Find the unknown length x in the following figures. - Vidyadip

Question

Find the unknown length x in the following figures.

✓

Answer

(i) Let given triangle be $\triangle A B C$ and $\triangle A B C$ is a right angled at $B$.
Again, let $A C=x$
Given, AB = 3 BC = 4

In $\triangle A B C$, by using Pythagoras property,
$(\text {Hypotenuse})^2=(\text {Base})^2+(\text {Perpendicular})^2$
i.e. $A C^2=A B^2+B C^2
\Rightarrow x^2=(3)^2+(4)^2=9+16$ $\Rightarrow \quad x^2=25 \Rightarrow x=\sqrt{25}=5$
Hence, the unknown length $(x)$ is 5.
(ii) Ans. $x=10$
(iii) Ans. $x=17 cm$
(iv) Ans. $x=25$ units
(v) Let given triangle be $\triangle A B C$ in which $A B=A C$. Then, it is an isosceles triangle and given $A D$ is perpendicular to $B C$, so $D$ is the mid-point of $B C$.

Given, $\quad A B=37=A C, A D=12, B C=x$
Now,$\quad$$B D=D C=\frac{B C}{2}=\frac{x}{2}$
In $\triangle A D B$, using Pythagoras property,
$A B^2 =B D^2+A D^2 $
$\Rightarrow (37)^2 =\left(\frac{x}{2}\right)^2+(12)^2 $
$\Rightarrow 1369 =\frac{x^2}{4}+144 $
$\Rightarrow 1369-144 =\frac{x^2}{4} \Rightarrow 1225=\frac{x^2}{4} $
$\Rightarrow 1225 \times 4 =x^2 $
$\Rightarrow x^2 =1225 \times 4 \Rightarrow x^2=4900 $
$\Rightarrow x =\sqrt{4900}=70$
Hence, the unknown length of $x$ is 70 units.
(vi) Let given triangle be $\triangle P R S$ which is divided into two $\triangle P Q R$ and $\triangle R Q S$.

In $\triangle P Q R$, using Pythagoras property,
$P R^2 =P Q^2+Q R^2 $
$\Rightarrow (12)^2 =x^2+(3)^2 \Rightarrow 144=x^2+9 $
$\Rightarrow 144-9 =x^2 $
$\Rightarrow x^2 =135 \Rightarrow x=\sqrt{135}$
Hence, the unknown length of $(x)$ is $\sqrt{135} cm$.

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