The required tangents can be constructed on the given circle as follows. Step 1 Draw a circle with the help of a bangle. Step 2 Take a point P outside this circle and take two chords QR and ST. Step 3 Draw perpendicular bisectors of these chords. Let them intersect each other at point O. Step 4 JoinRead more
The required tangents can be constructed on the given circle as follows.
Step 1
Draw a circle with the help of a bangle.
Step 2
Take a point P outside this circle and take two chords QR and ST.
Step 3
Draw perpendicular bisectors of these chords. Let them intersect each other at point O.
Step 4
Join PO and bisect it. Let U be the mid-point of OP. Taking U as centre, draw a circle at V and W. Join PV and PW.
PV and PW are the required tangents.
Justification
The construction can be justified by proving that PV and PW are the tangents to the circle. For this, first of all, it has to proved that O is the centre of the circle. Let us join OV and OW.
We know that perpendicular bisector of a chord passes through the centre. Therefore, the perpendicular bisector of chords QR and ST pass through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. ∠PVO is an angle in the semi-circle. We know that an angle in the semi- circle is a right angle.
∴ ∠PVO = 90°
⇒ OV ⊥ PV
Since OV is the radius of the circle, PV has to be a tangent of the circle. Similarly, PW is a tangent of the circle.
Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm. A ΔAB'C' whose sides are 3/2 times of ΔABC can be drawn as follows. Step 1 Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment wRead more
Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.
A ΔAB’C’ whose sides are 3/2 times of ΔABC can be drawn as follows.
Step 1
Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre.
Let these arcs intersect each other at O and O’. Join OO’. Let OO’ intersect AB at D.
Step 2
Taking D as centre, Draw an arc of 4 cm radius which cuts the extended line segment OO’ at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
Step 3
Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
Step 4
Locate 3 points (as 3 is greater between 3 and 2) A₁, A₂, and A₃ on AX such that AA₁ = A₁ A₂ = A₂ A₃
Step 5
Join BA₂ and draw a line through A₁ parallel to BA₂ to intersect extended line segment AB at point B’.
Step 6
Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C is the required triangle.
Justification
The construction can be justified by proving that
AB’ = 3/2 AB, B’C’ = 3/2 BC, AC’ = 3/2 AC
In ΔABC and ΔAB’C’,
∠ABC and ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (AA similarity criterion)
⇒ AB’/AB = B’C’/BC = A’C’/AC …(1)
In ΔAA₂B = ΔAA₃B’
∠AA₂B = ∠AA₃B’ (Common)
∠AA₂B = AA₃B’ (Corresponding angles)
∴ ΔAA₂B ∼ AA₃B’ (AA Similarity Criterion)
⇒ AB/AB’ = AA₂/AA₃ ⇒ AB/AB’ = 2/3 …(2)
On comparing equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 3/2
⇒ AB’ = 3/2 AB,B’C’ = 3/2 BC, AC’ = 3/2 AC
This justifies the construction.
It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other. The required triangle can be drawn as follows. Step 1 Draw a line segment AB = 4 cm. Draw a ray SA Making 90° with it. Step 2 Draw an arc of 3 cm radius while taking A as itRead more
It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other.
The required triangle can be drawn as follows.
Step 1
Draw a line segment AB = 4 cm. Draw a ray SA Making 90° with it.
Step 2
Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.
Step 3
Draw a ray AX making an acute angle with AB, opposite to vertex C.
Step 4
Locate 5 points (as 5 is greater in 5 and 3 ), A₁, A₂, A₃, A₄, A₅, on line segment AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅
Step 5
Join A₃B. Draw a line through A₅ parallel to AB intersecting extended line segment AB at B’.
Step 6
Through B’, Draw a line parallel to BC intersecting extended line segment AC at C’.
ΔAB’C’ is the required triangle.
Justification
The construction can be justified by proving that AB’ = 5/3 AB, B’C’ = 5/3 BC, AC’ = 5/3 AC
In ΔABC and ΔAB’C’,
∠ABC = ∠AB’C’ (corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔABC ∼ ΔAB’C’ (AA similarity criterion)
⇒ AB/AB’ = BC / B’C’ = AC/AC’ …(1)
In ΔAA₃B and ΔAA₅B’,
∠A₃AB = ∠A₅AB’ (common)
∠AA₃B = ∠AA₅B’ (corresponding angles)
∴ ΔAA₃B ∼ ΔAA₅B’ (AA similarity criterion)
⇒ AB/AB, = AA₃/AA₅ ⇒ AB/AB’ = 3/5 …(2)
On comparing equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 5/3
⇒ AB’ = 5/3 AB, B’C’ = 5/3 BC, AC’ = 5/3 AC
This justifies the construction.
