Question

find the missing length indicated. 64 9 12 100

Explanation:

Step1: Recall the geometric mean theorem (altitude-on-hypotenuse theorem) for a right triangle. The theorem states that in a right triangle, the altitude drawn to the hypotenuse is the geometric mean between the segments into which it divides the hypotenuse. Also, each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. In formula terms, if we have a right triangle with altitude \( h \) to the hypotenuse, dividing the hypotenuse into segments \( a \) and \( b \), then \( h^2 = a \times b \)? Wait, no, actually, the correct formula from the geometric mean theorem (also known as the leg - geometric mean theorem) is that if we have a right triangle, and we draw an altitude from the right angle to the hypotenuse, let the length of the altitude be \( h \), the length of one segment of the hypotenuse be \( x \) and the other be \( y \), and the length of the leg adjacent to segment \( x \) be \( l_1 \) and the leg adjacent to segment \( y \) be \( l_2 \), then \( l_1^2=x\times(x + y)\), \( l_2^2=y\times(x + y)\) and \( h^2=x\times y \). Wait, in our case, the triangle is a right triangle (since there is a right angle at the top), and we draw an altitude of length 12 to the hypotenuse, which is divided into segments \( x \) and 16. Also, the leg adjacent to the segment of length 16 has length equal to the hypotenuse of the smaller right triangle with legs 12 and 16? Wait, no, actually, the correct application here is that in a right triangle, when an altitude is drawn from the right angle to the hypotenuse, then \( (\text{altitude})^2=\text{segment}_1\times\text{segment}_2 \)? No, that's not correct. Wait, the correct formula is that each leg is the geometric mean of the hypotenuse and the adjacent segment. Wait, let's denote the right triangle as \( \triangle ABC \) with right angle at \( C \), and altitude \( CD \) to hypotenuse \( AB \), where \( D \) is on \( AB \), \( AD=x \), \( DB = 16 \), and \( CD = 12 \). Then, by the geometric mean theorem, \( CD^2=AD\times DB \)? No, that's not right. Wait, actually, \( AC^2=AD\times AB \) and \( BC^2=BD\times AB \), and \( CD^2=AD\times BD \). Wait, yes! The altitude to the hypotenuse of a right triangle is the geometric mean of the segments into which it divides the hypotenuse. So \( CD^2=AD\times BD \). Wait, no, in our case, the triangle is a right triangle, and the altitude is 12, one segment of the hypotenuse is \( x \) and the other is 16. Wait, no, maybe I got the triangle wrong. Wait, the triangle in the diagram: there is a right angle at the top, so the triangle is a right triangle, and we drop an altitude from the right angle to the hypotenuse, creating two smaller right triangles, each similar to the original triangle and to each other. So, let's call the original right triangle \( \triangle PQR \), right - angled at \( Q \), and altitude \( QS \) to hypotenuse \( PR \), with \( PS=x \), \( SR = 16 \), and \( QS=12 \). Then, \( \triangle QSR \sim\triangle PQS\sim\triangle PQR \). So, from the similarity of \( \triangle QSR \) and \( \triangle PQS \), we have \( \frac{QS}{PS}=\frac{SR}{QS} \), which gives \( QS^2=PS\times SR \). Wait, that's the correct formula! So \( 12^2=x\times16 \)? No, wait, no. Wait, if \( \triangle QSR \) and \( \triangle PQS \) are similar, then the ratio of corresponding sides is equal. So \( \frac{QS}{PS}=\frac{SR}{QS} \), so cross - multiplying, we get \( QS^2 = PS\times SR \). Wait, but in our case, is the leg adjacent to \( x \) equal to 12? No, the altitude is 12. Wait, maybe I mixed up the segments. Wait, let's re - express: Let the right triangle have legs \( a \) and \( b \), hypotenuse \( c \), and altitude \( h \) to hypotenuse, dividing hypotenuse into \( m \) and \( n \) ( \( m + n=c \) ). Then we have three relationships: \( h^2=m\times n \), \( a^2=m\times c \), \( b^2=n\times c \). In our problem, the altitude \( h = 12 \), one segment \( n = 16 \), and we need to find the other segment \( m=x \). Wait, but according to \( h^2=m\times n \), we would have \( 12^2=x\times16 \), but that would give \( x=\frac{144}{16}=9 \), but 9 is one of the options. Wait, but let's check again. Wait, maybe the triangle is such that the leg (not the altitude) is related. Wait, no, the diagram shows a right triangle at the top, with an altitude of 12 to the base (hypotenuse), dividing the base into \( x \) and 16. So, using the geometric mean theorem (altitude - on - hypotenuse theorem), the length of the altitude squared is equal to the product of the lengths of the two segments of the hypotenuse. Wait, no, that's not the correct statement. The correct statement is that in a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which it divides the hypotenuse. So \( h^2 = x\times16 \), where \( h = 12 \). So \( 12^2=x\times16 \), so \( 144 = 16x \), then \( x=\frac{144}{16}=9 \). Wait, but 9 is one of the options (the purple box). Wait, but let's confirm. Let's consider the two smaller right triangles. The triangle with legs 12 and 16 is similar to the triangle with legs \( x \) and 12. So the ratio of corresponding sides should be equal. So \( \frac{12}{x}=\frac{16}{12} \), cross - multiplying gives \( 12\times12 = 16\times x \), so \( 144 = 16x \), so \( x=\frac{144}{16}=9 \).

Step1: Apply the geometric mean theorem (leg - geometric mean or altitude - on - hypotenuse theorem)

In a right triangle, when an altitude is drawn from the right angle to the hypotenuse, the square of the length of the altitude is equal to the product of the lengths of the two segments of the hypotenuse. Wait, no, the correct formula from the similarity of triangles is: If we have a right triangle \( \triangle ABC \) with right angle at \( C \), and altitude \( CD \) to hypotenuse \( AB \) (where \( D \) is on \( AB \)), then \( \triangle ACD\sim\triangle CBD\sim\triangle ABC \). From \( \triangle ACD\sim\triangle CBD \), we have \( \frac{AD}{CD}=\frac{CD}{BD} \), which implies \( CD^{2}=AD\times BD \)

Let \( AD = x \), \( BD = 16 \), and \( CD = 12 \)

Substitute the values into the formula: \( 12^{2}=x\times16 \)

Step2: Solve for \( x \)

We have the equation \( 144 = 16x \)

To solve for \( x \), divide both sides of the equation by 16: \( x=\frac{144}{16}=9 \)

Answer:

9