Difference between revisions of "Find the Area of an Isosceles Triangle"
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== Steps == | == Steps == | ||
===Finding the Area from the Side Lengths=== | ===Finding the Area from the Side Lengths=== | ||
| − | #Review the area of a parallelogram. Squares and rectangles are parallelograms, as is any four-sided shape with two sets of parallel sides. All parallelograms have a simple area formula: area equals base multiplied by the height, or '''A = bh'''.<ref>http://www.mathgoodies.com/lessons/vol1/area_triangle.html</ref> If you place the parallelogram flat on a horizontal surface, the base is the length of the side it is standing on. The height (as you would expect) is how high it is off the ground: the distance from the base to the opposite side. Always measure the height at a right (90 degree) angle to the base. | + | #Review the area of a parallelogram. Squares and rectangles are parallelograms, as is any four-sided shape with two sets of parallel sides. All parallelograms have a simple area formula: area equals base multiplied by the height, or '''A = bh'''.<ref name="rf1">http://www.mathgoodies.com/lessons/vol1/area_triangle.html</ref> If you place the parallelogram flat on a horizontal surface, the base is the length of the side it is standing on. The height (as you would expect) is how high it is off the ground: the distance from the base to the opposite side. Always measure the height at a right (90 degree) angle to the base. |
#*In squares and rectangles, the height is equal to the length of a vertical side, since these sides are at a right angle to the ground. | #*In squares and rectangles, the height is equal to the length of a vertical side, since these sides are at a right angle to the ground. | ||
#Compare triangles and parallelograms. There's a simple relationship between these two shapes. Cut any parallelogram in half along the diagonal, and it splits into two equal triangles. Similarly, if you have two identical triangles, you can always tape them together to make a parallelogram. This means that the area of any triangle can be written as '''A = ½bh''', exactly half the size of a corresponding parallelogram. | #Compare triangles and parallelograms. There's a simple relationship between these two shapes. Cut any parallelogram in half along the diagonal, and it splits into two equal triangles. Similarly, if you have two identical triangles, you can always tape them together to make a parallelogram. This means that the area of any triangle can be written as '''A = ½bh''', exactly half the size of a corresponding parallelogram. | ||
#Find the isosceles triangle's base. Now you have the formula, but what exactly do "base" and "height" mean in an isosceles triangle? The base is the easy part: just use the third, unequal side of the isosceles. | #Find the isosceles triangle's base. Now you have the formula, but what exactly do "base" and "height" mean in an isosceles triangle? The base is the easy part: just use the third, unequal side of the isosceles. | ||
#*For example, if your isosceles triangle has sides of 5 centimeters, 5 cm, and 6 cm, use 6 cm as the base. | #*For example, if your isosceles triangle has sides of 5 centimeters, 5 cm, and 6 cm, use 6 cm as the base. | ||
| − | #*If your triangle has three equal sides (equilateral), you can pick any one to be the base. An equilateral triangle is a special type of isosceles, but you can find its area the same way.<ref>http://mathworld.wolfram.com/IsoscelesTriangle.html</ref> | + | #*If your triangle has three equal sides (equilateral), you can pick any one to be the base. An equilateral triangle is a special type of isosceles, but you can find its area the same way.<ref name="rf2">http://mathworld.wolfram.com/IsoscelesTriangle.html</ref> |
#Draw a line between the base to the opposite vertex. Make sure the line hits the base at a right angle. The length of this line is the height of your triangle, so label it ''h''. Once you calculate the value of ''h'', you'll be able to find the area. | #Draw a line between the base to the opposite vertex. Make sure the line hits the base at a right angle. The length of this line is the height of your triangle, so label it ''h''. Once you calculate the value of ''h'', you'll be able to find the area. | ||
#*In an isosceles triangle, this line will always hit the base at its exact midpoint. | #*In an isosceles triangle, this line will always hit the base at its exact midpoint. | ||
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#*The other short side is the height, ''h''. | #*The other short side is the height, ''h''. | ||
#*The hypotenuse of the right triangle is one of the two equal sides of the isosceles. Let's call it ''s''. | #*The hypotenuse of the right triangle is one of the two equal sides of the isosceles. Let's call it ''s''. | ||
| − | #[[Use | + | #[[Use the Pythagorean Theorem|Set up the Pythagorean Theorem]]. Any time you know two sides of a right triangle and want to find the third, you can use the Pythagorean theorem: (side 1)<sup>2</sup> + (side 2)<sup>2</sup> = (hypotenuse)<sup>2</sup> Substitute the variables we're using for this problem to get <math>(\frac{b}{2})^2 + h^2 = s^2</math>. |
#*You probably learned the Pythagorean Theorem as <math>a^2 + b^2 = c^2</math>. Writing it as "sides" and "hypotenuse" prevents confusion with your triangle's variables. | #*You probably learned the Pythagorean Theorem as <math>a^2 + b^2 = c^2</math>. Writing it as "sides" and "hypotenuse" prevents confusion with your triangle's variables. | ||
#Solve for ''h''. Remember, the area formula uses ''b'' and ''h'', but you don't know the value of ''h'' yet. Rearrange the formula to solve for ''h'': | #Solve for ''h''. Remember, the area formula uses ''b'' and ''h'', but you don't know the value of ''h'' yet. Rearrange the formula to solve for ''h'': | ||
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===Using Trigonometry=== | ===Using Trigonometry=== | ||
| − | #Start with a side and an angle. If you know some [[Use | + | #Start with a side and an angle. If you know some [[Use Right Angled Trigonometry|trigonometry]], you can find the area of an isosceles triangle even if you don't know the length of one of its side. Here's an example problem where you only know the following:<ref name="rf2" /> |
#*The length ''s'' of the two equal sides is 10 cm. | #*The length ''s'' of the two equal sides is 10 cm. | ||
#*The angle θ between the two equal sides is 120 degrees. | #*The angle θ between the two equal sides is 120 degrees. | ||
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#*<math>A = \frac{1}{2}bh</math><br><math>= \frac{1}{2}(2x)(10cos60)</math><br><math>= (10sin60)(10cos60)</math><br><math>= 100sin(60)cos(60)</math> | #*<math>A = \frac{1}{2}bh</math><br><math>= \frac{1}{2}(2x)(10cos60)</math><br><math>= (10sin60)(10cos60)</math><br><math>= 100sin(60)cos(60)</math> | ||
#*You can enter this into a calculator (set to degrees), which gives you an answer of about 43.3 square centimeters. Alternatively, use properties of trigonometry to simplify it to A = 50sin(120º). | #*You can enter this into a calculator (set to degrees), which gives you an answer of about 43.3 square centimeters. Alternatively, use properties of trigonometry to simplify it to A = 50sin(120º). | ||
| − | #Turn this into a universal formula. Now that you know how this is solved, you can rely on the general formula without going through the full process every time. Here's what you end up with if you repeat this process without using any specific values (and simplifying using properties of trigonometry):<ref | + | #Turn this into a universal formula. Now that you know how this is solved, you can rely on the general formula without going through the full process every time. Here's what you end up with if you repeat this process without using any specific values (and simplifying using properties of trigonometry):<ref name="rf2" /> |
#*<math>A = \frac{1}{2}s^2sin\theta</math> | #*<math>A = \frac{1}{2}s^2sin\theta</math> | ||
#*s is the length of one of the two equal sides. | #*s is the length of one of the two equal sides. | ||
