{1}{2} \times 8 \times 14 = 56 [/tex]

Now, to find the shaded area part, we just have to subtract all of them.

[tex] \displaystyle \underset{\text{shaded part}}{\text{Area}} = 56 - 28 = 28 [/tex]


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