ABC is a right triangle with C as the right triangle and sides AC=6 cm and BC= 8 cm. line segment CD is drawn perpendicular to both AC and BC at C. if CD= 12 cm, find the distance from D to midpointS of AB
CS=5, so the distance is √(25+144) = 13
To find the distance from point D to the midpoints of AB, we need to find the lengths of AB, AD, and DB.
1. Let's start by finding the length of AB, the hypotenuse of the right triangle ABC. We can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
AC^2 + BC^2 = AB^2
Substituting the given values, we have:
6^2 + 8^2 = AB^2
36 + 64 = AB^2
100 = AB^2
Taking the square root of both sides, we find:
AB = √100
AB = 10 cm
2. Next, let's find the length of AD. AD is the vertical segment from D to the midpoint of AB.
Since CD is perpendicular to AC, and AB is the hypotenuse, AD is the height of the right triangle ABC. We can use similar triangles to find AD.
Applying the theorem of similar triangles, we have:
AD / AC = CD / BC
Substituting the given values, we have:
AD / 6 = 12 / 8
(AD * 8) / 6 = 12
AD * 8 = 72
AD = 9 cm
3. Finally, let's find the length of DB. Since D is the perpendicular from C to AB, DB is the horizontal segment from D to the midpoint of AB.
Since ABC is a right triangle, and CD is perpendicular to AB, DB is the width of the right triangle ABC. We can use similar triangles again to find DB.
Applying the theorem of similar triangles, we have:
DB / BC = CD / AC
Substituting the given values, we have:
DB / 8 = 12 / 6
(DB * 6) / 8 = 12
DB * 6 = 96
DB = 16 cm
4. Now that we have the lengths of AB, AD, and DB, we can find the distance from D to the midpoints of AB.
Since the midpoint of AB divides AB into two equal segments, the distance from D to the midpoints of AB is equal to half of AD plus half of DB.
(Distance from D to midpoints of AB) = (AD / 2) + (DB / 2)
= (9 cm / 2) + (16 cm / 2)
= 4.5 cm + 8 cm
= 12.5 cm
Therefore, the distance from D to the midpoints of AB is 12.5 cm.