Let PA and PB are two tangents of circle with centre O. Join OA and OB. We know that the radius is perpendicular to tangent. therefore ∠OAP = 90° and ∠OBP = 90°. In quadrilateral OAPB, ∠OPA + ∠PBO + ∠BOA = 360° 90° + ∠APB + 90° + ∠BOA = 360° ∠APB + ∠BOA = 180° Hence, it is proved that the angle betwRead more
Let PA and PB are two tangents of circle with centre O. Join OA and OB.
We know that the radius is perpendicular to tangent. therefore
∠OAP = 90° and ∠OBP = 90°.
In quadrilateral OAPB, ∠OPA + ∠PBO + ∠BOA = 360°
90° + ∠APB + 90° + ∠BOA = 360° ∠APB + ∠BOA = 180°
Hence, it is proved 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.
ABCD is a parallelogram, circumscribing a circle is a rhombus. ABCD is a parallelogram, therefore AB = CD ...(1) BC = AD ...(2) We know that the tangent drawn from some external point to circle are equal. so, DR = DS [Tangents drawn from point D] ...(3) CR = CQ [Tangents drawn from point C] ...(4) BRead more
ABCD is a parallelogram, circumscribing a circle is a rhombus.
ABCD is a parallelogram, therefore
AB = CD …(1)
BC = AD …(2)
We know that the tangent drawn from some external point to circle are equal. so,
DR = DS [Tangents drawn from point D] …(3)
CR = CQ [Tangents drawn from point C] …(4)
BP = BQ [Tangents drawn from point B] …(5)
AP = AS [Tangents drawn from point A] …(6)
Adding (3), (4), (5) and (6), we have
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CB) + BP + AP ) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC …(7)
From (1), (2) and (7), we have
AB = BC = CD = DA
Hence, ABCD is a rhombus.
Quadrilateral ABCD is circumscribing a circle with centre O, touching at point P, Q, R and S. join the points P, Q, R and S from the centre O. In ΔOAP and ΔOAS, OP = OS [radill of same circle] AP = AS [Tangents drawn from point A] AO = AO [Common] ΔOPA ≅ ΔOCA [SSS Congruency rule] Hence, ∠POA = ∠SOARead more
Quadrilateral ABCD is circumscribing a circle with centre O, touching at point P, Q, R and S.
join the points P, Q, R and S from the centre O.
In ΔOAP and ΔOAS,
OP = OS [radill of same circle]
AP = AS [Tangents drawn from point A]
AO = AO [Common]
ΔOPA ≅ ΔOCA [SSS Congruency rule]
Hence, ∠POA = ∠SOA or ∠1 = ∠8
Similarly, ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7
Sum of all angles at point O is 360°. Therefore
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°
⇒ 2∠1 + 2∠2 + 2∠5 + 2∠6 = 360
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180°
⇒ ∠AOB + ∠COD = 180°
Similarly, we can prove ∠BOC + ∠DOA = 180°
Hence, the opposite sides subtend supplementary angles at the centre.
A line segment of lenght 7.6 cm can be divided in the ratio of 5:8 as follow. Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an ancute angle with line segment AB. Step 2 Locate 13 (= 5 + 8) points A₁, A₂, A₃, A₄ ...... A₁₃, On AX such that A₁A₂ = A₂A₃ and so on. Step 3 Join BA₁₃. SteRead more
A line segment of lenght 7.6 cm can be divided in the ratio of 5:8 as follow.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an ancute angle with line
segment AB.
Step 2 Locate 13 (= 5 + 8) points A₁, A₂, A₃, A₄ …… A₁₃, On AX such that A₁A₂ = A₂A₃ and so on.
Step 3 Join BA₁₃.
Step 4 Through the point As, draw a line paraller to BA₁₃ (by making an angle equal to ∠AA₁₃
B) at A₅ intersection AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ration of 5:8. The lenghts of
AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.
Justification
The construction can be justified by proving that AC/ CB = 5/8
By construction, we have A₅C II A₁₃B. By applying Basic proportionality theorem for
the triangle AA₁₃B, we obtain
AC/CB = AA₅/A₅A₁₃ …(1)
From the figure, it can be observed that AA₅ and A₅A₁₃ contain 5 and 8 equal divisions of line
segments respectively.
