1.) Triangle Abc has a right triangle at C. CD is the altitude to AB. If CD = 6cm, AD = 3cm and DB= 5x - 3cm. Find x, AB, AC and BC.
2.) In right triangle ABC, CD is the altitude to the hypotenuse AB. If CD = 3cm and DB exceeds AD by 8cm. Find AD, DB, AB, BC and AC.
3.) A man travels 16km north and the 8km west. How far is he from the starting point?
4.) Find the side of a rhombus whose diagonals are 10cm and 26cm.
5,) The length of a rectangle is 7cm more than its width. If a diagonal measures 17cm. Find the dimensions of the
7.) A right triangle has legs a and b and hypotenuse c. Prove that the altitude h to the hypotenuse is h = ab / c.
+129840 1.) Triangle ABC has a right triangle at C. CD is the altitude to AB. If CD = 6cm, AD = 3cm and DB= 5x - 3cm. Find x, AB, AC and BC.
ΔADC ~ ΔACB
AD/DC = AC/CB
AC = √[AD^2 + DC^2 [ = √[ 3^2 + 6^2] = √45 = 3√5 ...so....
AD/DC = AC/CB
3/6 = [3√5] / [6√5] so BC = 6√5 = √180
AC^2 + BC^2 = AB^2
45 + 180 = AB^2
225 = AB^2
15 = AB
AD + DB = 15
3 + DB = 15 → DB = 12
DB = 12 = 5x - 3 → 15 = 5x → x = 3
+129840 7.) A right triangle has legs a and b and hypotenuse c. Prove that the altitude h to the hypotenuse is h = ab / c.
The area = a * b / 2 = h * c / 2 → a*b = c*h → h = ab / c
+129840 5,) The length of a rectangle is 7cm more than its width. If a diagonal measures 17cm. Find the dimensions of the
W^2 + [ W + 7]^2 = 17^2
W^2 + W^2 + 14W + 49 = 289
2W^2 + 14W − 240 = 0
W^2 + 7W − 120 = 0
(W − 8) (W + 15) = 0
W − 8 = 0 → W = 8cm and L = W + 7 = 8 + 7 = 15cm
