Question

Given: △CDE m∠D=90°, DK ⊥ CE CD:DE=3:5, KE=CK+8 Find: CE

Answer 1

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 2

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

$\dfrac{CD}{DE}=\dfrac{CK}{DK}$

$\dfrac{3}{5}=\dfrac{y}{\sqrt{9x^2-y^2}}$

Squaring both sides

$\dfrac{9}{25}=\dfrac{y^2}{9x^2-y^2}$

Cross multiply and get rid of denominator

$81x^2-9y^2=25y^2$

$81x^2=34y^2$-------------(1)

In ΔCDK , Using Pythagorean theorem

$CD^2+DE^2=CE^2$

$9x^2+25x^2=(2y+8)^2$

$34x^2=(2y+8)^2$-------------(2)

Divide eq(1) and eq(2)

$\dfrac{81}{34}=\dfrac{34y^2}{(2y+8)^2}$

$81(4y^2+32y+64)=1156y^2$

$324y^2+2592y+5184=1156y^2$

$832y^2-2592y-5184=0$

Using quadratic formula solve for y and we get

$y=4.5$

CE=2y+8 = 2(4.5) + 8 = 17

Hence, The length of CE=17