Given that : Volume of cube = 64 cm³ ⇒ (Side)³ = 64 cm³ ⇒ Side = 4 cm The sides of cuboids formed by joining the cubes are 4 cm, 4 cm and 8 cm. Surface area of resulting cuboid = 2 (Ib + bh +hl) =2(4×4 + 4×8 +4×8)cm² = 2(16+32+32)cm² = 2(80) cm² = 160 cm² Hence, the surface area of the resulting cubRead more
Given that : Volume of cube = 64 cm³
⇒ (Side)³ = 64 cm³ ⇒ Side = 4 cm
The sides of cuboids formed by joining the cubes are 4 cm, 4 cm and 8 cm.
Surface area of resulting cuboid
= 2 (Ib + bh +hl)
=2(4×4 + 4×8 +4×8)cm²
= 2(16+32+32)cm²
= 2(80) cm²
= 160 cm²
Hence, the surface area of the resulting cuboid 160 cm².
Radius of cylinder = 7 cm Height of cylinder = 13 - 7 = 6 cm Radius of hemi- sphere = 7 cm Inner surface area of the vessel = CSA of Cylinder + CSA of hemisphere = 2πrh +2πr² = 2 × 22/7 ×7×6 + 2 ×22/7 ×7² = 44(6 + 7) = 44 × 13 = 572cm² Hence, the inner surface area of the vessel is 572 cm²
Radius of cylinder = 7 cm
Height of cylinder = 13 – 7 = 6 cm
Radius of hemi- sphere = 7 cm
Inner surface area of the vessel = CSA of Cylinder + CSA of hemisphere
= 2πrh +2πr²
= 2 × 22/7 ×7×6 + 2 ×22/7 ×7²
= 44(6 + 7)
= 44 × 13
= 572cm²
Hence, the inner surface area of the vessel is 572 cm²
The tangents can be constructed on the given circles as follows. Step 1 Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius. Step 2 Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect theRead more
The tangents can be constructed on the given circles as follows.
Step 1
Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius.
Step 2
Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AS, And AR. These are the required tangents.
Justification
The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.
∠ASB is an angle in the semi-circle. We Know that an angle in a semi-circle is a right angle.
∴ ∠ASB = 90° ⇒ BS ⊥AS
Since BS is the radius of the circle, As has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.
The tangents can be constructed in the following manner: Step 1 Draw a circle of radius 5 cm and with centre as O. Step 2 Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A. Step 3 Draw a radius OB, making an angle of 120° (180° - 60°) with OA. StepRead more
The tangents can be constructed in the following manner:
Step 1
Draw a circle of radius 5 cm and with centre as O.
Step 2
Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.
Step 3
Draw a radius OB, making an angle of 120° (180° – 60°) with OA.
Step 4
Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°.
Justification
The construction can be justified by proving that ∠APB = 60°
By our construction
∠OAP = 90° ∠OBP = 90° and ∠AOB = 120°
We know that the sum of all interior angles of a quadrilateral = 360°
∠OAP + ∠AOB + ∠OBP + ∠ABP = 360° ⇒ 90° + 120° + 90° + ∠APB = 360°
∠APB = 60°
This justifies the construction.
The tangent can be constructed on the given circle as follow. Step 1 Taking any point O on the given plane as centre, draw a circle of 3 cm radius. Step 2 Take one of its diameters, PQ and extent it on both sides. Locate two points on this diameter surch that OR = OS = 7 cm Step 3 Bisect OR and OS.Read more
The tangent can be constructed on the given circle as follow.
Step 1
Taking any point O on the given plane as centre, draw a circle of 3 cm radius.
Step 2
Take one of its diameters, PQ and extent it on both sides. Locate two points on this diameter surch that OR = OS = 7 cm
Step 3
Bisect OR and OS. Let T and U be the mid-points of OR and OS respectively.
Step 4
Taking T and U as its centre and with TO and UO as radius, draw two circles. These two circles will intersect the circle at point V, W, X, Y respectively. join RV, RW, SX, and SY. These are the required tangents.
Justification
The construction can be justified by proving that RV, RW, SY, and SX are the tangents to the circle (whose centre is O and radius 3 cm). For this, join OV, OW, OX, and OY.
∠RVO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.
∴ ∠RVO = 90° ⇒ OV⊥RV
Since OV is the radius of the circle, RV has to be a tangent of the circle. Similarly, OW, OX, and OY are the tangents of the circle.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
The required tangents can be constructed on the given circle as follows. Step 1 Draw a circle with the help of a bangle. Step 2 Take a point P outside this circle and take two chords QR and ST. Step 3 Draw perpendicular bisectors of these chords. Let them intersect each other at point O. Step 4 JoinRead more
The required tangents can be constructed on the given circle as follows.
