Answer

Verified

410.4k+ views

Hint:- Apply Pythagoras Theorem.

Let the two legs of the triangle be \[x\] and \[y\].

And let the smaller leg be \[x\].

And the larger leg be \[y\].

As, we are given with the initial length of hypotenuse of the right-angled triangle.

So, according to Pythagoras theorem.

Sum of squares of two sides of a right-angled triangle is equal to the square of its hypotenuse.

So, \[{\left( x \right)^2} + {\left( y \right)^2} = {\left( {3\sqrt {10} } \right)^2}\]

\[ \Rightarrow {\left( x \right)^2} + {\left( y \right)^2} = 90\] (1)

Now, going to the other condition,

Smaller leg is tripled. So, the smaller leg becomes \[3x\].

Larger leg is doubled. So, the larger leg becomes \[2y\].

And, as given in the question, that new hypotenuse is \[9\sqrt 5 \].

So, again applying Pythagora's theorem.

\[{\left( {3x} \right)^2} + {\left( {2y} \right)^2} = {\left( {9\sqrt 5 } \right)^2}\]

\[ \Rightarrow 9{x^2} + 4{y^2} = 405\] (2)

Now, we had to find the value of\[x\] and \[y\] using equation 1 and 2.

So, splitting \[9{x^2}\] from equation 2. We get,

\[ \Rightarrow 5{x^2} + 4({x^2} + {y^2}) = 405\]

Now, putting the value of \[{x^2} + {y^2}\] from equation 1 to above equation we get.

\[

\Rightarrow 5{x^2} + 360 = 405 \\

\Rightarrow 5{x^2} = 405 - 360 = 45 \\

\Rightarrow {x^2} = 9 \\

\]

As, \[x\] and \[y\] are the length of sides of a triangle. So, they can have only positive values.

So, \[x = 3\]

Now, putting the value of \[x\] in equation 1. We get,

\[

\Rightarrow {y^2} + 9 = 90 \\

\Rightarrow {y^2} = 81 \\

\Rightarrow y = 9 \\

\]

Hence, sides of the right-angled triangle are 9cm and 3cm.

Note:-In these types of problems first find all the equations for the given conditions

using Pythagoras theorem. And then find the value of each side by solving the quadratic

equation. And remember that the length of the side should always be positive.

Let the two legs of the triangle be \[x\] and \[y\].

And let the smaller leg be \[x\].

And the larger leg be \[y\].

As, we are given with the initial length of hypotenuse of the right-angled triangle.

So, according to Pythagoras theorem.

Sum of squares of two sides of a right-angled triangle is equal to the square of its hypotenuse.

So, \[{\left( x \right)^2} + {\left( y \right)^2} = {\left( {3\sqrt {10} } \right)^2}\]

\[ \Rightarrow {\left( x \right)^2} + {\left( y \right)^2} = 90\] (1)

Now, going to the other condition,

Smaller leg is tripled. So, the smaller leg becomes \[3x\].

Larger leg is doubled. So, the larger leg becomes \[2y\].

And, as given in the question, that new hypotenuse is \[9\sqrt 5 \].

So, again applying Pythagora's theorem.

\[{\left( {3x} \right)^2} + {\left( {2y} \right)^2} = {\left( {9\sqrt 5 } \right)^2}\]

\[ \Rightarrow 9{x^2} + 4{y^2} = 405\] (2)

Now, we had to find the value of\[x\] and \[y\] using equation 1 and 2.

So, splitting \[9{x^2}\] from equation 2. We get,

\[ \Rightarrow 5{x^2} + 4({x^2} + {y^2}) = 405\]

Now, putting the value of \[{x^2} + {y^2}\] from equation 1 to above equation we get.

\[

\Rightarrow 5{x^2} + 360 = 405 \\

\Rightarrow 5{x^2} = 405 - 360 = 45 \\

\Rightarrow {x^2} = 9 \\

\]

As, \[x\] and \[y\] are the length of sides of a triangle. So, they can have only positive values.

So, \[x = 3\]

Now, putting the value of \[x\] in equation 1. We get,

\[

\Rightarrow {y^2} + 9 = 90 \\

\Rightarrow {y^2} = 81 \\

\Rightarrow y = 9 \\

\]

Hence, sides of the right-angled triangle are 9cm and 3cm.

Note:-In these types of problems first find all the equations for the given conditions

using Pythagoras theorem. And then find the value of each side by solving the quadratic

equation. And remember that the length of the side should always be positive.

Recently Updated Pages

The branch of science which deals with nature and natural class 10 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Give simple chemical tests to distinguish between the class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Differentiate between lanthanoids and actinoids class 12 chemistry CBSE

Classify weak and strong ligands out of a OH b F c class 12 chemistry CBSE

Define Vant Hoff factor How is it related to the degree class 12 chemistry CBSE