if a triangle has a height of 14 inches and a base of 9 inches, what is it's area?
609578
definitely 63 square inches. The formula to find the area of a triangle is A=(1/2) x base x height. That is 0.5 times base times height. So, if you plug in the numbers, you get\n0.5 x 14 x 9 square inches\n(0.5 x 14) x 9 square inches\n(7) x 9 square inches\n63 square inches\n\nMake sure that whenever you look for the area of the triangle, the problem tells you that the numbers are the height and base.\n\nYes, 24 is correct.
323670
measure each diagonal, come up with the area of one triangle X it by 4, then work out the area of the base simple!
344325
Area of a trapezoid = (1/2)*(base1 + base2)*altitude\nHence area of the trapezoid in question = (1/2)(3+11)(8) = 56 square inches
398958
The area of the equilateral triangle is:\n\nA = 0.433 s^2\n\nA = 0.433 (10 ft)^2\n\nA = 43.3 ft^2\n\nThe volume of the pyramid is:\n\nV = 1/3 A h\n\nV = 1/3 (43.3 ft^2) (6 ft)\n\nV = 86.6 ft^3
204855
Looks to me like you are trying to figure out a "PERIMETER" rather than an "AREA".  If so, you need to ADD the lengths of all the sides together.  If you ARE looking for an area, it is simply Length x Width (for a rectangle or square), 1/2 the base x height (for a triangle), or Pi x radius squared (Pi is approximately =3.1415926) for a circle.  If the "shape" of the object you are trying to find the area of is irregular, then you must divide it up into these familiar shapes first, find EACH individual area, then add those areas together for the total area.
566178
base=7cm x height=5cm/2 equals the area.\n\narea=17.5cm squared
619584
Do you mean inside area of triangle ?\n\nArea= 1/2 (base)(height)
599733
I believe it would be one-half times the length of each diagonal.\n\nIn other words, if the diagonals are 4 & 6 the area would be:\n\nA = 1/2 * 4 * 6 = 12\n\nIf you don't know the diagonals, but you know the length of the sides, you could construct the diagonals and calculate their dimensions using the Pythagorean theorem:  c^2 = a^2 + b^2, and then solve from there.  Keep in mind that you'll have to do some extra geometric analysis to find the length of each leg of the right triangles that are created by the intersection of the diagonals before you can use the theorem.
1183842
find the are  (base x height = area) and time that by 3........   OR  ........ take each of the length of the sides and times it by 3...
572751
Lateral Area\nThe lateral area is the surface area of a 3D figure, but excluding the area of any bases. Lateral Area is often abbreviated L.A. Imagine a soup can. Now cut down the side of the can and roll it flat. What was the circumference of the base is now the length of a rectangle. To find the area of this rectangle which is the same as the lateral area, multiply this length by the width, which was the height of the can.\nLateral Area of a prism: perimeter × height \nLateral Area of a cylinder: circumference × height. \nLateral Area of a regular pyramid: ½ perimeter × slant height. \nLateral Area of a right cone: ½ perimeter × slant height. \n\n\nThe lateral area of a regular pyramid or right cone is similar to that of prisms, but since each face is a triangle (or triangle-like), there is a factor of one half. The lateral area is thus half the slant height times the perimeter. The slant height is the distance from the vertex to the edge of the base where it is halfway between the base's vertices. If the pyramid is irregular and certainly if the cone is oblique, the surface area might not be calculatable using elementary techniques (which is a fancy way to say you may need calculus). It depends on if you can obtain the altitude (slant height) of each triangular face.\n\n\nSurface Area\nThe surface area of a figure is the sum of the area of all surfaces of a figure. Surface Area is often abbreviated S.A. The surface area of a prism or cylinder is the lateral area plus the area of each base. Since the bases for a prism or cylinder are congruent, this is often expressed as twice the area of the base.\nSurface Area = Lateral Area + n × Bases\nn = 2 for prisms/cylinders; n = 1 for pyramids/cones; n = 0 for spheres. \n\n\nThe surface area of a pyramid or cone is the lateral area plus the area of the single base.