ABC is a right angled triangle at A , AD perpendicular on BC making right angle , AB = 4 cm , CD = 5 cm , find ACxAD =
Since ABC is a right-angled triangle, we can use the Pythagorean theorem to find the length of the hypotenuse AC.
According to the Pythagorean theorem, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
So, AC^2 = AB^2 + BC^2.
We know that AB = 4 cm. Let's find the value of BC using the Pythagorean theorem as well.
BC^2 = AC^2 - AB^2.
We are given that CD = 5 cm and AD is perpendicular to BC.
Since AD is perpendicular to BC, it divides the triangle into two smaller right-angled triangles: ACD and ADB.
Hence, AC = AD + CD.
We also know that BC = BD + DC, where BD is the remaining part of BC.
Since ABC is a right-angled triangle, the angles at D and C are also right angles.
Therefore, CD = DC.
So, BC = BD + DC = BD + CD = BD + 5.
Now substituting the value of BC in the equation BC^2 = AC^2 - AB^2, we get:
(BD + 5)^2 = AC^2 - 4^2.
Expanding the equation, we get:
BD^2 + 10BD + 25 = AC^2 - 16.
Now, let's consider the right-angled triangle ADB.
We know that AD is perpendicular to BC, so ADB is also a right-angled triangle.
Using the Pythagorean theorem in triangle ADB, we get:
AB^2 = BD^2 + AD^2.
Substituting the values, we get:
4^2 = BD^2 + AD^2.
16 = BD^2 + AD^2.
Rearranging the equation, we get:
BD^2 = 16 - AD^2.
Substituting this value of BD^2 in the previous equation BD^2 + 10BD + 25 = AC^2 - 16, we get:
16 - AD^2 + 10BD + 25 = AC^2 - 16.
Adding AD^2 to both sides, we get:
16 + 10BD + 25 + AD^2 = AC^2 - 16 + AD^2.
Combining like terms, we get:
41 + 10BD = AC^2.
Now let's substitute the value of AC from the equation AC = AD + CD:
41 + 10BD = (AD + CD)^2.
Expanding the equation, we get:
41 + 10BD = AD^2 + 2AD*CD + CD^2.
Substituting the given values, we get:
41 + 10BD = AD^2 + 2AD*5 + 5^2.
Simplifying the equation, we get:
41 + 10BD = AD^2 + 10AD + 25.
Rearranging the equation, we get:
0 = AD^2 + 10AD + 25 - 10BD - 41.
Simplifying further, we get:
0 = AD^2 + 10AD - 10BD - 16.
Now let's substitute the value of BD from the equation BD = BC - 5:
0 = AD^2 + 10AD - 10(BC - 5) - 16.
Expanding the equation, we get:
0 = AD^2 + 10AD - 10BC + 50 - 16.
Combining like terms, we get:
0 = AD^2 + 10AD - 10BC + 34.
Now let's substitute the value of BC from the equation BC = BD + 5:
0 = AD^2 + 10AD - 10(BD + 5) + 34.
Expanding the equation, we get:
0 = AD^2 + 10AD - 10BD - 50 + 34.
Combining like terms, we get:
0 = AD^2 + 10AD - 10BD - 16.
This equation is identical to the previous equation we derived.
Therefore, the equation is satisfied for all values of AD and BD.
Since the equation is valid for all values of AD and BD, there is no unique solution for AC x AD.