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प्रश्न
If AB and CD are the common tangents in the circles of two unequal (different) radii, then show that seg AB ≅ seg CD.
उत्तर
Given: AB and CD are tangents to both circles.
To prove: seg AB ≅ seg CD
Construction: Extend seg AB and seg CD to intersect each other at point E, such that A – B – E, C – D – E.
Proof: In above figure,
`{:("AE" = "CE"),("BE" = "DE"):}}` ......(i) [Tangent segment theorem]
Consider, AE = CE
∴ AB + BE = CD + DE ......[A – B – E, C – D – E]
∴ AB + DE = CD + DE ......[From (i)]
∴ AB = CD
∴ Seg AB ≅ seg CD
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Seg XZ is a diameter of a circle. Point Y lies in its interior. How many of the following statements are true ? (i) It is not possible that ∠XYZ is an acute angle. (ii) ∠XYZ can’t be a right angle. (iii) ∠XYZ is an obtuse angle. (iv) Can’t make a definite statement for measure of ∠XYZ.
In the given figure, M is the centre of the circle and seg KL is a tangent segment.
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In the following figure ‘O’ is the centre of the circle.
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Use the information and fill in the boxes with proper numbers.
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The figure ΔABC is an isosceles triangle with a perimeter of 44 cm. The sides AB and BC are congruent and the length of the base AC is 12 cm. If a circle touches all three sides as shown in the figure, then find the length of the tangent segment drawn to the circle from point B.
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Proof: In ∆RMO and ∆RNO,
∠RMO ≅ ∠RNO = 90° ......[`square`]
hypt OR ≅ hypt OR ......[`square`]
seg OM ≅ seg `square` ......[Radii of the same circle]
∴ ∆RMO ≅ ∆RNO ......[`square`]
∠MOR ≅ ∠NOR
Similairy ∠MRO ≅ `square` ......[`square`]
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