Circles case study questions.
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[tex]\large\underline{\sf{Solution-}}[/tex]
Given that, ABC is a right-angled triangle right-angled at B.
Further given that, AB = 3 m and BC = 4 m
Now, In right-angle triangle ABC
By using Pythagoras Theorem, we get
[tex]\sf\: {AC}^{2} = {AB}^{2} + {BC}^{2} \\ [/tex]
[tex]\sf\: {AC}^{2} = {3}^{2} + {4}^{2} \\ [/tex]
[tex]\sf\: {AC}^{2} = 9 + 16 \\ [/tex]
[tex]\sf\: {AC}^{2} = 25 \\ [/tex]
[tex]\sf\: {AC}^{2} = {5}^{2} \\ [/tex]
[tex]\implies\bf\:AC = 5 \:m \\ [/tex]
Now, Further given that, a pit is dug out such that it touches the sides AC, BC, AB at P, Q, R respectively.
We know, Tangent and radius are perpendicular to each other.
So, Using this result, we get
[tex]\implies\sf\:OQ \: \perp \: BC \: \: and \: \: OR \: \perp \: AB \\ [/tex]
Also, it is given that
[tex]\sf\:AB \: \perp \: BC \: \\ [/tex]
Now, in quadrilateral OQBR
[tex]\sf\: \angle O + \angle Q + \angle B + \angle R = {360}^{ \circ} \\ [/tex]
[tex]\sf\: \angle O +91^{ \circ} + 90^{ \circ} + 90^{ \circ} = {360}^{ \circ} \\ [/tex]
[tex]\sf\: \angle O +270^{ \circ} = {360}^{ \circ} \\ [/tex]
[tex]\implies\sf\: \angle O = {90}^{ \circ} \\ [/tex]
[tex]\implies\sf\:OQBR \: is \: a \: rectangle \\ [/tex]
[tex]\implies\sf\:OQ = BR = r \: \: and \: \: QB = OR = r \\ [/tex]
Now,
[tex]\sf\: AR = AB - BR = 3 - r \\ [/tex]
[tex]\sf\: QC = AC - QB = 4 - r \\ [/tex]
Further, we know that length of tangents drawn from external point are equal.
[tex]\implies\sf\:AP = AR \: \: and \: \: PC = QC \\ [/tex]
[tex]\implies\sf\:AP = 3 - r \: \: and \: \: PC = 4 - r \\ [/tex]
Now,
[tex]\sf\: AC = AP + PC \\ [/tex]
[tex]\sf\: 5 = 3 - r + 4 - r \\ [/tex]
[tex]\sf\: 5 = 7 - 2r \\ [/tex]
[tex]\sf\: 5 - 7 = - 2r \\ [/tex]
[tex]\sf\: - 2 = - 2r \\ [/tex]
[tex]\implies\sf\:r = 1 \: m \\ [/tex]
Now,
[tex]\sf\: (i) \: \: AR = 3 - r = 3 - 1 = 2 \: m \\ [/tex]
[tex]\sf\: (ii) \: \: BQ = r = 1 \: m \\ [/tex]
[tex]\sf\: (iii) \: \: CQ = 4 - r =4 - 1 = 3\: m \\ [/tex]
[tex]\sf\: (iv) \: \: Radius\:of\:pit, \: r =1\: m \\ [/tex]