Answer :
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
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
In the parallelogram CDKM the value of the line segment CD and DK is 10 units and 17 units.
What is Pythagoras theorem?
Pythagoras theorem says that in a right angle triangle the square of hypotenuse side is equal to the sum of square of other two legs of right angle triangle.
The quadrilateral CDKM is a parallelogram. In this parallelogram, the side DA is perpendicular to the side CK.
The difference of the side DK and CD is,
[tex]DK - CD = 7\\DK=7+CD[/tex]
The length of the line segment CA and AK is 6 units and 15 units respectively.Then by the Pythagoras theorem,
[tex](AD)^2=(CD)^2-(AC)^2\\(AD)^2=(CD)^2-(6)^2[/tex] ....1
Again by using the Pythagoras theorem
[tex](AD)^2=(Dk)^2-(AK)^2\\(AD)^2=(DK)^2-(15)^2[/tex]
Put the value of (AD)², from equation 1 in the above equation as,
[tex](CD)^2-6^2=(DK)^2-(15)^2\\(CD)^2-36=(DK)^2-225[/tex]
Put the value of DK in the above equation as,
[tex](CD)^2-36=(7+CD)^2-225\\(7+CD)^2-(CD)^2=225-36\\(CD)^2+14CD+49-(CD)^2=189\\CD=10[/tex]
Hence, the value of DK is,
[tex]DK=7+10DK=17[/tex]
Hence, In the parallelogram CDKM the value of the line segment CD and DK is 10 units and 17 units.
Learn more about the Pythagoras theorem here;
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