<u>ΔACB</u> <u>ΔCDA</u>
AC² + BC² = AB² AD² + CD² = AC²
BC² = AB² - AC² BC² + CD² = AC² (AD=BC is given)
BC² = AC² - CD²
AB² - AC² = AC² - CD² (both sides were = to BC²)
AB² + CD² = 2AC²
(3)² + (√2)² = 2AC² (AB=3 and CD=√2 were given)
9 + 2 = 2AC²
11 = 2AC²
= AC²
= AC
= AC
30 because 5x6= 30. All you have to do if the fractions are already simplified, is multiply the denominators together.
Given :
A right angle triangle ABC .
To Find :
The perimeter of ABC .
Solution :
Since, triangle ABC is right angled and angle ∠ABC is 46° .
So, AC = AB cos 46° = 7.64 units.
Also, CB = AB sin 46° = 7.91 units.
Therefore, the perimeter of ΔABC is :
P = 11 + 7.64 + 7.91 units
P = 26.55 units
Hence, this is the required solution.
