The difference between the sides at the right angled triangle is 14 cm. The area of the triangle is 120 ${cm}^{2}$. Calculate the perimeter of the triangle.
Answer
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Hint: In the right angled triangle we have a hypotenuse, a perpendicular and a base. We can use Pythagoras theorem in right angled triangle:
${Hypotenuse}^2 ={Base}^2+ {Perpendicular}^2$
Area of a right angled triangle is given as: (½)x Base x Perpendicular
Perimeter of any triangle is given as: Sum of all the three sides.
Complete step-by-step answer:
Let ABC is a right angled triangle right angled at A.
Let the two sides AB=x cm and AC=y cm.
According to the question, the difference between the sides is 14 cm.
\[ \Rightarrow (x - y) \Rightarrow x = \left( {y + 14} \right)cm\], where (x > y).
Area of triangle =120(given)
\[ \Rightarrow \dfrac{1}{2} \times AB \times AC = 120\]
\[ \Rightarrow x \times y = 240\]
\[ \Rightarrow y \times (y + 14) = 240.....(\because y = x + 14)\]
\[ \Rightarrow {y^2} + 14y - 240 = 0\]
\[ \Rightarrow {y^2} + 24y - 10y - 240 = 0\]
\[ \Rightarrow y(y + 24) - 10(y + 24) = 0\]
\[ \Rightarrow (y + 24)(y - 10) = 0\]
\[ \Rightarrow y = 10;y = - 24\]
Neglecting \[y = - 24\]as sides cannot be negative.
One side of the right angled triangle = 10 cm.
Other side of the right angled triangle = \[y = 10 + 14 = 24\]cm
Now, for the perimeter of the triangle we must know the length of the third side.
Using Pythagoras theorem in right angle triangle ABC;
\[ \Rightarrow {(AC)^2} = {(x)^2} + {(y)^2}\]
\[ \Rightarrow {(AC)^2} = {(24)^2} + {(10)^2}\]
\[ \Rightarrow {(AC)^2} = 576 + 100\]
\[ \Rightarrow {(AC)^2} = 676\]
\[ \Rightarrow AC = 26\]cm.
Required perimeter is the sum of all sides of the triangle.
\[ \Rightarrow 24 + 10 + 26\]
\[ \Rightarrow 60\]cm.
Note: In this question we have already given that there is a difference of 14 cm between two sides which means that the given triangle will be either isosceles or scalene triangle.
An isosceles triangle has both two equal sides and two equal angles.
A scalene triangle has no sides equal.
An equilateral triangle has all three sides equal and hence, all angles equal to 60 degree.
${Hypotenuse}^2 ={Base}^2+ {Perpendicular}^2$
Area of a right angled triangle is given as: (½)x Base x Perpendicular
Perimeter of any triangle is given as: Sum of all the three sides.
Complete step-by-step answer:
Let ABC is a right angled triangle right angled at A.
Let the two sides AB=x cm and AC=y cm.
According to the question, the difference between the sides is 14 cm.
\[ \Rightarrow (x - y) \Rightarrow x = \left( {y + 14} \right)cm\], where (x > y).
Area of triangle =120(given)
\[ \Rightarrow \dfrac{1}{2} \times AB \times AC = 120\]
\[ \Rightarrow x \times y = 240\]
\[ \Rightarrow y \times (y + 14) = 240.....(\because y = x + 14)\]
\[ \Rightarrow {y^2} + 14y - 240 = 0\]
\[ \Rightarrow {y^2} + 24y - 10y - 240 = 0\]
\[ \Rightarrow y(y + 24) - 10(y + 24) = 0\]
\[ \Rightarrow (y + 24)(y - 10) = 0\]
\[ \Rightarrow y = 10;y = - 24\]
Neglecting \[y = - 24\]as sides cannot be negative.
One side of the right angled triangle = 10 cm.
Other side of the right angled triangle = \[y = 10 + 14 = 24\]cm
Now, for the perimeter of the triangle we must know the length of the third side.
Using Pythagoras theorem in right angle triangle ABC;
\[ \Rightarrow {(AC)^2} = {(x)^2} + {(y)^2}\]
\[ \Rightarrow {(AC)^2} = {(24)^2} + {(10)^2}\]
\[ \Rightarrow {(AC)^2} = 576 + 100\]
\[ \Rightarrow {(AC)^2} = 676\]
\[ \Rightarrow AC = 26\]cm.
Required perimeter is the sum of all sides of the triangle.
\[ \Rightarrow 24 + 10 + 26\]
\[ \Rightarrow 60\]cm.
Note: In this question we have already given that there is a difference of 14 cm between two sides which means that the given triangle will be either isosceles or scalene triangle.
An isosceles triangle has both two equal sides and two equal angles.
A scalene triangle has no sides equal.
An equilateral triangle has all three sides equal and hence, all angles equal to 60 degree.
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