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# In an isosceles triangle with a base of $8cm$ and altitude of $4{\text{ }}cm$. The length of the equal sides is: (A)$4cm$ (B)$4.3cm$ (C)$5.7cm$ (D)$6.4cm$

Last updated date: 17th Jun 2024
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Hint: For solving these types of question, students must know about the figure. Use the Pythagoras theorem to find the base of all right angled triangle i.e. ${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Height} \right)^2}$. In isosceles triangle two sides and two angles are same and equal

Given, $\vartriangle ABC$is an isosceles triangle
$AD$ is an altitude of triangle
$AD = 4cm$$,BC = 8cm BD = DC = 4cm In \vartriangle ABD Let AB$$ = xcm$
By using Pythagoras theorem
$A{B^2} = A{D^2} + B{D^2}$
Put the values of $AB,AD$and $BD$
$\Rightarrow {x^2} = {4^2} + {4^2}$
${x^2} = 16 + 16$
Now we get ${x^2} = 32$
Now take square root on both sides
$\Rightarrow$ $\sqrt {{x^2}} = \sqrt {32}$
$x = \sqrt {32}$
So we get $x = 5.7$
$\Rightarrow AB = 5.7cm$
We know that $\vartriangle ABC$is an Isosceles triangle.
$\Rightarrow AB = AC = 5.7cm$
If ${a^2}{\text{ }} + {\text{ }}{b^2}{\text{ }} = {\text{ }}{c^2}$, then the triangle is right.
If ${a^2}{\text{ }} + {\text{ }}{b^2}{\text{ > }}{c^2}$, then the triangle is acute.
If ${a^2}{\text{ }} + {\text{ }}{b^2}{\text{ < }}{c^2}$, then the triangle is obtuse.
Where $c$ represents the length of the hypotenuse, $a$ and $b$ represent the lengths of the other two sides.