Www.drfrostmaths.com



Geometry Worksheet 2 – Lengths and Areas[Source: IMC] Three equilateral triangles of unit radius are placed side by side. Determine the distance AB.AB[Source: UKMT Mentoring] ABCD is a square with side length 1. P is the midpoint of CD and Q is the intersection of AP and BD. Find the area of quadrilateral BCPQ.Find the proportion of the following shapes that are shaded.abcd[Source: SMC] The diagram shows a pattern found on a floor tile in the cathedral in Spoleto, Umbria. A circle of radius 1 surrounds four quarter circles, also of radius 1, which enclose a square. The pattern has four axes of symmetry. What is the side length of the square?One large circle and 3 smaller circles are inscribed inside a triangle. If the radius of the larger circle is 1, determine the radius of the smaller circles.[Source: SMC] The diagram shows two different semicircle inside a square with sides of length 2. The common centre of the semicircles lies on the diagonal of the square. What is the total shaded area?AπB6π3-22Cπ2D3π2-2E8π22-3504063040005In the diagram, the circle and the two semicircles have radius 1. What is the perimeter of the square?A 6+42 B 2+42+23 C 32+43 D 4+22+26 E 12[Source: SMC] [Difficult!] Three circles are contained within two parallel lines. The radii of the smallest and next biggest circles is s and t respectively. What is the radius of the big circle in terms of s and t?Geometry Worksheet 2 - ANSWERSThe horizontal distance between the two points is 5/2. Splitting a triangle into two right-angled ones and using Pythagoras, the height of the triangle is 3/2. Then by forming a triangle using these two lengths and using Pythagoras, the distance AB is 28/2=7.We can split the quadrilateral up into separate shapes:The shaded square clearly has area 14 and the top-right triangle area 18. The key is finding where the two diagonal lines intersect. Suppose we imagined the point D as the origin, and DC the x-axis. Then the line AP has equation y=1-2x and the line DB has equation y=x. They intersect when 1-2x=x, so x=13. Since P has an x value of 12, the width of the left shaded triangle is 12-13=16 and the height 12. So it has an area of 12×16×12=124.The total area therefore is 124+14+18=512.The ‘model solutions’ instead used the fact that DPQ and ABQ are similar (a justification of why the angles are the same would be required), with lengths in the ratio 2:1. This provides an alternative way of establishing that Q is a third of the way between lines DC and AB.A) 48387017653000Let the radius of the circle be 1. Then we can see using simple trigonometry that the distance from the centre of the circle to the corners of the triangle is 2. The two smaller triangles in the diagram combined make an isosceles triangle with two lengths of 2 and an angle of 120° between them. Thus the area of this is 12absinC=12×2×2×sin120=2sin60=3. The big diagram consists of 3 of these, so the total area is 33. The area of the circle is π, so the proportion is π33.B) 609600202819000We can use a similar approach. Let the radius of the circle be 1. Then a third of the triangle is one with two sides 1 and an angle of 120° between them. This has the area 12×1×1×sin120=12sin60=34. So that total area of the triangle is 334, and the proportion shaded is clearly 334π.365760-4254500C)Let the radius of the larger circle be 1. Then by Pythagoras, the radius of the smaller circle is ?. The proportion is thus:π122π×12=12D) Let the radius of the big circle be say 1. Then the distance from the centre of this circle to the corner of the square is 2. Let the radius of the smaller circle be r. Then the distance from the centre of the big circle to the corner can be expressed in terms of r as 1+r+r√2. So if 2=1+r(1+2), then r=2-11+2. Thus the proportion of the shape shaded is πr24=2-1241+22 (since the square has area 4).Since OP=1, by Pythagoras, OQ=2. Since QR is 1, OR=2-1. Thus the entire diagonal of the square will be 22-2, and hence the side length 22-12=2-2.Adding forms a suitable right-angled triangle, and since the large triangle is equilateral, ∠DEF=30°. Let r be the radius of the smaller circle.Using basic trigonometry and the triangle ABE, we find the length AE=2. But we can also find this length in terms of r. Clearly AE=1+r+DE. Using trigonometry again on the triangle DEF (using the known length DF), we find that DE=23r. Thus 2=1+r+23r, thus rearranging, r=33+2.Let the radius of the smaller semicircle be a and the radius of the larger semicircle b. Then a+b=2.We also know that the length OX=b=a2 (using the “√2 trick”) since OX is the diagonal of a square with length a. Substituting the second equation into the first, a+a2=2, so a=21+2, and b=221+2. The total area of the two semicircles is therefore π221+22+π2221+22=6π3+22=6π3-22.5193665-14097000[UKMT model solution] Let the vertices of the square be A, B, C, D and the centres of the circle and two semicircles be P, Q, R as shown. The midpoint of QR is S. By symmetry, P and S both lie on diagonal BD of square ABCD and the whole figure is symmetrical about BD.As P is distance 1 from both AD and DC, the length of DP is 2. As the circles and semicircles are mutually tangent, PQR is an equilateral triangle of side 2, so the length of PS is 3. As angles QBS and BSQ are 45° and 90° respectively, triangle SBQ is isosceles, so SB=SQ=1. Hence the length of BD is 2+3+1. Now the length of the side of the square is BD÷2 so the perimeter of the square is 4×BD÷2, that is 22×BD.So the perimeter is 222+3+1, that is 4+26+22.As always with these kinds of problems, the strategy is to add lines to form right-angled triangles, then use Pythagoras to generate some equations.Let x be the radius of the biggest circle. Then XY=x+s, and we can form similar expressions for XC and YC. Drawing a box around the triangle (formed by joining the centres of the circles) creates a number of handy right-angled triangles.We can see that AY=x-s and similarly, YD=x-t. Using Pythagoras on the triangle AXY, we find that AX=2xs. Similarly, DC=2xt by using triangle YDC. This means that XB=DC-AX=2xt-2xs.We can find BC in two different ways. Since BC=AD=AY+YD=2x-s-t, by using Pythagoras on triangle XBC we can form the equation s+t2-2xt-2xs2=2x-s-t2. Almost everything cancels out, leaving x=2st. ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download