Tangents on the given circle can be drawn as follow. Step 1 Draw a circle of 4 cm radius with centre as O on the given plane. Step 2 Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle a join OP. Step 3 Bisect OP. Let M be the mid-point of PO. Step 4 Taking M as itsRead more
Tangents on the given circle can be drawn as follow.
Step 1
Draw a circle of 4 cm radius with centre as O on the given plane.
Step 2 Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle a join OP.
Step 3
Bisect OP. Let M be the mid-point of PO.
Step 4
Taking M as its centre and MO as Its radius, draw a circle. Let it intersect the given circle at the points Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 4.47 cm each.
In ΔPQO,
since PQ is a tangent,
∠PQO = 90°
PO = 6 cm
QO = 4 CM
Applying Pythagoras theorem in ΔPQO, we obtain
PQ² + QO² = PQ² ⇒ PQ² + (4)² = (6)² ⇒ PQ² + 16 = 36
PQ² = 36 – 16 ⇒ PQ² = 20 ⇒ PQ = 2√5
PQ = 4.47 cm
Justification
The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ and OR.
∠PQO is an angle in the semi- circle. We know that angle in a semi-circle is a right angle.
∴ ∠PQO = 90° ⇒ OQ ⊥PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.
A ΔA’BC’ whose sides are 3/4 of the corresponding sides of Δ ABC can be drawn as follows. Step 1 Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Step 2 Draw a ray BX making an acute angle with BC on the opposite side of vertex A. Step 3 Locat 4 points (as 4 is greater in 3 and 4), B₁, B₂,Read more
A ΔA’BC’ whose sides are 3/4 of the corresponding sides of Δ ABC can be drawn as follows.
Step 1
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
Step 2
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 3
Locat 4 points (as 4 is greater in 3 and 4), B₁, B₂, B₃, B₄, on line segment BX.
Step 4
Join B₄C and Draw a line through B₃, parallel to B₄C intersecting BC AT C’.
Step 5
Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle.
Justification
The construction can be justified by proving A’B = 3/4 AB, BC’ = 3/4 BC, A’C’ = 3/4 AC
In ΔA’BC’ and ΔABC,
∠A’C’B = ∠ACB (Corresponding angles)
∠A’BC’ = ∠ABC (Common)
∴ ΔA’BC’ ∼ ΔABC (AA similarity criterion)
⇒ A’B/AB = BC’/BC = A’C’/AC …(1)
In ΔBB₃C’ and ΔBB₄C,
∠B₃BC’ = ∠B₄BC (Common)
∠BB₃C’ = ∠BB₄C ( Corresponding angles)
∴ ΔBB₃C ∼ ΔBB₄C (AA similarity criterion)
⇒ BC’/BC = BB₃/BB₄ ⇒ BC’/BC = 3/4 …(2)
From equations (1) and (2), we obtain
A’B/AB = BC’/BC = A’C’/AC = 3/4
⇒ A’B = 3/4 AB, BC’ = 3/4 BC A’C’ = 3/4 AC
This justifies the construction.
A Pair to tangents to the given circle can be constructed as follow. Step 1: Takking any point 0 of the given plane as centre, draw a circle of 6 cm radius. Locat point P, 10 cm away from O. Join OP. Step 2 Bisect OP. Let M be the mid-point of PO. Step 3 Taking M as centre and MO as dadius, draw a cRead more
A Pair to tangents to the given circle can be constructed as follow.
Step 1:
Takking any point 0 of the given plane as centre, draw a circle of 6 cm radius. Locat point P, 10 cm away from O. Join OP.
Step 2
Bisect OP. Let M be the mid-point of PO.
Step 3
Taking M as centre and MO as dadius, draw a circle.
Step 4
Let this circle intersect the previous circle at point Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents. The lengths of tangents PQ and PR are 8 cm each.
Justification
The construction can be justified by proving that PQ and PR are the tangents to the circle ( whose centre is O and radius is 6 cm ). For this, join OQ and OR.
∠PQO is an angle in the semi- circle. We know that angle in a semi- circle is a right angle.
∴ ∠PQO = 90° ⇒ OQ ⊥PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.