AA₅/A₅A₁₃ = 5/8 …(2)
On comparing equations (1) and (2), we obtain AC/CB = 5/8
This justifies the construction.
Step 1 Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5cm radius Similarly, taking point B as its centre, draw an arc of 6 cm and ΔABC is the required triangle. Step 2 Draw a ray AX making an acute angle with line AB on the opposite side of vertex C. Step 3 Locate 3 points ARead more
Step 1
Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5cm radius Similarly, taking point B as its centre, draw an arc of 6 cm and ΔABC is the required triangle.
Step 2
Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
Step 3
Locate 3 points A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁ = A₁A₂ = A₂A₃.
Step 4
Join BA₃ and draw a line through A₂ parallel to BA₃ to intersect AB at point B’
Step 5
Draw a line through B’ parallel to the line BC to intersect AC at C’.
ΔAB’C is the required triangle.
Justification
The construction van be justified by proving that
AB’ = 2/3 AB, B’C’ = 2/3 BC, AC’ = 2/3 AC
By construction, we have B’C’ II BC
∴ ∠AB’C’ = ∠ABC (corresponding angles)
In ΔAB’C’ and ΔABC,
∠AB’C’ = ∠ABC (Proved above)
∠B’AC’ = ∠ BAC (common)
∴ Δ AB’C’ ∼ ΔABC (AA similarity criterion)
⇒ AB’/AB = B’C’/BC = AC/AC …(1)
In Δ AA₂B’ and ΔAA₃B,
∠A₂AB’ = ∠A₃AB (common)
∠AA₂B’ = ∠AA₃B (corresponding angles)
∴ ΔAA₂B’ ∼ ΔAA₃B (AA similarity criterion)
⇒ AB’/AB = AA₂/AA₃
⇒ AB’/AB = 2/3 …(2)
From equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 2/3
⇒ AB’ = 2/3 AB, B’C’ = 2/3 BC, AC’ = 2/3 AC
This justifies the construction.
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.
Let PA and PB are two tangents of circle with centre O. Join OA and OB. We know that the radius is perpendicular to tangent. therefore ∠OAP = 90° and ∠OBP = 90°. In quadrilateral OAPB, ∠OPA + ∠PBO + ∠BOA = 360° 90° + ∠APB + 90° + ∠BOA = 360° ∠APB + ∠BOA = 180° Hence, it is proved that the angle betwRead more
Let PA and PB are two tangents of circle with centre O. Join OA and OB.
See lessWe know that the radius is perpendicular to tangent. therefore
∠OAP = 90° and ∠OBP = 90°.
In quadrilateral OAPB, ∠OPA + ∠PBO + ∠BOA = 360°
90° + ∠APB + 90° + ∠BOA = 360° ∠APB + ∠BOA = 180°
Hence, it is proved 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.
Prove that the parallelogram circumscribing a circle is a rhombus.
ABCD is a parallelogram, circumscribing a circle is a rhombus. ABCD is a parallelogram, therefore AB = CD ...(1) BC = AD ...(2) We know that the tangent drawn from some external point to circle are equal. so, DR = DS [Tangents drawn from point D] ...(3) CR = CQ [Tangents drawn from point C] ...(4) BRead more
ABCD is a parallelogram, circumscribing a circle is a rhombus.
See lessABCD is a parallelogram, therefore
AB = CD …(1)
BC = AD …(2)
We know that the tangent drawn from some external point to circle are equal. so,
DR = DS [Tangents drawn from point D] …(3)
CR = CQ [Tangents drawn from point C] …(4)
BP = BQ [Tangents drawn from point B] …(5)
AP = AS [Tangents drawn from point A] …(6)
Adding (3), (4), (5) and (6), we have
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CB) + BP + AP ) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC …(7)
From (1), (2) and (7), we have
AB = BC = CD = DA
Hence, ABCD is a rhombus.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the Centre of the circle.
Quadrilateral ABCD is circumscribing a circle with centre O, touching at point P, Q, R and S. join the points P, Q, R and S from the centre O. In ΔOAP and ΔOAS, OP = OS [radill of same circle] AP = AS [Tangents drawn from point A] AO = AO [Common] ΔOPA ≅ ΔOCA [SSS Congruency rule] Hence, ∠POA = ∠SOARead more
Quadrilateral ABCD is circumscribing a circle with centre O, touching at point P, Q, R and S.
