Area

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#15019

Areas

Square $$A=s^2$$	Rectangle $$A=L\times\;w$$	Triangle $$A=\frac{1}{2}\times \;b\times \;h$$
Circle $$A=\pi r^2$$ $$C=\pi d$$ or $$C=2\pi r$$	Annulus the annulus is the area between two circles, the ‘donut’ $$A=\pi(R^2-r^2)$$ $$R=\sqrt{\frac{A}{\pi}+r^2}$$ $$r=\sqrt{R^2-\frac{A}{\pi}}$$	Trapezium $$A=\frac{h}{2}(a+b)$$
Parallelogram $$A=b\;\times\;h$$	Rhombus $$A=\frac{1}{2}\times\; x\times\;y$$	Kite $$A=\frac{1}{2}\times\; x\times\; y$$

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#15020

A circle with radius of 6cm is inscribed in a circle as shown in the diagram. which expression represents the shaded region?
a. $$\pi\times 12^2 – 12^2$$
b. $$\pi\times 6^2 – 12^2$$
c. $$\pi\times 6^2 – 6^2$$
d. $$\pi\times 6^2 – 12\times 6$$

first we need to find the base and height of the square, using Pythagoras. Let each side be x and the hypotenuse will be 12cm (as the radius is 6cm, so the diameter will be 12 which is the hypotenuse)

x² + x² = 12²

2x² = 144

x² = 72

 

The area of the square will be: A = x × x = x²
∴ A = 72cm²

Area of the circle = π × 6²
Shaded area = circle – square

= π × 6² – 72

since 6 × 12 = 72, D is the correct answer

D

[Admin]

#18797

Find the shaded area.

Outer square area = 300 × 300

= 90 000 mm²

The four quadrants that have been cut out will form a circle with radius 100mm.
Area of circle = π × 100²

= 31 415.93 mm²

Shaded area = 90,000 – 31,415.93

= 58,584.07mm²

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#18798

If the radius of a circle is 10cm, find the diameter, area and circumference.
the diameter = 2 × 10 = 20cm	r = 10cm diameter is twice the radius
A = $$\pi\times10^2$$ = 314.16 cm2	A = $$\pi r^2$$ – remember to always use the radius not the diameter for area
C = 2 x $$\pi$$ × 10 = 62.83 cm	C = 2$$\pi$$r

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#18799

If the diameter of a circle is 15cm, find the radius, area and circumference:
r = 1/2 × 15 = 7.5cm	d = 15cm
A = $$\pi\times7.5^2$$ = 176.71 cm2	A = $$\pi r^2$$
C = $$\pi$$ × 15 = 47.12 cm	C = $$\pi$$d

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#18800

If the area of a circle is 150cm2, find the radius, diameter and circumference.
150 = $$\pi$$r2	A = $$\pi r^2$$
$$\frac{150}{\pi}= r^2$$	(divide both sides by $$\pi$$)
r2	(square root both sides)
r = √47.75 = 6.9 cm
If r = 6.9 then d = 2 × 6.9 d = 13.8cm	diameter is twice the radius
C = 2 × $$\pi$$ × 6.9 = 43.4 cm	C = 2$$\pi$$r

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#18801

If the circumference of a circle is 200cm, find the radius, diameter and area.
200 = 2$$\pi$$r	C = 2$$\pi$$r
100 = $$\pi$$r	divide both sides by 2
r = 31.83cm	divide both sides by $$\pi$$
d = 2 × 31.8 = 63.7 cm	d = 2r
A = $$\pi$$ × 31.832 = 3183 cm2	A = $$\pi$$r2

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#18802

Find the unknown parallel side if the area is 43cm2, the height is 3cm and the other parallel side is 2cm.
43 = 3/2 × (a + 2)	Area = h/2(a + b) A = 43, h = 3, b = 9
86 = 3(a + 2)	multiply both sides by 2
86 = 3a + 6	expand the brackets
80 = 3a	subtract 6 from both sides
a = 26.67 cm	divide both sides by 3

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#18803

Find the height if the area of a trapezium is 500cm2 and the parallel sides are 12cm and 9cm.
500 = h/2 × (12 + 9) 500 = h/2 × 21	Area = h/2(a + b) A = 500, a = 12, b = 9
1000 = h × 21	multiply both sides by 2
h = 47.6 cm	divide both sides by 21

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#18804

Find the area between two circles, with inside radius of 10cm and an outer radius of 15 cm.
$$A =\pi\times(15^2-10^2)$$ A = 392.7 cm2	R = 15, r = 10

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#18805

Find the size of the outer radius if the area of an annulus is 200cm2 and the inner radius is 5cm.
$$R=\sqrt{\frac{200}{\pi}+5^2}$$ R = 9.4cm	$$R=\sqrt{\frac{A}{\pi}+r^2}$$ A = 200, r = 5

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#18806

Find the size of the inner radius if the area of an annulus is 370cm2 and the outer radius is 13cm.
$$r=\sqrt{13^2-\frac{370}{\pi}}$$ r = 7.2cm	$$r=\sqrt{R^2-\frac{A}{\pi}}$$ A = 370, R = 13

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#18807

Find the area of a triangle with sides 10cm, 15cm and 17cm, correct to two decimal places.
$$S=\frac{10 + 15 + 17}{2}$$ $$=\frac{42}{2}$$ = 21	First we find s: half the perimeter
$$A=\sqrt{21(21-10)(21-15)(21-17)}$$ $$A=\sqrt{21\times11\times6\times4}$$ $$A=\sqrt{5544}$$ = 74.46cm2	Using the formula we substitute where a = 10, b = 15, c = 17 and s = 21 $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ (it doesn’t matter what order the a, b and c are in- as long as all three sides get used)
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