Let, O be the centre and AB is tangent at B. Given : OA = 5cm and AB = 4 Cm We Know that the radius is Perpendicular to tangent. Therefore, In ΔABO, OB ⊥AB. In ΔABO, by pythagoras theorem, AB² +BO² = OA² ⇒ 4² + BO² = 5² ⇒ 16 + BO² = 25 ⇒ BO² = 9 ⇒ BO = 3 ⇒ Therefore, the radius of circle is 3 cm.
Let, O be the centre and AB is tangent at B.
Given : OA = 5cm and AB = 4 Cm
We Know that the radius is Perpendicular to tangent.
Therefore, In ΔABO, OB ⊥AB.
In ΔABO, by pythagoras theorem,
AB² +BO² = OA²
⇒ 4² + BO² = 5²
⇒ 16 + BO² = 25
⇒ BO² = 9
⇒ BO = 3
⇒ Therefore, the radius of circle is 3 cm.
Let O be the centre of two concentric circles with radius 5 cm (OP) and 3 cm (OA). PQ is chord of larger circle which is a tangent to inner circle. We know that the radius is perpendicular to tangent. Therefore, in ΔPQO, OA ⊥PQ. In ΔAPO, by Pythagoras theorem, OA² + AP² = OP² ⇒ 3² + AP² = 5² ⇒ 9 + ARead more
Let O be the centre of two concentric circles with radius 5 cm (OP) and 3 cm (OA). PQ is chord of larger circle which is a tangent to inner circle.
We know that the radius is perpendicular to tangent. Therefore, in ΔPQO, OA ⊥PQ.
In ΔAPO, by Pythagoras theorem,
OA² + AP² = OP²
⇒ 3² + AP² = 5²
⇒ 9 + AP² = 25 ⇒ AP² = 16 ⇒ AP = 4
In ΔOPQ, OA ⊥ PQ
AP = AQ [Perpendicular from the centre bisects the chord]
So, PQ = 2AP = 2 × 4 = 8
Hence, the length of chord of larger circle is 8 cm.
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.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Tangents on the given circle can be drawn as follow. Step 1 Draw a circle of 4 cm radius with centre as O on the given plane. Step 2 Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle a join OP. Step 3 Bisect OP. Let M be the mid-point of PO. Step 4 Taking M as itsRead more
Tangents on the given circle can be drawn as follow.
See lessStep 1
Draw a circle of 4 cm radius with centre as O on the given plane.
Step 2 Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle a join OP.
Step 3
Bisect OP. Let M be the mid-point of PO.
Step 4
Taking M as its centre and MO as Its radius, draw a circle. Let it intersect the given circle at the points Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 4.47 cm each.
In ΔPQO,
since PQ is a tangent,
∠PQO = 90°
PO = 6 cm
QO = 4 CM
Applying Pythagoras theorem in ΔPQO, we obtain
PQ² + QO² = PQ² ⇒ PQ² + (4)² = (6)² ⇒ PQ² + 16 = 36
PQ² = 36 – 16 ⇒ PQ² = 20 ⇒ PQ = 2√5
PQ = 4.47 cm
Justification
The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ and OR.
∠PQO is an angle in the semi- circle. We know that angle in a semi-circle is a right angle.
∴ ∠PQO = 90° ⇒ OQ ⊥PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and angle B = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
A ΔA’BC’ whose sides are 3/4 of the corresponding sides of Δ ABC can be drawn as follows. Step 1 Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Step 2 Draw a ray BX making an acute angle with BC on the opposite side of vertex A. Step 3 Locat 4 points (as 4 is greater in 3 and 4), B₁, B₂,Read more
A ΔA’BC’ whose sides are 3/4 of the corresponding sides of Δ ABC can be drawn as follows.
See lessStep 1
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
Step 2
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 3
Locat 4 points (as 4 is greater in 3 and 4), B₁, B₂, B₃, B₄, on line segment BX.
Step 4
Join B₄C and Draw a line through B₃, parallel to B₄C intersecting BC AT C’.
Step 5
Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle.
Justification
The construction can be justified by proving A’B = 3/4 AB, BC’ = 3/4 BC, A’C’ = 3/4 AC
In ΔA’BC’ and ΔABC,
∠A’C’B = ∠ACB (Corresponding angles)
∠A’BC’ = ∠ABC (Common)
∴ ΔA’BC’ ∼ ΔABC (AA similarity criterion)
⇒ A’B/AB = BC’/BC = A’C’/AC …(1)
In ΔBB₃C’ and ΔBB₄C,
∠B₃BC’ = ∠B₄BC (Common)
∠BB₃C’ = ∠BB₄C ( Corresponding angles)
∴ ΔBB₃C ∼ ΔBB₄C (AA similarity criterion)
⇒ BC’/BC = BB₃/BB₄ ⇒ BC’/BC = 3/4 …(2)
From equations (1) and (2), we obtain
A’B/AB = BC’/BC = A’C’/AC = 3/4
⇒ A’B = 3/4 AB, BC’ = 3/4 BC A’C’ = 3/4 AC
This justifies the construction.
Draw a circle of radius 6 cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths.
A Pair to tangents to the given circle can be constructed as follow. Step 1: Takking any point 0 of the given plane as centre, draw a circle of 6 cm radius. Locat point P, 10 cm away from O. Join OP. Step 2 Bisect OP. Let M be the mid-point of PO. Step 3 Taking M as centre and MO as dadius, draw a cRead more
A Pair to tangents to the given circle can be constructed as follow.
See lessStep 1:
Takking any point 0 of the given plane as centre, draw a circle of 6 cm radius. Locat point P, 10 cm away from O. Join OP.
Step 2
Bisect OP. Let M be the mid-point of PO.
Step 3
Taking M as centre and MO as dadius, draw a circle.
Step 4
Let this circle intersect the previous circle at point Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents. The lengths of tangents PQ and PR are 8 cm each.
Justification
The construction can be justified by proving that PQ and PR are the tangents to the circle ( whose centre is O and radius is 6 cm ). For this, join OQ and OR.
∠PQO is an angle in the semi- circle. We know that angle in a semi- circle is a right angle.
∴ ∠PQO = 90° ⇒ OQ ⊥PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.
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.
Let, O be the centre and AB is tangent at B. Given : OA = 5cm and AB = 4 Cm We Know that the radius is Perpendicular to tangent. Therefore, In ΔABO, OB ⊥AB. In ΔABO, by pythagoras theorem, AB² +BO² = OA² ⇒ 4² + BO² = 5² ⇒ 16 + BO² = 25 ⇒ BO² = 9 ⇒ BO = 3 ⇒ Therefore, the radius of circle is 3 cm.
Let, O be the centre and AB is tangent at B.
See lessGiven : OA = 5cm and AB = 4 Cm
We Know that the radius is Perpendicular to tangent.
Therefore, In ΔABO, OB ⊥AB.
In ΔABO, by pythagoras theorem,
AB² +BO² = OA²
⇒ 4² + BO² = 5²
⇒ 16 + BO² = 25
⇒ BO² = 9
⇒ BO = 3
⇒ Therefore, the radius of circle is 3 cm.
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.
Let O be the centre of two concentric circles with radius 5 cm (OP) and 3 cm (OA). PQ is chord of larger circle which is a tangent to inner circle. We know that the radius is perpendicular to tangent. Therefore, in ΔPQO, OA ⊥PQ. In ΔAPO, by Pythagoras theorem, OA² + AP² = OP² ⇒ 3² + AP² = 5² ⇒ 9 + ARead more
Let O be the centre of two concentric circles with radius 5 cm (OP) and 3 cm (OA). PQ is chord of larger circle which is a tangent to inner circle.
See lessWe know that the radius is perpendicular to tangent. Therefore, in ΔPQO, OA ⊥PQ.
In ΔAPO, by Pythagoras theorem,
OA² + AP² = OP²
⇒ 3² + AP² = 5²
⇒ 9 + AP² = 25 ⇒ AP² = 16 ⇒ AP = 4
In ΔOPQ, OA ⊥ PQ
AP = AQ [Perpendicular from the centre bisects the chord]
So, PQ = 2AP = 2 × 4 = 8
Hence, the length of chord of larger circle is 8 cm.
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.