See lessjoin the points P, Q, R and S from the centre O.
In ΔOAP and ΔOAS,
OP = OS [radill of same circle]
AP = AS [Tangents drawn from point A]
AO = AO [Common]
ΔOPA ≅ ΔOCA [SSS Congruency rule]
Hence, ∠POA = ∠SOA or ∠1 = ∠8
Similarly, ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7
Sum of all angles at point O is 360°. Therefore
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°
⇒ 2∠1 + 2∠2 + 2∠5 + 2∠6 = 360
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180°
⇒ ∠AOB + ∠COD = 180°
Similarly, we can prove ∠BOC + ∠DOA = 180°
Hence, the opposite sides subtend supplementary angles at the centre.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
A line segment of lenght 7.6 cm can be divided in the ratio of 5:8 as follow. Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an ancute angle with line segment AB. Step 2 Locate 13 (= 5 + 8) points A₁, A₂, A₃, A₄ ...... A₁₃, On AX such that A₁A₂ = A₂A₃ and so on. Step 3 Join BA₁₃. SteRead more
A line segment of lenght 7.6 cm can be divided in the ratio of 5:8 as follow.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an ancute angle with line
segment AB.
Step 2 Locate 13 (= 5 + 8) points A₁, A₂, A₃, A₄ …… A₁₃, On AX such that A₁A₂ = A₂A₃ and so on.
Step 3 Join BA₁₃.
Step 4 Through the point As, draw a line paraller to BA₁₃ (by making an angle equal to ∠AA₁₃
B) at A₅ intersection AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ration of 5:8. The lenghts of
AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.
Justification
The construction can be justified by proving that AC/ CB = 5/8
By construction, we have A₅C II A₁₃B. By applying Basic proportionality theorem for
the triangle AA₁₃B, we obtain
AC/CB = AA₅/A₅A₁₃ …(1)
From the figure, it can be observed that AA₅ and A₅A₁₃ contain 5 and 8 equal divisions of line
segments respectively.
See lessAA₅/A₅A₁₃ = 5/8 …(2)
On comparing equations (1) and (2), we obtain AC/CB = 5/8
This justifies the construction.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
Step 1 Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5cm radius Similarly, taking point B as its centre, draw an arc of 6 cm and ΔABC is the required triangle. Step 2 Draw a ray AX making an acute angle with line AB on the opposite side of vertex C. Step 3 Locate 3 points ARead more
Step 1
See lessDraw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5cm radius Similarly, taking point B as its centre, draw an arc of 6 cm and ΔABC is the required triangle.
Step 2
Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
Step 3
Locate 3 points A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁ = A₁A₂ = A₂A₃.
Step 4
Join BA₃ and draw a line through A₂ parallel to BA₃ to intersect AB at point B’
Step 5
Draw a line through B’ parallel to the line BC to intersect AC at C’.
ΔAB’C is the required triangle.
Justification
The construction van be justified by proving that
AB’ = 2/3 AB, B’C’ = 2/3 BC, AC’ = 2/3 AC
By construction, we have B’C’ II BC
∴ ∠AB’C’ = ∠ABC (corresponding angles)
In ΔAB’C’ and ΔABC,
∠AB’C’ = ∠ABC (Proved above)
∠B’AC’ = ∠ BAC (common)
∴ Δ AB’C’ ∼ ΔABC (AA similarity criterion)
⇒ AB’/AB = B’C’/BC = AC/AC …(1)
In Δ AA₂B’ and ΔAA₃B,
∠A₂AB’ = ∠A₃AB (common)
∠AA₂B’ = ∠AA₃B (corresponding angles)
∴ ΔAA₂B’ ∼ ΔAA₃B (AA similarity criterion)
⇒ AB’/AB = AA₂/AA₃
⇒ AB’/AB = 2/3 …(2)
From equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 2/3
⇒ AB’ = 2/3 AB, B’C’ = 2/3 BC, AC’ = 2/3 AC
This justifies the construction.