Answer: 211 in²
Step-by-step explanation:
To find the area of the shaded region, you want to find the area of the rectangle, then subtract the area of the circle and the triangle. First, we need to find the area of the triangle. We know that there is a circle because 2 halves make a full circle.
Area of Circle
A=πr²
A=π(6)² [exponent]
A=36π [multiply]
A=113
-------------------------------------------------------------------------------------------------------
Area of Triangle
A=bh/2
A=(18)(12)/2 [multiply]
A=216/2 [divide]
A=108
Now that we have the area of the shapes, let's find the area of the entire rectangle.
A=lw
A=(36)(12) [multiply]
A=432
With the area of the rectangle, we can subtract the areas.
432-108-113=211
The area of the shaded region is 211 in².
Answer:
The value of x in centimeters is 22.5 cm
Step-by-step explanation:
The corresponding sides of the similar triangles are proportion which means have equal ratios
∵ Δ ABC is similar to Δ FGH
∵ AB and FG are corresponding sides
∵ BC and GH are corresponding sides
∵ AC and FH are corresponding sides
→ That means the ratios between them are equal
∴
=
= ![\frac{AC}{FH}](https://tex.z-dn.net/?f=%5Cfrac%7BAC%7D%7BFH%7D)
∵ AB = 6 cm and FG = 15 cm
∵ BC = 5 cm and GH = 12.5 cm
∵ AC = 9 cm and FH = x
→ Substitute them in the rule above to find x
∴
=
= ![\frac{9}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7Bx%7D)
→ By using cross multiplication with the 1st and 3rd ratios
∵ 6 × x = 15 × 9
∴ 6x = 135
→ Divide both sides by 6 to find x
∵
= ![\frac{135}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B135%7D%7B6%7D)
∴ x = 22.5 cm
∴ The value of x in centimeters is 22.5 cm
Answer:
c) 110°
Step-by-step explanation:
Arc FP is twice the measure of the marked angle, so is 70°. If FD is supposed to be a diameter, then arc FPD is a semicircle (180°) and arc PD is 180° -70° = 110°.
Step-by-step explanation:
first of all we have to change kg into ton= 48500kg into ton =53.462ton(48500/907.168)
after that we add all the weight of 4 container = 28+45+16.8+53.462
total will be his total mass of tones he had =143.262