Question

In the given figure, a circle with centre 'O' is touching the side ST of APST at U and also touches PS and PT produced in Q and R respectively. If overline OP=25 cm and radius of circle is 7 cm, then find the perimeter of triangle PST .​

Answer 1

Final Answer:

The perimeter of triangle PST, where a circle with centre 'O' is touching the side ST of APST at U and also touches PS and PT produced in Q and R respectively, is 48cm.

Given:

In the given figure, a circle with centre 'O' is touching the side ST of APST at U and also touches PS and PT produced in Q and R respectively.

The overline OP=25 cm and the radius of the circle is 7 cm.

To Find:

The perimeter of triangle PST

Explanation:

The radii of the circle drawn to the tangents at the points of their contact are always perpendicular to the tangents.

Step 1 of 5

Join the following pairs of points

• O and Q
• O and R
• O and P

Thus the two right-angled triangles $\Delta QOP$ and $\Delta ROP$ are formed.

Step 2 of 5

The two right-angled triangles $\Delta QOP$ and $\Delta ROP$ are congruent by the RHS (Right angle Hypotenus Side) congruency, as evident from the following parameters.

• Hypotenuse OP is common
• Side OQ =side OR = 7cm [radii of the same circle]
• $\angle OQP =\angle ORP =90^0$

Step 3 of 5

Since the corresponding parts of the congruent triangles are equal, the following is inferred for the two right-angled triangles $\Delta QOP$ and $\Delta ROP$.

• Side PQ = Side PR

Similarly, by joining the points O and S, O and T, the congruent right-angled triangles are obtained, from which the following is inferred.

• Side SU = Side SQ
• Side UT = side TR

Step 4 of 5

Applying the Pythagoras theorem, in the two right-angled triangles $\Delta QOP$ and $\Delta ROP$, the following data are obtained.

Side PQ

$=\sqrt{OP^2-OQ^2}\\ =\sqrt{25^2-7^2}\\ =\sqrt{625-49} \\=\sqrt{576} \\=24cm$

Side PR

$=\sqrt{OP^2-OR^2}\\ =\sqrt{25^2-7^2}\\ =\sqrt{625-49} \\=\sqrt{576} \\=24cm$

Step 5 of 5

The perimeter of the triangle PST

= Side PS + Side ST + Side PT

=PS + (SU + UT) + PT

=PS + SQ +TR + PT

=(PS + SQ) + (PT + TR)

=PQ + PR

=24 + 24

=48cm

The required perimeter of triangle PST, where a circle with centre 'O' is touching the side ST of APST at U and also touches PS and PT produced in Q and R respectively, is 48cm.

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Answer 2

Answer:

The perimeter of triangle PST, where a circle with center 'O' is touching the side ST of APST at U and also touches PS and PT produced in Q and R respectively, is 48cm.

Explanation:

Given:

In the given figure, a circle with center 'O' is touching the side ST of APST at U and also touches PS and PT produced in Q and R respectively.

The overline OP=25 cm and the radius of the circle is 7 cm.

To Find:

The perimeter of triangle PST

The radii of the circle drawn to the tangents at the points of their contact are always perpendicular to the tangents.

Join the following pairs of points

O and Q

O and R

O and P

Therefore the two right angled triangles are formed.

The two right angled triangles are congruent by the RHS congruency, as evident from the following parameters.

Hypotenuse OP is common

Side OQ is equal to side OR that is 7cm

And angle Q and R are equal.

Since the corresponding parts of the congruent triangles are equal the following is inferred for the two right angles triangles.

Side PQ= Side PR

Similarly by joining the points O and S and T, the congruent right angled triangles are obtained.

Applying Pythagoras theorem, in two right angles triangles then,

$\sqrt{OP^{2} -OQ^{2} }$ after substituting the values Side PQ is 24 cm.

$\sqrt{OP^{2}-OR^{2} }$ after substituting the values Side PR is 24 cm.

The perimeter of the triangle is 48 cm.

The required perimeter of triangle PST, where a circle with center O is is 48 cm.

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