Question

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Two circles of radii 8cm and 5cm with their centres A and B touching each other externally is shown in the figure below. The length of direct common tangent PQ is :​

Answer 1

Answer:

The length of common tangent PQ is $\sf\:4\sqrt{10}\:cm$.

Step-by-step-explanation:

NOTE : Kindly refer to the attachment first.

Draw $\sf\:BM\:\perp\:AP$ ... [ Construction ]

A - E - B ...

[ Therorm of touching circles ]

$\left\begin{tabular}{c}\sf\:AE\:=\:AP\:=\:8\: cm\\\sf\:BE\:=\:BQ\:=5\:cm\end{tabular}\right\}\sf... [\:Radii\:of\:same\:circle\:]$

$\sf\:m\angle\:APQ\:=\:m\angle\:BQP\:=\:90^{\circ}$ ... [ Tangent theorem ] ( 1 )

In $\sf\:\square\:BMPQ,\\\\m\angle\:MPQ+m\angle\:PQB+m\angle\:QBM+m\angle\:BMP=360^{\circ}$ ... [ Sum of all angles of quadrilateral is 360° ]

$\therefore\:90^{\circ}+90^{\circ}+m\angle\:QBM+90^{\circ}=360^{\circ}$ ...

[ From construction, ( 1 ) and A - M - P ]

$\therefore\:270^{\circ}+m\angle\:QBM=360^{\circ}\\\\\therefore\:m\angle\:QBM=360^{\circ}-270^{\circ}\\\\\therefore\:m\angle\:QBM=90^{\circ}$

$\therefore\sf\square\:BMPQ\:is\:a\:rectangle$ ..

[ All angles are of 90° ]

$\therefore\:BQ=PM=5 cm$ ...

[ Opposite sides of rectangle are congruent. ]

Now,

AM + MP = AP ... [ A - M - P ]

$\therefore\sf\:AM+5=8\\\\\therefore\sf\:AM=8-5\\\\\therefore\boxed{\sf\:AM=3\:cm}$

Now,

AB = AE + BE ... [ A - E - B ]

$\therefore\sf\:AB=8+5\\\\\therefore\boxed{\sf\:AB=13\:cm}$

Now, in $\triangle\:AMB,:m\angle\:AMB=90^{\circ}$ ...

[ Construction ]

$\therefore\sf\:AB^{2}=AM^{2}+BM^{2}$ ....

[ Pythagoras Theorem ]

$\therefore\sf\:(13)^{2}=(3)^{2}+BM^{2}\\\\\implies\sf\:BM^{2}=(13)^{2}-(3)^{2}\\\\\implies\sf\:BM^{2}=169-9\\\\\implies\sf\:BM^{2}=160\\\\\implies\sf\:BM=\sqrt{160}\:\:\:...[\:Taking\:square\:roots\:on\:both\:sides\:]\\\\\implies\sf\:BM=\sqrt{16\:\times\:10}\\\\\implies\sf\:BM=4\sqrt{10}\:cm$

Now,

BM = PQ ...

[ Opposite sides of rectangle are congruent. ]

$\sf\boxed{\sf\:PQ=4\sqrt{10}\:cm}$

Ans.: The length of tangent PQ is $\sf\:4\sqrt{10}\:cm$