Question

Tangent AT = 15 cm, radius
OA = 8 cm. Find OT, BT.

WARNING ⚠
PLEASE ANSWER THE QUESTION IF YOU KNOW THE ANSWER IF YOUR ANSWER IS IRRILATIVE I'LL REPORT THAT ANSWER
​​​​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Pythagoras Theorem and Theorems of Tangent and Radius has been used. We see that we are given the length of Radius. By theorem, we know that Radius of Circle is perpendicular to its tangent which is drawn from an external point. This means the triangle becomes Right angled and by the usage of Pythagoras Theorem, we can apply value and find the answer. Again using second Theorem we can find length of BT.

Let's do it !!

_______________________________________________

★ Formula Used :-

$\\\;\boxed{\sf{(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}}$

_______________________________________________

★ Solution :-

Given,

» Radius of the Circle = OA = OB = 8 cm

» Length of AT = 15 cm

» Tangents from external point T = AT and BT

• Theorem 1 - When a Tangent is drawn from an external point which touches the circle, the radius is perpendicular to the circle if the radius is extended to the point where Tangent Touches the circle.

According to this theorem,

✒ OA ⊥ AT and ∠OAT = 90° (since OA is radius and AT is tangent to the circle)

Similarly,

✒ OB ⊥ BT and ∠OBT = 90° (since OB is radius and BT is tangent to the circle)

This means, ∆OAT is a Right - Angled Triangle. Similarly, ∆OBT is a Right - Angled Triangle.

• Theorem 2 - If two tangents are drawn from a external point to a circle, then both the tangents are equal in length.

According to this, AT = BT

The reason because AT and BT are tangents drawn from an external point T to the circle.

_______________________________________________

~ For the length of OT ::

In ∆OAT , we have,

• Hypotenuse = OT

• Base = OA = 8 cm

• Height = AT = 15 cm

So, according to the Pythagoras Theorem and Theorem 1, we get,

$\\\;\;\sf{:\rightarrow\;\;(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(OT)^{2}\;=\;\bf{(OA)^{2}\;+\;(AT)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(OT)^{2}\;=\;\bf{(8)^{2}\;+\;(15)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(OT)^{2}\;=\;\bf{64\;+\;225}}$

$\\\;\;\sf{:\rightarrow\;\;(OT)^{2}\;=\;\bf{289}}$

$\\\;\;\sf{:\rightarrow\;\;(OT)\;=\;\bf{\sqrt{289}}}$

$\\\;\;\bf{:\rightarrow\;\;(OT)\;=\;\bf{17\;\;cm}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;length\;\;of\;\;OT\;=\;\bf{17\;\;cm}}}}$

_______________________________________________

~ For the length of BT ::

Using Theorem 2, we get that,

✒ AT = BT

This means, BT = AT = 15 cm

This can be verified using Pythagoras Theorem like earlier as ∠OBT = 90°. Then,

$\\\;\;\sf{:\rightarrow\;\;(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(OT)^{2}\;=\;\bf{(OB)^{2}\;+\;(BT)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(17)^{2}\;=\;\bf{(8)^{2}\;+\;(BT)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(BT)^{2}\;=\;\bf{(17)^{2}\;-\;(8)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(BT)^{2}\;=\;\bf{289\;-\;64}}$

$\\\;\;\sf{:\rightarrow\;\;(BT)^{2}\;=\;\bf{225}}$

$\\\;\;\sf{:\rightarrow\;\;(BT)\;=\;\bf{\sqrt{225}}}$

$\\\;\;\bf{:\rightarrow\;\;(BT)\;=\;\bf{15\;\;cm}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;length\;\;of\;\;BT\;=\;\bf{15\;\;cm}}}}$

_______________________________________________

★ More Formulas to Know :-

$\\\;\sf{\leadsto\;\;Area\;of\;Sector\;=\;\dfrac{\pi r^{2} \theta}{360^{\circ}}}$

$\\\;\sf{\leadsto\;\;Length\;of\;Arc\;=\;\dfrac{2\pi r \theta}{360^{\circ}}}$

$\\\;\sf{\leadsto\;\;Area\;subtended\;by\;chord\;=\;\dfrac{\pi r^{2} \theta}{360^{\circ}}\;-\;\dfrac{1}{2}\:r^{2}\:\sin \theta}$

$\\\;\sf{\leadsto\;\;Area\;of\;Circle\;=\;\pi r^{2}}$

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Circle\;=\;2\pi r}$

Answer 2

Answer:

thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks