Finding the Area of a Similar Right Triangle

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To find the area of a larger right triangle that is similar to a smaller one, the relationship between their sides is crucial. If the hypotenuse of the larger triangle is twice as long, then all corresponding sides are also twice as long, maintaining the proportion. The area formula for a triangle is A = 1/2(bh), and for the larger triangle, it becomes A = 1/2(2b)(2h), resulting in an area four times greater than the smaller triangle. This demonstrates that the area of similar triangles scales with the square of the ratio of their corresponding sides. Understanding these proportional relationships is key to solving area problems involving similar triangles.
grace77
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For example: if it was given that two right triangles are similar triangles and that the hypotenuse of one is twice as long than the other how would you find the area of the triangle with the twice as long hypotenuse given the area of the other?
Similar right triangles means they are the same corresponding angle? And area of a triangle is 1/2(bxh)
Would really appreciate it if someone could point me in the right direction. Thanks
 
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Similar means the angles are congruent between the two figures, and the corresponding sides are in proportion.Imagine two similar right triangles and the sides of the bigger one are each k times the sides of the smaller one.

a=(1/2)bh, and A=(1/2)(kb)(kh).
How do a and A compare?
 
symbolipoint said:
Similar means the angles are congruent between the two figures, and the corresponding sides are in proportion.Imagine two similar right triangles and the sides of the bigger one are each k times the sides of the smaller one.

a=(1/2)bh, and A=(1/2)(kb)(kh).
How do a and A compare?
The base and height would be double due to the ratio right? Therefore the area would be 1/2(2b x2h)
 
grace77 said:
The base and height would be double due to the ratio right? Therefore the area would be 1/2(2b x2h)

For your example, yes. My discussion is general. Your example uses, "twice the lengths" but my generalization uses "k times the lengths".
 

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