\n\nThe surface area of a sphere is equal to 4r2. Analogous to the unit circle is the unit sphere. Similarly, just as there are 2 radians of angle in one revolution, there are 4 steradians of solid angle in all directions.\n\nExample: Consider a right pyramid A-BCDE with vertex A and square base BCDE of length 20" on each side and a slant height of 26". What are its lateral and surface areas?\n\nAnswer: We don't need the height for this calculation, but we will calculate it anyway to stress the difference between slant height and height. The slant height is the hypotenuse of a right triangle where the height is one leg and 20"/2 = 10" is the other leg. Thus 102 + h2 = 262 or 100 + h2 = 676. Thus h2 = 576 or h = 24". The lateral surfaces are all triangles with a base of 20" and a height (the slant height) of 26". There are four of them. Thus the lateral area is 4×½×20"×26" = 1040 in2. The base is 20" square or 400 in2. Thus the [total] surface area is 1440 in2.\n\n  Understanding surface area may be clearer if you refer back to the net associated with the object. At left is a net for a cube and at right a portion of a net for a sphere. Each of these portions of a sphere is called a gore.\n\n Now is a good time to review something learned in algebra, namely (x + y)2 = x2 + xy + xy + y2 = x2 + 2xy + y2. The diagram at the right should clarify this further, help you remember the FOIL method, as well as give a physical basis for this relationship. (Remember also, the square root of (x2 + y2) does NOT equal x + y.) Consider extending the FOIL method first into trinomials: (a + b + c)(d + e + f) = ad + ae + af + bd + be + bf + cd + ce + cf. The distributive property is another way to consider this situation. Here the box method is useful.
654733
If a pyramid is of 'n' number of sides then its base area 'A' is given by\nA = n*a^2*{cot(pi/n)}/4 where the base is a regular polygon of each side 'a'.\nIf the height is 'h' and the pyramid be right (i.e. the axis of symmetry is perpendicular to the base) then the volume of the pyramid 'V' is given by V = h*A/3.\nTherefore we can write h = 3*V/A.\nSo h = 3*V/[n*a^2*{cot(pi/n)}/4]\nor h = 12*V/[n*a^2*{cot(pi/n)}]\nThis is a relation from which height can be calculated if the volume no. of sides and length of each side of the pyramid be known.\n\nHowever, if the slant height 'L' is known then the height can be expressed as\nh = sqrt[L^2 - {0.5*a*cot(pi/n)}^2]\n\nBut, if the length of the slant surface edge 'E' is mentioned then the height is to be calculated from\nh = sqrt[E^2 - {0.5*a*cosec(pi/n)}^2]\n\nHope this helps.
182640
I'm not sure if this is what you're asking, but if...\nd = length\nd-3 = width\n\nthen the area is d(d-3), a.k.a. d^2 - 3d .\n\nHope this helps!
899919
16.72 inches squared.\n\n\nDivide an equilateral triangle into two equal right triangles (meaning, draw the height).  The right triangles will have angles of 30, 60, 90, right (an equilateral triangle has 3 60 degree angles)?\n\nThe ratio of the sides in such a right triangle (the smallest leg being a) is a:a(sq(3)):2a, where 2a is the hypotenuse and a(sq(3)) is the other leg.\n\nThe "other leg" in this case is the height.  So a, half of the length of the side of the whole equilateral triangle, is 5.382/sq(3).\n\nMultiply this by 2 to get the whole length of the side.  We have our base and height.\n\n\nThe area of an equilateral triangle is 1/2bh.  b=2(5.382)/sq(3)\nh=5.382\n\nPlug those in and you get 16.72.
375022
That's the formula for the area of a triangle.  A = area, b = length of the base, h = height.  So let's say you have a triangle that's 12 cm long at it's base, and 5 cm high, then you just plug those numbers in, and get\n\nA = 1/2 * 12 * 5 = 30 (which is measured in square cm)
326809
The area of a trapezoid is the height times the average of the bases, or h*(b1+b2)/2.  In your case, your bases are the top and bottom of your figure (parallel to the x-axis), and your height is along the y-axis.  So,\n\nh = a\nb1 = c\nb2 = b\n\nA = a(b+c)/2
975485
H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72
579191
Area of a triangle:\n\nA = 1/2 (B)(A)\nwhere B = base which could be any of the three sides\nand A = altitude (or height), which is the perpendicular distance to the base B from the opposite vertex.