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.
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.