the length of two sides of a right triangle containing the right angle differ by 2cm.if the area if the triangle is 24cm²,find the perimeter of the triangle
Answers
Answer:
area = 1/2 bh
b = 2 + h
area = 2 + h × h×1/2
= ( 2h + h² )/2
24 = ( 2h + h² )/2
48 = 2h + h²
h² + 2h - 48 = 0
lets take h as x
x² + 2x - 48 = 0
by factorization ,
x = +3 or x = -11
since the value of a side cant be negative , we take x = 3 cm
h = x = 3cm
b = 2 + 3
= 5 cm
AC² = AB² + BC²
AC = √AB² + BC²
= √9 + 25
= √34
perimeter = 3 + 5 + √34
= 8 + √34
Hope it helped :)
ANSWER:-
Given:
The length of two sides of a right ∆ containing the right angle differ by 2cm, If the area of the ∆ is 24cm².
To find:
The perimeter of the triangle.
Solution:
Let the sides at right angles be a & b.
&
the third side be c.
⚫a-b = 2cm
⚫Area of ∆= 24cm²
Therefore,
=) a- b= 2
=) a= 2+b..........(1)
We know that, area of triangle:
=) 1/2 × base × height.
=) 24= 1/2 × b × (2+b)
=) 24 × 2= b² + 2b
=) b² + 2b= 48
=) b² + 2b -48 =0
=) b² + 8b -6b -48 =0
=) b(b+ 8) -6(b+8)=0
=) (b+8)(b-6)=0
=) b+8= 0 or b-6=0
=) b= -8 or b= 6
Since,
The side of the ∆ can't be negative.
So,
b= 6cm
Therefore,
Putting the value of b in equation (1), we get;
=) a= 2+ 6
=) a= 8cm
Now,
Using Pythagoras Theorem:
(Hypotenuse)²=(base)² +(perpendicular)²
=) c² = a² + b²
=) c² = 8² + 6²
=) c² = 64 + 36
=) c² = 100
=) c = √100
=) c = 10cm
Now,
Perimeter of triangle=side + side + side
=) Perimeter of ∆= 8cm+ 6cm + 10cm
=) Perimeter of ∆= 24cm
Thus,
The perimeter of the ∆ is 24cm.