Lesson Miscellaneous geometric problems
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<H2>Miscellaneous geometric problems</H2> <H3>Problem 1</H3>What is the distance between the tips of the minute hand and the hour hand of a clock at 1:35 pm, where the length of the minute hand is 14 cm and the length of the hour hand is 9 cm. <B>Solution</B> <pre> At 1:35 pm, the angle between the minute hand and vertical direction up is {{{(35/60)*2pi}}} = {{{(35/30)pi}}} = {{{(7/6)pi}}} radians (since the minute hand makes a full rotation in 60 minutes). The angle between the hour hand and vertical direction up is {{{(1/12)*2pi}}} + {{{(35/(12*60))*2pi}}} = {{{(1/6)pi}}} + {{{(7/72)pi}}} = {{{(19/72)pi}}}. (since the hour hand makes a full rotation in 12 hours = 12*60 minutes). The difference between these angles is {{{(7/6)pi}}} - {{{(19/72)pi}}} = {{{(84/72)pi}}} - {{{(19/72)pi}}} = {{{(65/72)pi}}}. Thus the angle between the two hands at 1:35 pm is {{{(65/72)pi}}} radians. It is more than the right angle {{{pi/2}}} = {{{(36/72)pi}}}, but less than the straight angle of {{{pi}}} = {{{(72/72)pi}}}. Now, we have an obtuse triangle with the sides of 14 cm and 9 cm and the angle of {{{(65/72)pi}}} radians between them. To find the distance between the tips of the hands, apply the cosine law equation d = {{{sqrt(14^2+9^2-2*14*9*cos((65/72)*3.14159265))}}} = 22.745 cm. <U>ANSWER</U>. The distance between the tips is about 22.745 cm. </pre> <H3>Problem 2</H3>Two circles with the centers at A and B intersect and share a common chord CD. The radius of circle A is 10 in, the radius of circle B is 16 in. The distance between centers is 22 in. Find CD. <B>Solution</B> <pre> {{{ drawing( 600, 400, -30, 30, -20, 20, circle(-11,0,.35),circle(11,0,.35), circle(-11,0,10),circle(11,0,16), locate(-11,2,A), locate(11,2,B), circle(-39/11,(16sqrt(21))/11,.35),circle(-39/11,-(16sqrt(21))/11,.35), locate(-39/11-0.5,(16sqrt(21))/11+2,C),locate(-39/11-0.5,-(16sqrt(21))/11-0.5,D), blue(line(-11,0, -39/11,(16sqrt(21))/11)), blue(line( 11,0, -39/11,(16sqrt(21))/11)), blue(line(-39/11,(16sqrt(21))/11,-39/11,-(16sqrt(21))/11)), graph( 600, 400, -30, 30, -20, 20, 0)) }}} Draw the radii of the circles from their centers to the intersection point C. Connect the centers by the straight line AB. You will get a triangle ABC with the sides 10 in, 16 in and 22 in. Find its area using the Heron's formula area = {{{sqrt(s*(s-10)*(s-16)*(s-22))}}}, where s = {{{(10+16+22)/2)}}} = 24 is the semi-perimeter. Substituting this number into the formula, you will get area = {{{sqrt(24*(24-10)*(24-16)*(24-22))}}} = {{{sqrt(24*14*8*2)}}} = {{{sqrt(3*8^2*7*4)}}} = = {{{8*2*sqrt(21)}}} = {{{16*sqrt(21)}}} = 73.32121112... square inches. Write the formula for the area of triangle ABC in other way, using the base AB = 22 in and the height h from point C to the base area = {{{(1/2)*22*h}}} = 11*h square inches. You will get an equation 11*h = {{{16*sqrt(21)}}}. Hence, h = {{{(16/11)*sqrt(21))}}}. This value of h is half of the length CD; so CD = {{{(32/11)*sqrt(21)}}} = 13.331 inches (rounded). At this point, the problem is just solved. <U>ANSWER</U>. The length CD is {{{(32/11)*sqrt(21)}}} = 13.331 inches (rounded). </pre> <H3>Problem 3</H3>A paper cone of circular base has the diameter of 10 cm and the height of 12 cm. It is flattened out into the sector of a circle. What is the angle of the sector? <B>Solution</B> This problem is nice. Solving it step by step, I will try to explain you why it is so. <pre> When we flatten out a cone into the sector of a circle, the radius of this circle is the slant height of the cone, while the arc of this sector is the circumference of the base circle of the cone. So, we should find the slant height and the circumference of the base circle. The slant height is the hypotenuse of the right triangle, whose legs are the radius of the base circle and the height of the cone. So, in our case, the slant height is h = {{{sqrt((10/2)^2+12^2)}}} = {{{sqrt(5^2+12^2)}}} = {{{sqrt(25+144)}}} = {{{sqrt(169)}}} = 13 cm. The circumference of the base circle is C = {{{pi*d}}} = {{{3.14159265*10}}} = 31.4159265 cm (approximately). Now the angle of the sector {{{alpha}}} is connected with the slant height and the arc length by equation C = {{{h*alpha}}}, where {{{alpha}}} is in radians. Here "h" plays the role of the radius of the arc. So, we find {{{alpha}}} = {{{C/h}}} = {{{31.4159265/13}}} = 2.416609731 radians. If you want to have it in degrees, convert radians to degrees {{{alpha}}} = {{{2.416609731*(360/(2*pi))}}} = {{{2.416609731*(360/(2*3.14159265))}}} = 138.4615 degrees (approximately). <U>ANSWER</U>. The angle of the sector is about 2.416609731 radians or 138.4615 degrees. </pre> At this point, the problem is solved in full. The problem is nice, because different conceptions of geometry play together to provide the solution. <H3>Problem 4</H3>Gibb’s Hill Lighthouse, Southampton, Bermuda, in operation since 1846, stands 117 feet high on a hill 245 feet high, so its beam of light is 362 feet above sea level. Find the distance from the top of the lighthouse to the horizon. <B>Solution</B> <pre> Make a sketch. In the sketch, you will see the circle representing the Earth, with the radius of 3,958.8 miles. Also, you will see the right-angled triangle with the hypotenuse length of R + h = 3958.8 + {{{362/5280}}} = 3958.8 + 0.06856 = 3958.87 miles and one leg of the length R = 3958.8 miles. So, you can find the other leg of this triangle, which is a tangent to the Earth surface L = {{{sqrt((R+h)^2-R^2)}}} = {{{sqrt(3958.87^2-3958.8^2)}}} = 23.5 miles (rounded). It is the distance to the horizon from the top of the lighthouse. </pre> <H3>Problem 5</H3>Regular hexagon ABCDEF has sides of length 2. The point P is the midpoint of AB. Q is the midpoint of BC and so on. Find the area of the hexagon PQRSTU. <B>Solution</B> <pre> The area of the regular hexagon ABCDEF is 6 times the area of an equilateral triangle with the side length 2. So, the area of ABCDEF is {{{6*2^2*(sqrt(3)/4)}}} = {{{6*sqrt(3)}}} square units. To get the area of the regular hexagon PQRSTU, we should subtract from the area ABCDEF the area of triangle PBQ 6 times. PBQ is an isosceles triangle with the lateral side length of 1 and the concluded angle of 120 degrees. Therefore, its area is {{{(1/2)*1*1*sin(120^o)}}} = {{{sqrt(3)/4}}}. Thus the area of PQRSTU is {{{6*sqrt(3)}}} - {{{6*(sqrt(3)/4)}}} = {{{(6-6/4)*sqrt(3)}}} = {{{4.5*sqrt(3)}}} = 7.794228634 square units. <U>ANSWER</U> </pre> <H3>Problem 6</H3>An archery target is constructed of five concentric circles such that the area of the inner circle is equal to the area of each of the four rings. If the radius of the outer circle is 12 m, find the width of the band between the second and the third circle. <B>Solution</B> <pre> We have one circle and four rings: in all, 1 + 4 = 5 shapes of equal areas. The area of the outer circle is {{{pi*12^2}}} = {{{144*pi}}}. The area of each shape is 1/5 of the total area, since the partial areas of all shapes are equal to each other. They want you find the width of the band for the ring between the third and the second circles (counting from the center). For the area of the second circle you have {{{(2/5)*144*pi}}} = {{{pi*r[2]^2}}}, where {{{r[2]}}} is the radius of the second circle. So, from this equation {{{(2/5)*144}}} = {{{r[2]^2}}}, {{{r[2]}}} = {{{sqrt((2/5)*144)}}} = {{{12*sqrt(2/5)}}} meters. For the area of the third circle you have {{{(3/5)*144*pi}}} = {{{pi*r[3]^2}}}, where {{{r[3]}}} is the radius of the third circle. So, from this equation {{{(3/5)*144}}} = {{{r[3]^2}}}, {{{r[3]}}} = {{{sqrt((3/5)*144)}}} = {{{12*sqrt(3/5)}}} meters. Now the width of the band " a " is the difference a = {{{r[3]}}} - {{{r[2]}}} = {{{12*(sqrt(3/5)-sqrt(2/5))}}} = 1.706 meters (rounded). <U>ANSWER</U> </pre> <H3>Problem 7</H3>The sum of the distance from a point P to (4,0) and (-4,0) is 9. If the abscissa of P is 1, find its ordinate. <B>Solution</B> <pre> Let A, C and P be the given points A = (-4,0), C = (4,0). Let P = (1,y) be the point, for which we want to find its coordinate 'y' such that |AP| + |CP| = 9. (1) Notice that line AC is horizontal (lies on x-axis of the (x,y) coordinate system). Draw a perpendicular PD from vertex