# Given: cdkm is a parallelogram, da perpendicular to ck , dk – cd = 7 ca = 6, ak = 15 find: cd and

Given: cdkm is a parallelogram, da perpendicular to ck , dk – cd = 7 ca = 6, ak = 15 find: cd and dk

## This Post Has 4 Comments

1. bradleycawley02 says:

We have this parallelogram shown in the picture (just ignore the letters). Let's  DK (in the picture it is AC) with x and CD (in the picture it is AD) with x-7. Applying Pythagoras theorem, we can write that DA (in the problem, in the picture it is AE) is . Solving this equation, we find the value of x and it is 106/7. That's the value of DK. And CD=106/7-7=57/7
$Cdkm is a parallelogram, da ⊥ ck , dk – cd = 7 ca = 6, ak = 15 find: cd and dk$

2. colin5514 says:

We will use Pythagorean theorem and solve the system of equations:CD² + DA² = 6²DK² + DA² = 15²DK = CD + 7CD² + DA² = 36   / · ( - 1 )DK² + DA² = 225 CD² - DA² = - 36+  DK² + DA² = 225DK² - CD² = 189( CD + 7 )² - CD² = 189CD² + 14 CD + 49 - CD² = 18914 CD = 189 - 4914 CD = 140CD = 10
DK = 10 + 7 = 17  CD = 10,  DK = 17

3. hfleysher says:

If I understand the question correctly, then we have 2 right triangles;
ΔDAC and ΔDAK. To make our lives easier, we denote CD as x, DK as x+7, and DA as y. Since CA=6 and AK=15, we use the Pythagorean Theorem:
ΔDAC; CA²+DA²=CD²=   6²+y²=x²
ΔDAK; CA²+AK²=DK²=    15²+y²=(x+7)²
Using these two equations, we find that CD equals 10 and DK equals 17.
Hope it helps!

4. fjdkj says:

With the information given, the triangle CAD and KAD are both right angled triangle and they share the base AD.

So we can form 2 equation and solve them simultaneously.

(AD)² = (CD)² - 6²                (i)

(AD)² = (DK)² - 15²          (ii)

But DK - CD = 7. So, DK=7+CD
Now let CD=x
From the 2 equations above,