Prove that the parallelogram circumscribing a circle is a rhombus
Solution:
ABCD is a parallelogram. Therefore, opposite sides are equal.
AB = CD
BC = AD
According to Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
Therefore,
BP = BQ (Tangents from point B)…… (1)
CR = CQ (Tangents from point C)…… (2)
DR = DS (Tangents from point D)…… (3)
AP = AS (Tangents from point A)……. (4)
Adding (1) + (2) + (3) + (4)
BP + CR + DR + AP = BQ + CQ + DS + AS
On re-grouping,
BP + AP + CR + DR = BQ + CQ + DS + AS
AB + CD = BC + AD
Substitute CD = AB and AD = BC since ABCD is a parallelogram, then
AB + AB = BC + BC
2AB = 2BC
AB = BC
∴ AB = BC = CD = DA
This implies that all the four sides are equal.
Therefore, the parallelogram circumscribing a circle is a rhombus.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 10
Cuemath is one of the world's leading math learning platforms that offers LIVE 1-to-1 online math classes for grades K-12. Our mission is to transform the way children learn math, to help them excel in school and competitive exams. Our expert tutors conduct 2 or more live classes per week, at a pace that matches the child's learning needs.
Video Solution:
Prove that the parallelogram circumscribing a circle is a rhombus.
NCERT Solutions Class 10 Maths Chapter 10 Exercise 10.2 Question 11
Summary:
It has been proved that the parallelogram ABCD circumscribing a circle with center O is a rhombus.
☛ Related Questions:
- 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.
- Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
- 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(A) 7 cm(B) 12 cm(C) 15 cm(D) 24.5 cm
- In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠ POQ = 110°, then ∠ PTQ is equal to(A) 60°(B) 70°(C) 80°(D) 90°