Question

in the given figure, value of x(in cm) is
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Answer 1

In ∆ABP & ∆QRP ,

/ ABP = / QRP (coordinate angle)

/ APB = / QPR (common angle)

:. ∆ABP ≈ ∆QRP (by A.A. axiom)

Ratio of the sides of the triangles are equal

$\frac{AB}{QR} = \frac{AP}{QP} = \frac{PB}{PR}$

=> $\frac{2}{x} = \frac{2.4}{3.6+2.4} = \frac{3.2}{4.8+3.2}$

=> $\frac{2}{x} = \frac{2.4}{6} = \frac{3.2}{8}$

=> $\frac{2}{x} = \frac{2.4}{6}$

=> $x = \frac{6×2}{2.4}$

=> $x = \frac{12}{2.4}$

=> $x = 10$

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Answer 2

Given: $PB=3.2cm,BR=4.8cm,AQ=3.6cm,AP=2.4cm,AB=2cm, QR=x cm$.

To find: The value of x.

Solution:

Observe that, In ∆ABP & ∆QRP,

$\[\angle ABP = \angle QRP\]$ (coordinate angle)

$\[\angle APB = \angle QPR\]$

$\[\angle APB = \angle QPR\]$ (common angle)

:. $\[\Delta ABP \simeq \Delta QRP\]$ (by A.A. axiom)

Know that, Ratio of the sides of the triangles are equal.

Therefore, find the value of x.

$\[ \Rightarrow \frac{2}{x} = \frac{{2.4}}{{3.6 + 2.4}} = \frac{{3.2}}{{4.8 + 3.2}}\]$

$\[ \Rightarrow \frac{2}{x} = \frac{{2.4}}{6} = \frac{{3.2}}{8}\]$

$\[ \Rightarrow \frac{2}{x} = \frac{{2.4}}{6}\]$

$\[ \Rightarrow x = \frac{{6 \times 2}}{{2.4}}\]$

$\[ \Rightarrow x = \frac{{12}}{{2.4}}\]$

$\[ \Rightarrow x = 10\]$

Hence, the value of x is 10 cm.