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

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

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

[tex]Cdkm is a parallelogram, da ⊥ ck , dk – cd = 7 ca = 6, ak = 15 find: cd and dk[/tex]

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

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!

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-(AC)²

(AD)² = (CD)² - 6² (i)

(AD)² = (DK)² - (AK)²

(AD)² = (DK)² - 15² (ii)

But DK - CD = 7. So, DK=7+CD

Now let CD=x

From the 2 equations above,

(AD)²=x²-36 (i)

(AD)²=(7+x)²-225 (ii)

x²-36=49+14x+x²-225

14x=140

x=10

CD = 10.

DK = 7+CD

= 7+10

= 17