what is the area of triangle fgh? round your answer to the nearest tenth of a square centimeter.

April 11, 2022 Post a Comment

Grade 8 Problems and Questions on Circles with Detailed Solutions

Detailed solutions and full explanations to grade eight problems and questions on circles are presented.

The iii circles C1, C2 and C3 accept their centers O1, O2 and O3 on the line 50 and are all tangent at the same point. If the diameter of the largest circle is xx units, what is the ratio of the area of the largest circle to the area of the smallest circle?

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Solution
The diameter of circumvolve C1 is equal to xx units and therefore its radius is equal to x units. The expanse A of the largest circle C1 is equal to
A = π (ten)²
The bore of circumvolve C2 is equal to the radius of circle C1 which is ten unis. The diameter of circle C3 is equal to the radius of circumvolve C2 which is five units. The radius of circumvolve C3 is equal to 2.5. We at present calculate the area B of the smallest circle C3.
B = π (two.v)²
The ratio of A to B is given past
A / B = π (10)² / π (2.5)²
Simplify
= (10)² / (2.5)²
= (ten / 2.v)² = 4² = 16

Mrs Parkinson's garden is made up of 4 squares and two semicircles as shown beneath. Each small-scale square has an area of 4 square meters. Detect the total area of the garden.

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Solution
The garden is made up of 4 squares and 2 semi circles. The full expanse of the four squares is equal to
4 � 4 = 16 square meters Since the area of a small-scale square is equal to iv square meters, the side of a modest foursquare is equal to 2 meters. Also the radius of a semicircle is equal to length of the side of the square which ii meters. The area of the surface within the 2 semicircles is equal to the expanse inside 1 whole circle and is equal to
π (two)² = 4 π
The total area of the garden is equal to
16 + 4 π = 28.56 square meters. (with π = iii.14)
A water sprinkler can spray water at a maximum altitude of 12 one thousand in all directions. What area of the garden tin can this sprinkler irrigate? round your answer to the nearest square meter.
Solution
One full rotation of the sprinkler would irrigate an area enclosed by a circle of radius 12 m. Hence the area of the garden that the sprinkler tin can irrigate is given by
π 12² = 144 π = 452 foursquare meters
A circular garden with a diameter of 10 meters is surrounded by a walkway of width ane meter. Observe the area of the walkway (shaded function).

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Solution
The walkway is enclosed betwixt a smaller circle of radius 10 meters and a larger circle of radius eleven meters and therefore the area of the walkway is equal to the area enclosed by the larger circle minus the surface area enclosed by the smaller circumvolve and is equal to
π � 11ⁱⁱ - π � ten² = 121 π - 100 π = 21 π square meters
A circular pizza costs $19.99. What is the price of one foursquare centimeter if the diameter of the pizza is 36 cm?
Solution
The $19.99 is the full cost of the whole pizza whose area is
Pi � (36/2)² = 1017 square cm
The price of 1 square cm is equal to
19.99 / 1017 = $0.02 = ii cents per foursquare cm
How much fencing is needed for the Robinsons round blossom garden that has an expanse of five square meters?(round your answer to the nearest meter.)
Solution
The fencing volition exist put around the circular garden and therefore the length of the fencing is equal to its circumference. The radius r of the garden is found using the area
Pi � r² = 5
rⁱⁱ = 5 / Pi
r = √ (five / Pi) = 1.26 meters
The circumference of the garden is equal to
2r � Pi= 2.52 Pi = 8 meters (rounded to the nearest meter)
The length of the fencing needed is equal to 8 meters
The radius of circular disk is increased past 20%. What is the percent increase in the area of the disk?
Solution
If r is the radius of the disk, its area (before increase) is equal to
Pi r²
If r is increased by 20% information technology becomes
r + 20% r = r + (xx/100) r = four + 0.2 r = ane.two r
and the area (later increase) of the disk becomes
Pi (1.two r)² = 1.44 Pi r²
Alter in area
Modify = Area after increment - Area before increment = 1.44 Pi rⁱⁱ - Pi r^two
= Pi r² (1.44 - 1) = 0.44 Pi r^two
Percent alter in surface area
(Modify / Area before change) � 100% = (0.44 Pi r²/ Pi r²) � 100%
= 0.44 � 100% = 44%
A circular table has a dimeter of 100 inches. A circular tablecloth hangs over the table 15 inches around the table. What is the area of the tablecloth?
Solution
If the table has a bore of 100 inches and the table fabric hangs 15 inches around the table, then the diameter of the tablecloth is equal to
100 + 15 + xv = 130 inches
The expanse of the tablecloth is equal to
Pi (130 / two) = iv,225 Pi = 13,267 square inches
ABCD is a square with one vertex at the center of the circumvolve and two vertices on the circle. What is the length of AC if the expanse of the circle is 100 foursquare cm?
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Solution
Note that the length of the side of the square AB has equal length as the radius r of the circle. Since the area is given, nosotros can write
100 = π r^two
Solve for rⁱⁱ
r² = 100 / π
We now utilize Pythagora'southward theorem in triangle ABC to observe the length of AC
Air conditioning² = AB² + AC²
Both lengths of AB and BC are equal to r. Hence
ACⁱⁱ = r² + r²
= 2 r² = 2 (100 / π) = 200 / π
AC = √ (2 � 100 / π) = √ (100) � √ (two / π)
= 10 √ (2 / π) cm
The ratio of the perimeter of circle A to the perimeter of circle B is 3:1. What is the ratio of the area of circle A to the area of circle B?
Solution
Permit Ra the radius of circle A and Rb the radius of circumvolve B. The ratio of perimeter of circle A to perimeter of circle B gives
(π 2 Ra) / (π 2 Rb) = 3/1 = Ra / Rb
which gives
Ra = 3 Rb
Nosotros at present express the areas Aa and Ab of the two cricles
Aa = π Ra² = π (3 Rb)²
Ab = π Rbⁱⁱ
The ratio of the areas is given past
Aa / Ab = [ π (3 Rb)^two ] / [ π Rb² ]
= π 9 Rb^two / π Rb²
= 9
The ratio is 9:i