In right triangle ABC, altitude CD is drawn to hypotenuse AB. If AD equals 4 and AC exceeds DB by 1, find DB.
It is from a worksheet about proportions in a right triangle.
I assume you made a diagram, or were given one.
Angle C should be 90
let BD = x
then AC = x+1
I see three similar triangles, but the important one is
ADC similar to
ACB
then AD/AC = AC/AB
4/(x+1) = (x+1)/(x+4) which turns into
x^2 - 2x - 15 = 0
(x-5)(x+3) = 0
x = 5 or x = -3, but clearly x > 0
so x = 5
BD = 5
defrgt
To find the length of DB, we can make use of the properties of similar triangles in a right triangle.
In triangle ADC, we have AD = 4.
Let's denote DB as x.
Since AC exceeds DB by 1, then AC = x + 1.
Since CD is an altitude, it divides the right triangle ABC into two smaller, similar triangles, ADC and BDC.
By the property of similar triangles, the corresponding sides of these triangles are proportional.
So, we have the following proportion:
AD/AC = CD/BC
Replacing the given values, we have:
4/(x + 1) = x/BC
Cross-multiplying, we get:
4BC = x(x + 1)
Expanding the expression:
4BC = x^2 + x
To solve for x, we need to find the value of BC.
Let's consider triangle ABC. By the Pythagorean theorem, we have:
AC^2 + BC^2 = AB^2
Substituting the given values:
(x + 1)^2 + BC^2 = AB^2
Expanding the expression:
x^2 + 2x + 1 + BC^2 = AB^2
Since AB is the hypotenuse, we know that AB is the largest side of the triangle. Thus, we can say:
AB^2 > BC^2
Or, in other words:
x^2 + 2x + 1 > BC^2
Combining these equations, we have:
x^2 + x = 4BC
x^2 + 2x + 1 > BC^2
Now, we need to find the value of BC. We can proceed by trial and error or solve it algebraically, depending on the precision required.
Let's start by finding a value of x that satisfies both equations. For example, let's try x = 2.
Plugging in x = 2 into the first equation, we get:
4(BC) = 2^2 + 2 = 8
BC = 2
Plugging in x = 2 into the second equation, we get:
2^2 + 2(2) + 1 > BC^2
4 + 4 + 1 > BC^2
9 > BC^2
BC^2 < 9
BC < 3
Since BC cannot be greater than 3, we can conclusively say that BC = 2.
Now, plugging in BC = 2 into the first equation, we get:
4(2) = x^2 + x
8 = x^2 + x
Rearranging the equation:
x^2 + x - 8 = 0
Factoring the quadratic equation:
(x - 2) (x + 4) = 0
Setting each factor equal to zero:
x - 2 = 0 or x + 4 = 0
x = 2 or x = -4
Since we are dealing with lengths, x cannot be negative. Thus, x = 2.
Therefore, DB = x = 2.
To find DB in the right triangle ABC, we can use the concept of proportions.
Let's first label the given information on the triangle:
Triangle ABC with right angle at C
AD = 4
AC exceeds DB by 1
Now, let's use the concept of similar triangles to find the value of DB:
Step 1: Notice that triangle ABC is similar to triangle ADB since they both share angle A. This means that the ratios of corresponding sides in the triangles are equal.
Step 2: Let's denote the length of DB as x. Since AC exceeds DB by 1, then AC = x + 1.
Step 3: Now, we can set up the proportion using the lengths of corresponding sides in the two triangles:
AC/AB = AD/DB
Plugging in the given values:
(x + 1)/(x + (x + 1)) = 4/x
Step 4: Simplify the equation and solve for x:
(x + 1)/(2x + 1) = 4/x
Cross-multiplying:
x(x + 1) = 4(2x + 1)
x^2 + x = 8x + 4
Rearranging the equation:
x^2 + x - 8x - 4 = 0
x^2 - 7x - 4 = 0
Step 5: Now, we can solve this quadratic equation. Factoring might not be possible, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = -7, and c = -4. Plugging in these values:
x = (-(-7) ± √((-7)^2 - 4 * 1 * (-4))) / (2 * 1)
x = (7 ± √(49 + 16)) / 2
x = (7 ± √65) / 2
Step 6: Now we have two solutions for x. We'll discard the negative value since length cannot be negative. So we have:
x = (7 + √65) / 2
Therefore, DB = (7 + √65) / 2, which is the final answer.
Summary:
Using the concept of proportions in the right triangle ABC, we set up a proportion between corresponding sides in triangle ABC and ADB. Solving this proportion equation using algebraic manipulation and the quadratic formula, we find that DB equals (7 + √65) / 2.