How do you solve this? Without going into anything to complicated as this should be Year 10 maths
Answers
Given : ABCD is a square with each side 5cm .
To Find : The area of the shaded region .
Solution : On observing the figure we can see two quadrants , quadrant ADC & quadrant ABC .
If we join A to C , then it will be common for triangles ADC & ABC . And they will be congruent by SSS congruence condition.
Therefore the area of both quadrants will also be equal . Now we can find area of quadrant as ;
[tex]\large\boxed{\red{\bf Area_{quadrant}=\dfrac{\pi r^2}{4}}}[/tex]
Here radius will be equal to 5cm .
⇒ Area = πr² / 4 .
⇒ Area = π (5cm)² / 4 .
⇒ Area = 22/7 × 25 × 4 cm².
⇒ Area = 19.64 cm² .
So , total area of both quadrants = 39.28 cm² .
Also , area of square will be :
[tex]\large\boxed{\bf{\red{Area_{square}=(side)^2}}}[/tex]
⇒ Area = 5cm × 5cm .
⇒ Area = 25 cm².
Now , subtract area of one quadrant from the area of square = 25cm² - 19.64 cm² = 5.36 cm².
Similarly area of other white region = 5.36cm² .
And the areas sum will be = 5.36cm² × 2 = 10.72cm² .
Now , from the figure it's clear that ,
⇒ Area of unshaded region + Area of shaded region = 25cm².
⇒ 10.72cm² + ar( Shaded region ) = 25cm².
⇒ ar ( Shaded region ) = 25cm² - 10.72cm².
⇒ ar ( Shaded region ) = 14.28 cm².
Hence the area of shaded region is 14.28 cm².
[tex]\large\boxed{\red{\bf Answer = 14.28cm^2}}[/tex]
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[tex] \red { \orange {\boxed {\boxed{Answer}}}}[/tex]
[tex] = \frac{5}{6} \times \frac{6}{7} [/tex][tex] = \frac{5 \times 6}{6 \times 7} [/tex][tex] = \frac{30}{42} [/tex][tex] = \frac{30 \div 6}{42 \div 6} [/tex][tex] = \frac{5}{7} [/tex]#IfWrongPleaseReport
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Answers
Answer:
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Step-by-step explanation:
Given;
The ratios of the angle in the triangle = 4:5:6;
Unknown:
The angle of the triangle = ?
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We must understand that the sum of angles in a triangle is 180°
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
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Answers
Answer:
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Step-by-step explanation:
Una empresa dedicada a la venta de buzos en La Victoria, dispone para la confección de 500 m de tela algodón y 520 m de tela poliéster. Cada pantalón del buzo necesita 1 m de algodón y 2 m de poliéster; para cada casaca se necesitan 2 m de algodón y 1 m de poliéster. La ganancia por cada pantalón es de S/ 9 y la ganancia por cada casaca es de S/ 12. El fabricante sabe que sus máquinas en promedio demoran 4 minutos en confeccionar un pantalón y 10 minutos en confeccionar una casaca; él promete entregar un pedido en una semana trabajando sus máquinas de lunes a viernes 8 horas diarias.
a) Modele la función objetivo y elabore el SIL para las restricciones.
b) ¿Cuánto será el pedido máximo de buzos (casaca y pantalón) que podrá entregar en una semana de trabajo? ¿Cuánto será su ganancia máxima?
Answers
Answer:
x₁ = 60 pantalón de buzo
x₂ = 0 casaca
z(max) = 540 unidades de monedas
Step-by-step explanation:
El problema se resolverá usando un software de solución de problemas de optimización (integer linear programming solver)
z = 9*x₁ + 12*x₂ a maximizar
Sujeto a:
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1*x₁ + 2*x₂ ≤ 500
2)Segunda restricción disponibilidad de poliester 520 m de tela
2*x₁ + 1*x₂ ≤ 520
3) Tercera restricción tiempo disponible 8 h/d * 5 d/s = 40 h/s
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4)Restricciones generales x₁ ≥ 0 ; x₂ ≥ 0 ambas solo deben tomar valores enteros
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x₁ = 60
x₂ = 0
z(max) = 540