[Maths]
(Not a challenge, but urgent :D)
The length of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of other side by 1 cm. Find the length of each side. Also find the perimeter and the area of the triangle.
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Answers
Step-by-step explanation:
Given :-
The length of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of other side by 1 cm.
To find :-
Find the length of each side. Also find the perimeter and the area of the triangle?
Solution :-
Let the sides of a right angled triangle be a,b and c units
Let c be the hypotenuse.
Given that
The length of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm
=> c = (a+2) cm
=> a = (c-2) cm -------(1)
and
The length of the hypotenuse of a right-angled triangle exceeds the length of twice the length of other side by 1 cm
=> c = (2b+1 ) cm
=> 2b+1 = c
=> 2b = c-1
=> b = (c-1)/2 cm---------(2)
The three sides of the right angled triangle are (c-2) cm , (c-1)/2 cm and c cm
We know that
By Pythagoras Theorem,
In a right angled triangle The square of the hypotenuse is equal to the sum of the two other sides.
=> c² = a²+b²
=> c² = (c-2)²+[(c-1)/2]²
=> c² = (c²-2(c)(2)+2²) + [(c²-2(c)(1)+1²)/4]
Since (a-b)² = a²-2ab+b²
=> c² = (c²-4c+4)+[(c²-2c+1)/4]
=> c² = [4(c²-4c+4)+(c²-2c+1)]/4
=> c² = (4c²-16c+16+c²-2c+1)/4
=> c² = (5c²-18c+17)/4
=> 4c² = 5c²-18c+17
=> 5c²-18c+17-4c² = 0
=> c²-18c+17 = 0
=> c²-1c-17c+17 = 0
=> c(c-1)-17(c-17) = 0
=> (c-1)(c-17) = 0
=> c-1 = 0 or c-17 = 0
=> c = 1 or c = 17
c can not be equal to 1 according to the given problem
Therefore, c = 17 cm
Hypotenuse = 17 cm
On Substituting the value of c in (1) then
a = 17-2 = 15 cm
On Substituting the value of c in (2) then
b = (17-1)/2
=> b = 16/2
=> b = 8 cm
Therefore, a = 15 cm , b = 8 cm ,c = 17 cm
Perimeter of a triangle = Sum of the lengths of the all sides
=> P = a+b+c
=>P = 15+8+17
=> P = 40 cm
We know that
Area of a right angled triangle = (1/2)ab sq.units
On Substituting the values of a and b then
Area = (1/2)×15×8 cm²
=> A = 15×4 cm²
=> A = 60 cm²
or
We know that Heron's formula
Area of a triangle whose sides are a,b,c is √[(S(S-a)(S-b)(S-c)] sq.units
Where , S = (a+b+c)/2 units
Now S = 40/2 = 20 cm
Area =√[20(20-15)(20-8)(20-17)]
=> A =√[20(5)(12)(3)]
=> A =√(3600)
=> A = √(60)²
=> A = 60 cm²
Answer:-
The lengths of the given right angled triangle are 8 cm , 15 cm and 17 cm
Perimeter of the given right angled triangle is 40 cm
Area of the right angled triangle is
60 cm²
Check :-
Sides are 8 cm , 16 cm and 17 cm
8² = 8×8 = 64
15² = 15×15 = 225
17² = 17×17 = 289
8²+15² = 64+225
8²+15² = 289
8²+15² = 17²
They are the sides of the right angled triangle.
Verified the given relations in the given problem.
Used formulae:-
Pythagoras Theorem:-
In a right angled triangle The square of the hypotenuse is equal to the sum of the two other sides.
- Perimeter of a triangle = Sum of the lengths of the all sides
- Area of a right angled triangle = (1/2)ab sq.units
- Heron's formula
- Area of a triangle whose sides are a,b,c is √[(S(S-a)(S-b)(S-c)] sq.units
- Where , S = (a+b+c)/2 units