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Q11:
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution :
Given: A parallelogram $ABCD$ circumscribing a circle with center $O$. The sides $AB$, $BC$, $CD$, and $DA$ touch the circle at points $P$, $Q$, $R$, and $S$ respectively.
To Prove: Parallelogram $ABCD$ is a rhombus (i.e., $AB = BC = CD = DA$).
Step 1: Applying the Tangent Theorem
We know that the lengths of tangents drawn from an external point to a circle are equal. [Theorem: Tangents from an external point to a circle are equal in length.]
From point $A$: $AP = AS$ --- (i)
From point $B$: $BP = BQ$ --- (ii)
From point $C$: $CQ = CR$ --- (iii)
From point $D$: $DR = DS$ --- (iv)
Step 2: Summing the Equations
Adding equations (i), (ii), (iii), and (iv):
$(AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CQ)$
Rearranging the terms based on the sides of the parallelogram:
$(AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CQ)$
$AB + CD = AD + BC$ --- (v)
Step 3: Utilizing Parallelogram Properties
Since $ABCD$ is a parallelogram, we know that opposite sides are equal:
$AB = CD$ and $BC = AD$ [Property of a parallelogram]
Step 4: Substitution and Simplification
Substitute $CD = AB$ and $AD = BC$ into equation (v):
$AB + AB = BC + BC$
$2AB = 2BC$
$AB = BC$
Step 5: Conclusion
Since $AB = BC$ and we already know $AB = CD$ and $BC = AD$ (opposite sides of a parallelogram), it follows that:
$AB = BC = CD = DA$.
A parallelogram with all sides equal is defined as a rhombus. Therefore, $ABCD$ is a rhombus.
Final Answer: Since all sides of the parallelogram $ABCD$ are equal ($AB = BC = CD = DA$), the parallelogram is proven to be a rhombus.
More Questions from Class 10 Mathematics Circles EXERCISE 10.2
- Q1: Choose the correct option and give justification. From a point $Q$, the length of the tangent to a circle is $24$ cm and the distance of $Q$ from the centre is $25$ cm. The radius of the circle is
- Q10: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
- Q12: A triangle $ABC$ is drawn to circumscribe a circle of radius $4$ cm such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8$ cm and $6$ cm respectively (see Fig. 10.14). Find the sides $AB$ and $AC$.
- Q13: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
- Q2: Choose the correct option and give justification. In Fig. 10.11, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^{\circ}$, then $\angle PTQ$ is equal to
- Q3: Choose the correct option and give justification. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^{\circ}$, then $\angle POA$ is equal to
- Q4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
- Q5: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
- Q6: The length of a tangent from a point $A$ at distance $5$ cm from the centre of the circle is $4$ cm. Find the radius of the circle.
- Q7: Two concentric circles are of radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle.
- Q8: A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that $AB + CD = AD + BC$.
- Q9: In Fig. 10.13, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^{\circ}$.
