Question

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Answer 1

★ Concept ::

Here the concept of Pythagoras Theorem has been used. We see that we are given a triangle where we have to find the value of x. The value of x here is the length component of triangle. Even we see that ∠B is 90° according to the figure. So Pythagoras Theorem can be applied. Firstly we can find the values required according to Pythagoras Theorem and then find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}}}$

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★ Solution :-

Given,

» Length of AC = 41 cm

» Length of AB = 40 cm

» Length of BC = x cm

We see that in a Right Angled Triangle,

• Hypotenuse = The opposite side to Right Angle.

• Base = Side containing right angle on which triangle stands.

• Height = Side forming the right triangle which is perpendicular to the base.

From all these we can get,

• Hypotenuse = AC = 41 cm

• Base = BC = x cm

• Height = AB = 40 cm

According to Pythagoras Theorem, we know that the square of Hypotenuse is equal to the sum of squares of the other two sides of right triangle.

Then, we know the mathematical formulation of Pythagoras Theorem, as

$\\\;\sf{\Longrightarrow\;\;(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}$

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;(41)^{2}\;=\;\bf{(x)^{2}\;+\;(40)^{2}}}$

$\\\;\sf{\Longrightarrow\;\;(1681)\;=\;\bf{(x)^{2}\;+\;(1600)}}$

$\\\;\sf{\Longrightarrow\;\;1681\;=\;\bf{(x)^{2}\;+\;1600}}$

$\\\;\sf{\Longrightarrow\;\;x^{2}\;=\;\bf{1681\;-\;1600}}$

$\\\;\sf{\Longrightarrow\;\;x^{2}\;=\;\bf{81}}$

$\\\;\sf{\Longrightarrow\;\;x\;=\;\bf{\sqrt{81}}}$

$\\\;\sf{\Longrightarrow\;\;x\;=\;\bf{\sqrt{9\:\times\:9}}}$

$\\\;\bf{\Longrightarrow\;\;x\;=\;\bf{\green{9\;\;cm}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Required\;\:value\;\:of\;\:x\;=\;\bf{\purple{9\;\;cm}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Perimeter\;of\;\Delta\;=\;Sum\;of\;all\;Sides}$

$\\\;\sf{\leadsto\;\;Area\;of\;\Delta\;=\;\dfrac{1}{2}\;\times\;Base\:\times\:Height}$

$\\\;\sf{\leadsto\;\;Area\;of\;\Delta\;=\;\sqrt{s(s\:-\:a)(s\:-\:b)(s\:-\:c)}}$

$\\\;\sf{\leadsto\;\;Semi\:-\:Perimeter\;of\;\Delta,\;s\;=\;\dfrac{Sum\;of\;all\;Sides}{2}}$

Answer 2

Given : A right angled triangle ABC with measurements.

To Find : Value of x

⠀⠀⠀⠀⠀⠀⠀⠀_______________

Here, ABC is a right angled triangle. As we know that, in a right angled triangle sum of the square of base and perpendicular is equal to the square of hypotenuse of the right angled triangle.

i.e. H² = P² + B² [ Pythagoras theorem ]

Where

• H denotes hypotenuse of the right angled triangle.
• P denotes perpendicular of the right angled triangle.
• B denotes base of the right angled triangle.

Here

• AC = 41cm = Hypotenuse of the right angled triangle.
• BC = x = Perpendicular the right angled triangle.
• AB = 40cm = Base of the right angled triangle.

We've to find the value of x

Now, we'll put the given values and find the value of x

=> AC² = BC² + AB²

=> (41)² = (x)² + (40)²

=> (41)² - (40)² = x²

=> 1681 - 1600 = x²

=> x² = 81

=> x = √81

=> x = 9cm

∴ Hence, the value of x is 9cm.