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)$6.4cm$
Answer
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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
Complete step by step answer:
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$
Which is the required answer.
So, the correct answer is “Option C”.
Note:
In this type of question, we should have to draw the figure first which is asked in the question. For solving these types of questions students should know all properties of the figure. Also, one should apply the Pythagoras theorem properly and don’t do any mistakes in calculation. Application on the Pythagorean Theorem is used to check whether the triangle is acute, obtuse, or right.
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.
Complete step by step answer:
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$
Which is the required answer.
So, the correct answer is “Option C”.
Note:
In this type of question, we should have to draw the figure first which is asked in the question. For solving these types of questions students should know all properties of the figure. Also, one should apply the Pythagoras theorem properly and don’t do any mistakes in calculation. Application on the Pythagorean Theorem is used to check whether the triangle is acute, obtuse, or right.
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.
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