Question

4. Find the value of x in a right-angled
triangle:​

Answer 1

★ How to do :-

Here, we are given with the measurement of the perpendicular side and the hypotenuse side of a right-angled triangle but, we are not given with the measurement of the base of the same triangle. We are asked to find the same. We can find the value of the base by a simple concept named as Pythagoras theorem. This property is only applicable for any right-angled triangles. This property is used to find the value of any side of a right-angled triangle, in which we should be given with two sides. So, let's solve!!

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➤ Solution :-

${\sf \leadsto \underline{\boxed{\sf {Base}^{2} + {Perpendicular}^{2} = {Hypotenuse}^{2}}}}$

Write the names of the sides given in triangle.

${\sf \leadsto {BC}^{2} + {AB}^{2} = {AC}^{2}}$

Substitute the values of the sides.

${\sf \leadsto {x}^{2} + {6}^{2} = {10}^{2}}$

Find the square numbers of both numbers given.

${\sf \leadsto {x}^{2} + 36 = 100}$

Shift the number 36 from LHS to RHS, changing it's sign.

${\sf \leadsto {x}^{2} = 100 - 36}$

Subtract the values on RHS.

${\sf \leadsto {x}^{2} = 64}$

Remove square on LHS and insert square root on RHS

${\sf \leadsto x = \sqrt{64}}$

Find the square root of 64 to get the answer.

${\sf \leadsto x = 8 \: \: cm}$

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${\red{\underline{\boxed{\bf So, \: the \: base \: measures \: 8 \: cm}}}}$

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Verification :-

${\tt \leadsto {x}^{2} + {6}^{2} = {10}^{2}}$

Substitute the value of x.

${\tt \leadsto {8}^{2} + {6}^{2} = {10}^{2}}$

Find the square numbers of all three numbers.

${\tt \leadsto 64 + 36 = 100}$

Add the values on LHS.

${\tt \leadsto 100 = 100}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!

Answer 2

Step-by-step explanation:

$ac {}^{2} = ab {}^{2} + bc { }^{2} \\ \\( 10) {}^{2} = (6) {}^{2} + (bc) {}^{2} \\ 100 = 36 + bc {}^{2} \\ 36 + bc {}^{2} = 100 \\ bc {}^{2} = 100 - 36 \\ bc {}^{2} = 64 \\ bc = \sqrt{} 64 \\ bc = 8cm$