Answer:
CE = 17
Step-by-step explanation:
Lets assume :
CK=x
KE=x+8
CE= 2x+8
CD= y
DE=5/3y
We can then use the thereom of altitudes to hypotnuse and geometric means to get the system of equation:
y= √(x)(2x+8)
5/3 y= √(2x+8)(x+8)
So the equation would be:
(5/3*(√2x^2+8x))^2= (√2x^2+24x+64)^2
25/9*(2x^2+8x)= 2x^2+24x+64
9(50/9x^2 +200/9x) =9(2x^2+24x+64)
50x^2+200x=18x^2+216x+576
32x^2-16x-576=0
16[2x^2-x-36]=0
x=-4; x=9/2
CE=2x+8
CE=2(9/2)+8=17
Answer:
The length of CE=17
Step-by-step explanation:
In △CDE, m∠D=90°, DK ⊥ CE CD:DE=3:5, KE=CK+8
Let length of CK be y
KE=CK+8 = y+8
CE=2y+8
Please see the attachment for figure.
In ΔCDK and ΔDEK
∠CKD=∠DKE (each 90°)
∠CDK=∠DEK (opposite angle of right angle triangle)
∴ ΔCDK ≈ ΔDEK by AA similarity
[tex]\dfrac{CD}{DE}=\dfrac{CK}{DK}[/tex]
[tex]\dfrac{3}{5}=\dfrac{y}{\sqrt{9x^2-y^2}}[/tex]
Squaring both sides
[tex]\dfrac{9}{25}=\dfrac{y^2}{9x^2-y^2}[/tex]
Cross multiply and get rid of denominator
[tex]81x^2-9y^2=25y^2[/tex]
[tex]81x^2=34y^2[/tex]-------------(1)
In ΔCDK , Using Pythagorean theorem
[tex]CD^2+DE^2=CE^2[/tex]
[tex]9x^2+25x^2=(2y+8)^2[/tex]
[tex]34x^2=(2y+8)^2[/tex]-------------(2)
Divide eq(1) and eq(2)
[tex]\dfrac{81}{34}=\dfrac{34y^2}{(2y+8)^2}[/tex]
[tex]81(4y^2+32y+64)=1156y^2[/tex]
[tex]324y^2+2592y+5184=1156y^2[/tex]
[tex]832y^2-2592y-5184=0[/tex]
Using quadratic formula solve for y and we get
[tex]y=4.5[/tex]
CE=2y+8 = 2(4.5) + 8 = 17
Hence, The length of CE=17
