Find the area of the shaded region.
1. Find the area of the shaded region.
Sector area AOB :
θ/360° × πr²
= 58°/360° × π(5)²
= 29°/180° × 25π
= 145π/36
= 12.65 cm²
Missing side (AT) of triangle TAO:
tan(58°) = AT/5
AT = 5tan(58°)
AT= 8 cm
Area of triangle TAO:
1/2 × base × height
= 1/2 × AO × AT
= 1/2 × 5 × 8
= 20 cm²
Shaded region:
TAO - AOB
= 20 cm² - 12.65 cm²
= 7.35 cm²
2. find the area of shaded region
Jawaban:
The example is wrong, It doesn't make sense
1. 9750 cm2
2.
Penjelasan dengan langkah-langkah:
1.
[tex] \frac{3}{4} \times ( \frac{22}{7} \times {35}^{2} ) \\ \frac{3}{4} \times ( \frac{22}{7} \times 3850 ) \\ \frac{3}{4} \times 12100 \\ = 9750 cm {}^{2} [/tex]
2.
[tex] = 3.14 \times {12}^{2} \\ = 3.14 \times 144 \\ 452.16cm {}^{2} [/tex]
[tex] = \frac{22}{7} \times {7}^{2} \\ = \frac{22}{7} \times 49 \\ = 154cm {}^{2} [/tex]
[tex]452.16 - 164 = 288.16cm {}^{2} [/tex]
Sorry if I'm wrong
3. find the total area of the shaded region in the figure
Jawaban:
The total area of the shaded reagion in the figure = 76 cm
Penjelasan dengan langkah-langkah:
Total area of the shaded region:
[tex] = \frac{(5 + 7 + 6) \times 8}{2} + \frac{(7 + 6) \times 8}{2} - 2 \times \frac{8 \times 6}{2}[/tex]
[tex] = 72 + 52 - 48 = 76 \: {cm}^{2} [/tex]
4. 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]
5. find the area of each shaded triangle
Jawaban:
jawaban: arti dari find the area of each shaded triangle adalah:tentukan luas masing-masing segitiga yang diarsir
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SEMOGAMEMBANTUJADIKANTECERDASYA
MAKASIH<3
NOTE:NOCOPASORANGYA
#IWCHU
6. 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]
7. Find the area of the shaded triangle.
Jawaban: 77
77
Penjelasan:
(11 x 14) : 2 = 77
maaf kalau salah
Jawaban:
Ini ya jawabannya.
Semoga membantu.
8. 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]
9. The figure shows circle of radius 14 cm with one quadrant removed, touching the sides of square. Find @ the perimeter of the unshaded region, (i) the area of the unshaded region, (iii) the area of the shaded region.
di antara nilai p berikut yang memenuhi proporsi p:7 21:49 adalah
10. 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
11. area of the shaded region
AE = 48 cm
AB = 48 : 4 = 12 cm
AC = 12 + 12 = 24 cm
big circle area = 3,14 x 24 x 24 = 1808,64 cm2
4 semiircles area = 4 x 1/2 x 3,14 x 6 x 6 = 226,08 cm2
area of the shaded = 1808,64 - 226,08 = 1582,56 cm2
GOOD LUCK
12. 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
13. 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
14. Find the area of each shaded triangle.
sebenarnya saya tuh gak tau maaf
15. 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! :)
16. if the area of small square is 1cm2,then the area of shaded region is....
jika luas kotak kecil adalah 1cm2, maka area yang diarsir adalah ....
PQ = √(2² + 4²)
PQ = √(4 + 16)
PQ = √20
PQ = 2√5 cm
L.persegi = PQ² = (2√5)² = 20 cm²
L.persegi kecil = 2² = 4 cm²
L.arsir = L.persegi - L.persegi kecil
L.arsir = 20 - 4
L.arsir = 16 cm²
semoga membantu ^^
17. Find the area of the blue region
Penjelasan dengan langkah-langkah:
Area of trapezium = (10 + 14) × 12 ÷ 2 cm²
Area of trapezium = 144 cm²
Area of triangle = 14 × 12 ÷ 2 cm²
Area of triangle = 84 cm²
Area of blue region = 144 cm² - 84 cm²
Area of blue region = 60 cm²
semoga membantu ya
18. the figure is made up of a semicircle in a quadrant. find the area of the shaded region
I hope this helps...................
19. 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 = ½sA = 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 mA = 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/CONLUSION4. Area = 49.2075 m²
5. Area = 63.25 m²
20. the figure is made up of two identical quadrants and a square. find the area of the shaded region.
[tex] \frac{1}{4} [/tex] x 21 + [(21 x 21) -
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