Answer
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Hint: Some measurements of a right triangle are given, using pythagoras theorem and relating to the given measurements leads us to the solution.
Complete step-by-step answer:
Given that\[AC = 41{\text{ }}cm\]
Area = 180 \[c{m^2}\]
Now we are going to use Pythagoras theorem.
Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides”.
So we have, \[A{B^2} + B{C^2} = A{C^2}\]
Given that a side of the right angled triangle\[AB{\text{ is }}41{\text{ }}cm\].
Now we substitute \[AB = 41{\text{ }}cm\] on \[A{B^2} + B{C^2} = A{C^2}\]
We have\[A{B^2} + B{C^2} = {41^2}\;\]……(1)
Now we are going to use the area of the triangle.
We already know that the area of the triangle\[ABC\],
\[Area\;of\;ABC = \]\[\dfrac{1}{2} \times BC \times AB\]
Given that, \[Area\;of\;ABC = 180{\text{ }}c{m^2}\]
Now substituting \[Area\;of\;ABC = 180{\text{ }}c{m^2}\] on\[Area\;of\;ABC = \]\[\dfrac{1}{2} \times BC \times AB\], we get\[
180 = \dfrac{1}{2} \times BC \times AB \\
BC \times AB = 180 \times 2 \\
\]
\[BC \times AB = 360\]……….(2)
Now consider the term, \[{\left( {AB - BC} \right)^2}\]
Now we are going to expand the term \[{\left( {AB - BC} \right)^2}\]
\[{\left( {AB - BC} \right)^2} = A{B^2} + B{C^2} - 2 \times AB \times BC\]
Now we are going to substitute (i) and (ii) to get the solution.
\[{\left( {AB - BC} \right)^2} = {41^2}\; - 2 \times 360\]
\[{\left( {AB - BC} \right)^2} = (41 \times 41)\; - 2 \times 360\]
\[{\left( {AB - BC} \right)^2} = 1681 - 720\]
\[{\left( {AB - BC} \right)^2} = 961\]
Now we are going to take square root on both sides.
\[AB - BC = 31\;cm\]
Here \[AB - BC\] is the difference between the legs of the given right angled triangle.
Hence we finally found that the difference between the legs of the given right angled triangle is 31 cm. So, Option (D) is the correct answer.
Note:
In the right angled triangle, the legs of the right triangle are the two sides that intersect to determine the right angle. The sides of this triangle have been named as perpendicular, base and hypotenuse. Here, the hypotenuse is the longest side and it is opposite to the angle \[{90^ \circ }\]
Complete step-by-step answer:
Given that\[AC = 41{\text{ }}cm\]
Area = 180 \[c{m^2}\]
Now we are going to use Pythagoras theorem.
Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides”.
So we have, \[A{B^2} + B{C^2} = A{C^2}\]
Given that a side of the right angled triangle\[AB{\text{ is }}41{\text{ }}cm\].
Now we substitute \[AB = 41{\text{ }}cm\] on \[A{B^2} + B{C^2} = A{C^2}\]
We have\[A{B^2} + B{C^2} = {41^2}\;\]……(1)
Now we are going to use the area of the triangle.
We already know that the area of the triangle\[ABC\],
\[Area\;of\;ABC = \]\[\dfrac{1}{2} \times BC \times AB\]
Given that, \[Area\;of\;ABC = 180{\text{ }}c{m^2}\]
Now substituting \[Area\;of\;ABC = 180{\text{ }}c{m^2}\] on\[Area\;of\;ABC = \]\[\dfrac{1}{2} \times BC \times AB\], we get\[
180 = \dfrac{1}{2} \times BC \times AB \\
BC \times AB = 180 \times 2 \\
\]
\[BC \times AB = 360\]……….(2)
Now consider the term, \[{\left( {AB - BC} \right)^2}\]
Now we are going to expand the term \[{\left( {AB - BC} \right)^2}\]
\[{\left( {AB - BC} \right)^2} = A{B^2} + B{C^2} - 2 \times AB \times BC\]
Now we are going to substitute (i) and (ii) to get the solution.
\[{\left( {AB - BC} \right)^2} = {41^2}\; - 2 \times 360\]
\[{\left( {AB - BC} \right)^2} = (41 \times 41)\; - 2 \times 360\]
\[{\left( {AB - BC} \right)^2} = 1681 - 720\]
\[{\left( {AB - BC} \right)^2} = 961\]
Now we are going to take square root on both sides.
\[AB - BC = 31\;cm\]
Here \[AB - BC\] is the difference between the legs of the given right angled triangle.
Hence we finally found that the difference between the legs of the given right angled triangle is 31 cm. So, Option (D) is the correct answer.
Note:
In the right angled triangle, the legs of the right triangle are the two sides that intersect to determine the right angle. The sides of this triangle have been named as perpendicular, base and hypotenuse. Here, the hypotenuse is the longest side and it is opposite to the angle \[{90^ \circ }\]
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