. In AABC, LB = 90° and BM is an altitude. If AM = 9 and CM=16, find the perimeter of AABC
Answers
Solution :-
given that,
→ ∠B = 90°
→ AM = 9
→ CM = 16
so,
→ Altitude BM = √(AM * CM) = √(9 * 16) = √144 = 12
then,
→ Area of ∆ABC = (1/2) * BM * AC = (1/2) * 12 * (AM + CM) = 6 * 25 = 150
also,
→ Area of ∆ABC = (1/2) * AB * BC
→ AB * BC = 2 * 150 = 300 ------ Eqn.(1)
and,
→ AB² + BC² = AC² { by pythagoras theorem }
→ AB² + BC² = 25² = 625 ----- Eqn.(2)
then,
→ (AB + BC)² = AB² + BC² + 2 * AB * BC
→ (AB + BC)² = 625 + 2 * 300
→ (AB + BC)² = 625 + 600 = 1225
→ (AB + BC) = 35 ------ Eqn.(3)
and,
→ (AB - BC)² = AB² + BC² - 2 * AB * BC
→ (AB - BC)² = 625 - 2 * 300
→ (AB - BC)² = 625 - 600 = 25
→ (AB - BC) = 5 ------ Eqn.(4)
adding Eqn.(3) and Eqn.(4)
→ 2AB = 35 + 5
→ AB = 40/2 = 20
putting value of AB in Eqn.(3),
→ BC = 35 - 20 = 15
therefore,
→ the perimeter of ∆ABC = AB + BC + AC = 20 + 15 + 25 = 60 (Ans.)
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Given : In ΔABC,∠B = 90° and BM is an altitude. If AM = 9 and CM=16,
To find : the perimeter of ΔABC
Solution:
Geometric Mean (Altitude) Theorem
The length of the altitude from the right angle to the hypotenuse in a right triangle is the geometric mean of the lengths of segments the altitude divides the hypotenuse into.
=> BM² = AM * CM
=> BM² = 9 * 16
=> BM = 12
Using Pythagorean theorem
BC² = BM² + CM²
=> BC² = 12² + 16²
=> BC = 20
AB² = BM² + AM²
=> AB² = 12² + 9²
=> AB = 15
AC = AM + CM = 9 + 16 = 25
perimeter of ΔABC = AB + BC + AC
= 15 + 20 + 25
= 60
perimeter of ΔABC is 60
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