A right triangle has legs of lengths 9 cm and 12 cm. calculate the length of the hypotenuse?
2 answers:
Let the legs of the triangle be a and b, and the hypotenuse c.
Your first instinct might tell you to use the Pythagorean theorem to go about solving this because
. This works, but it is slow.
The fastest way to solve this is to recognize that the right triangle is a special triangle where the ratio of the sides are 3:4:5. This means that if the legs are 9 and 12, then the hypotenuse is 15 because 3*3 is 9, 3*4 is 12, and 3*5 is 15.
Your first instinct might tell you to use the Pythagorean theorem to go about solving this because
The fastest way to solve this is to recognize that the right triangle is a special triangle where the ratio of the sides are 3:4:5. This means that if the legs are 9 and 12, then the hypotenuse is 15 because 3*3 is 9, 3*4 is 12, and 3*5 is 15.
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Answer:
A), B) and D) are true
Step-by-step explanation:
A) We can prove it as follows:
B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then
.
C) Consider . This set is orthogonal because
, but S is not orthonormal because the norm of (0,2) is 2≠1.
D) Let A be an orthogonal matrix in . Then the columns of A form an orthonormal set. We have that
. To see this, note than the component
of the product
is the dot product of the i-th row of
and the jth row of
. But the i-th row of
is equal to the i-th column of
. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then
E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.
In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set and suppose that there are coefficients a_i such that
. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then
then
.