Homeschool Math Blog

OK, maybe this is not the best time of year to mention this (I realize your minds may be on other things :), but if I don't do it now, I may forget it later : ( I have made a new worksheet generator... for areas of triangles, parallelograms, trapezoids, generic quadrilaterals, pentagons, and/or hexagons (you can choose the shape(s) to be used). It uses the coordinate grid. You can select either the first quadrant or all quadrants. http://www.homeschoolmath.net/worksheets/area_triangles_polygons.php Something like this: The answer key shows the polygon drawn in the coordinate grid, and of course the area in square units. This is a focus topic typically for 6th grade math, but may also appear in 5th and 7th grades.

This is a solution to the triangle problem I posted here . The picture is below; we are to find the area of the triangle. Since AC is tangent to the circle at D and AB is tangent to the circle at E, then the distances AD and AE are equal. That is, AE must be 2. I can't remember nor find a name for this theorem. It has to do with two tangents from the same point (and it is easy to prove using congruent triangles), saying that those distances are congruent. Similarly, CD and CF are congruent -- both are 3. And also by the same reasoning, BE = BF (because BA and BC are tangets to the circle). I will call BE and BF as x, it's easier to manipulate in an equation. Now, we can solve x fairly easily by applying the Pythagorean theorem to the right triangle ABC. (x + 2) 2 + (x + 3) 2 = (2 + 3) 2 x 2 + 4x + 4 + x 2 + 6x + 9 = 25 2x 2 + 10x + 13 = 25 2x 2 + 10x − 12 = 0 x 2 + 5x − 6 = 0 x = (−5 ± √( 25 − 4(−6))) / 2 x = (−5 ± 7) / 2 x = − 6 or x = ...

I'm reposting here a nice triangle problem with the permission of Antonio Gutierrez from GoGeometry.com The figure shows a right triangle ABC. The inscribed circle and the hypotenuse are tangent at D. If AD = 2, and CD = 3, find the area of triangle ABC. The solution is now posted here !

This is the solution for the triangle problem with equal areas that I posted earlier. We are given that the areas of the three right triangles are equal, that is the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD. We will make an equation based on that fact. For that, I like to use x as my variable, so I denote the longer side of the rectangle with a, the other side with b, the distance AE with x, and the distance BF with y. We are asked the ratio AE:EB, which is the same as x : (a − x) using my notation, and the ratio BF:FC, which is the same as y : (b − y) using my notation. The area of triangle ADE is its base times altitude divided by 2, or bx/2. The area of triangle EBF is its base times altitude divided by 2, or y(a − x)/2. The area of triangle CDF is its base times altitude divided by 2, or a(b − y)/2. And these three are equal. Basically you just make two equations from the above information, and manipulate your equations until y...

I hope Pat doesn't mind that I copied the image from his blog... He posted this triangle puzzle on his blog and I thought you might enjoy it, too! Basically, we have a triangle DFE inside a rectangle, dividing the rectangle into various triangles.  And, the three areas of fainter color are equal . That is, the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD. (Notice the image is not drawn to scale at all.) And, we're asked to solve the RATIOS AE : EB and BF : FC. The solution is here .

This is a set of three geometry videos dealing with angles. First, showing that vertical angles are equal: Next, finding out about the angles in a parallelogram . I start out with two parallel lines and a transversal (line that intersects them both). We explore the angles formed, which some of them are corresponding angles, some are vertical angles. I draw a new line, and get a parallelogram. Lastly, here is a short and easy proof about the angles in a triangle .

These are answers to my earlier pos t where I asked if certain figures necessarily are parallelograms. The question was: Does the given information in each diagram guarantee that each is a parallelogram? Figure 1: This one you can't get around; it ends up being a parallelogram, actually a dandy rhombus. Let's prove it. You can notice it has lots of sides of the same length. If we draw a diagonal, we get two triangles with all kinds of same sides: The two triangles ABD and BCD end up having all three sides the same. So by the SSS triangle congruence theorem, they are congruent triangles. Hence, their corresponding angles are the same. I've marked the corresponding angles with the same colors. Actually the triangles are even isosceles so the blue and purple angles are even congruent... but we don't need that fact. To prove ABCD is a parallelogram, we need to prove its two sides are parallel. And for that, it's often handy to use the corresponding angle theorem: if c...