P of triangle APC to its base AC, so point D = (1,0) is the intersection of the perpendicular with AC. Thus we have now triangle APC and two right-angled triangles ADP and CDP. Let 'a' be the length CP and 'c' be the length AP: a = |CP|, c = |AP|. (2) We have |AD| = 1 - (-4) = 5; |CD| = 4 - 1 = 3. (3) Perpendicular PD is the common leg of triangles ADP and CDP, so we can write, using Pythagorean equation |AP|^2 - |AD|^2 = |CP|^2 - |CD|^2. Substituting here from relations (2) and (3), we get this equation c^2 - 5^2 = a^2 - 3^2, which implies c^2 - a^2 = 5^2 - 3^2, c^2 - a^2 = 16. (4) So, now we have this system of equations (1) and (4) c + a = 9. (1') c^2 - a^2 = 16, (4') Now we are on a finish line to complete the solution. In equation (4'), factor left side as c^2 - a^2 = (c+a)*(c-a) and replace (c+a) by 9, based on equation (1'). Then instead the system (1'), (4') you will get the system a + c = 9. (1'') 9(c-a) = 16. (4'') Open parentheses in (4'') and multiply equation (1'') by 9 (both sides) 9c + 9a = 81. 9c - 9a = 16, Add the two last equations and get 18c = 81 + 16 = 97, c = 97/18. Subtract the two last equations and get 18a = 81 - 16 = 65, c = 65/18. Now we can find 'y' y^2 = |PD|^2 = |CP|^2 - |AD|^2 = a^2 - 3^2 = {{{(65/18)^2 - 9}}} = {{{(65^2-18^2*9)/18^2}}} = {{{1309/18^2}}}, y = {{{sqrt(1309)/18}}} = 2.010005835, or y = 2.01, approximately. <U>ANSWER</U>. y = {{{sqrt(1309)/18}}} = 2.010005835, or y = 2.01, approximately. </pre> My other additional lessons on miscellaneous Geometry problems in this site are - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Find-the-rate-of-moving-of-the-tip-of-a-shadow.lesson>Find the rate of moving of the tip of a shadow</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-radio-transmitter-accessibility-area.lesson>A radio transmitter accessibility area</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Miscellaneous-problems-on-parallelograms.lesson>Miscellaneous problems on parallelograms</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Remarkable-properties-of-triangles-into-which-diagonals-divide-a-quadrilateral.lesson>Remarkable properties of triangles into which diagonals divide a quadrilateral</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-trapezoid-divided-in-four-triangles-by-its-diagonals.lesson>A trapezoid divided in four triangles by its diagonals</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/A-problem-on-heptagon.lesson>A problem on a regular heptagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/The-area-of-a-regular-octagon.lesson>The area of a regular octagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/The-fraction-of-the-area-of-a-regular-octagon.lesson>The fraction of the area of a regular octagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Try-to-solve-this-nice-Geometry-problem.lesson>Try to solve these nice Geometry problems !</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Find-the-angle-between-sides-of-folded-triangle.lesson>Find the angle between sides of folded triangle</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-problems-on-three-spheres.lesson>A problem on three spheres</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-sphere-placed-in-an-inverted-cone.lesson>A sphere placed in an inverted cone</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/07-An-upper-level-Geometry-problem-on-special-%2815-30-135%29-triangle.lesson>An upper level Geometry problem on special (15°,30°,135°)-triangle</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-great-Math-Olympiad-level-Geometry-problem.lesson>A great Math Olympiad level Geometry problem</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Nice-geometry-problem-of-a-Math-Olympiad-level.lesson>Nice geometry problem of a Math Olympiad level</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/OVERVIEW-of-my-lessons-on-additional-misc-Geometry-problems.lesson>OVERVIEW of my additional lessons on miscellaneous advanced Geometry problems</A> To navigate over all topics/lessons of the Online Geometry Textbook use this file/link <A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.