Someone sent in this very simple question (a student?). Leg b of the right triangle is twice as long as the base a . If the area is 36 cm squared, what is the length in of the leg b ? A little bit of algebra helps in this problem. FIRST strive to make a picture. Need a right triangle, the leg twice as long as the base. Here in my picture things aren't exactly to the scale, but it suffices for illustration purposes: So we actually know that b = 2a. The area of a triangle here is base times height over 2, and remember the height is the other leg, and it's twice the base: area = ba/2 = (2a)(a)/2 , and this is 36 (given). So we get our equation: (2a)(a)/2 = 36 a 2 = 36 a = 6. The leg b is therefore 12 cm long. check: Legs are 12 and 6, so the area is 12 * 6 / 2 = 36.

Recently there's been a lot of talk about an essay written by mathematician/teacher John Lockhart, called Lockhart's Lament . Some people praise it, some are more skeptical. Lockhart's Lament makes for good reading and he raises some really interesting points, so I can heartily recommend reading it. Personally I don't fully agree with every statement he makes there. But his MAIN point concerns mathematics as an art, and how we should teach it. I went ahead and copied a part of the essay below. This is direct quote from the essay, presenting a VERY GOOD example with the triangle problem. So let me try to explain what mathematics is, and what mathematicians do. I can hardly do better than to begin with G.H. Hardy's excellent description: A mathematician, like a painter or poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas. So mathematicians sit around making patterns of ideas. What sort of patte...

I have another neat problem for you to solve. This one should be accessible to middle schoolers on up. The sides of triangles A and B measure 5, 5, 8 and 5, 5, 6 respectively. What is the ratio of the area of triangle A to that of triangle B? Express in simplest a:b form. This problem is original to John Morse. He is a local mathematics researcher/ author/tutor/computer programmer in Delmar, NY, and has written this problem to help learners use creative and problem-solving skills in various ways. And I think this problem can indeed help in that - it can be solved in many different ways. [update - solution follows] I like this problem because there are many ways to solve it and to use it with different grade-level students - such as is already mentioned in the comments. 1) You could use this as a drawing and measuring exercise with 6th or 7th graders who have learned to do compass and ruler constructions. Once they know how to construct a triangle given its three sides (or see here ...

Trigonometry: Finding the value of sine Pi/3. First we need to remember that the whole circle is 360° and in radians it is 2Pi. So then Pi is 180°, and Pi/3 is 60°. To find sine of Pi/3, you'd want to have a right triangle with one angle 60°. Fortunately that is easy to come by; just take an equilateral triangle and draw an altitude to it. You will have two identical 30°-60°-90° triangles. And yes this is one of the special triangles - also used in drafting, and there are rulers in this shape . Where on this picture is the 60° angle? Where's the 30° angle? Now, to get sine 60° one needs side lengths. I made the sides of this equilateral triangle ABC to be 2 units. The side CD is obviously just 1 unit (easy numbers thus far!) But what about the height h? Well, that's where we need to dig up the goold ole' Pythagoras. Can't forget him. You write the equation, h 2 + 1 2 = 2 2 h 2 = 2 2 − 1 2 = 3. So taking square roots... h = √3. Then, to the sine. Remember sine i...

My child is dealing identifying Triangles as SAS,SSS,AAS,ASA and HL Theroms. She is also having trouble with the Flow proofs and Column proofs on explaing why 2 triangles are congruent. All of the theorems about proving that triangles (or other shapes) are congruent can be "translated" into a drawing problem: If I have my 'secret' triangle and I give you THESE pieces of information, can you reproduce my triangle? Can you do that every time, no matter what my triangle? We can describe a triangle using 6 pieces of information: the legths of the three sides, and the measures of the three angles. But you don't need all of those to be able to draw my secret triangle. Can you draw a copy of my triangle if I tell you that.... my triangle has a side 5 cm long, another side 6 cm long, and the angle between those sides is 29 degrees (I've given you S - A - S)? my triangle has a 30 degree angle, a 60 degree angle, and a 90 degree angle (I've given you A - A - A)? I...