See lessStep 1
Draw a circle with the help of a bangle.
Step 2
Take a point P outside this circle and take two chords QR and ST.
Step 3
Draw perpendicular bisectors of these chords. Let them intersect each other at point O.
Step 4
Join PO and bisect it. Let U be the mid-point of OP. Taking U as centre, draw a circle at V and W. Join PV and PW.
PV and PW are the required tangents.
Justification
The construction can be justified by proving that PV and PW are the tangents to the circle. For this, first of all, it has to proved that O is the centre of the circle. Let us join OV and OW.
We know that perpendicular bisector of a chord passes through the centre. Therefore, the perpendicular bisector of chords QR and ST pass through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. ∠PVO is an angle in the semi-circle. We know that an angle in the semi- circle is a right angle.
∴ ∠PVO = 90°
⇒ OV ⊥ PV
Since OV is the radius of the circle, PV has to be a tangent of the circle. Similarly, PW is a tangent of the circle.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1(1/2) times the corresponding sides of the isosceles triangle.
Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm. A ΔAB'C' whose sides are 3/2 times of ΔABC can be drawn as follows. Step 1 Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment wRead more
Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.
See lessA ΔAB’C’ whose sides are 3/2 times of ΔABC can be drawn as follows.
Step 1
Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre.
Let these arcs intersect each other at O and O’. Join OO’. Let OO’ intersect AB at D.
Step 2
Taking D as centre, Draw an arc of 4 cm radius which cuts the extended line segment OO’ at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
Step 3
Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
Step 4
Locate 3 points (as 3 is greater between 3 and 2) A₁, A₂, and A₃ on AX such that AA₁ = A₁ A₂ = A₂ A₃
Step 5
Join BA₂ and draw a line through A₁ parallel to BA₂ to intersect extended line segment AB at point B’.
Step 6
Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C is the required triangle.
Justification
The construction can be justified by proving that
AB’ = 3/2 AB, B’C’ = 3/2 BC, AC’ = 3/2 AC
In ΔABC and ΔAB’C’,
∠ABC and ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (AA similarity criterion)
⇒ AB’/AB = B’C’/BC = A’C’/AC …(1)
In ΔAA₂B = ΔAA₃B’
∠AA₂B = ∠AA₃B’ (Common)
∠AA₂B = AA₃B’ (Corresponding angles)
∴ ΔAA₂B ∼ AA₃B’ (AA Similarity Criterion)
⇒ AB/AB’ = AA₂/AA₃ ⇒ AB/AB’ = 2/3 …(2)
On comparing equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 3/2
⇒ AB’ = 3/2 AB,B’C’ = 3/2 BC, AC’ = 3/2 AC
This justifies the construction.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other. The required triangle can be drawn as follows. Step 1 Draw a line segment AB = 4 cm. Draw a ray SA Making 90° with it. Step 2 Draw an arc of 3 cm radius while taking A as itRead more
It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other.
See lessThe required triangle can be drawn as follows.
Step 1
Draw a line segment AB = 4 cm. Draw a ray SA Making 90° with it.
Step 2
Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.
Step 3
Draw a ray AX making an acute angle with AB, opposite to vertex C.
Step 4
Locate 5 points (as 5 is greater in 5 and 3 ), A₁, A₂, A₃, A₄, A₅, on line segment AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅
Step 5
Join A₃B. Draw a line through A₅ parallel to AB intersecting extended line segment AB at B’.
Step 6
Through B’, Draw a line parallel to BC intersecting extended line segment AC at C’.
ΔAB’C’ is the required triangle.
Justification
The construction can be justified by proving that AB’ = 5/3 AB, B’C’ = 5/3 BC, AC’ = 5/3 AC
In ΔABC and ΔAB’C’,
∠ABC = ∠AB’C’ (corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔABC ∼ ΔAB’C’ (AA similarity criterion)
⇒ AB/AB’ = BC / B’C’ = AC/AC’ …(1)
In ΔAA₃B and ΔAA₅B’,
∠A₃AB = ∠A₅AB’ (common)
∠AA₃B = ∠AA₅B’ (corresponding angles)
∴ ΔAA₃B ∼ ΔAA₅B’ (AA similarity criterion)
⇒ AB/AB, = AA₃/AA₅ ⇒ AB/AB’ = 3/5 …(2)
On comparing equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 5/3
⇒ AB’ = 5/3 AB, B’C’ = 5/3 BC, AC’ = 5/3 AC
This justifies the construction.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Given that : Volume of cube = 64 cm³ ⇒ (Side)³ = 64 cm³ ⇒ Side = 4 cm The sides of cuboids formed by joining the cubes are 4 cm, 4 cm and 8 cm. Surface area of resulting cuboid = 2 (Ib + bh +hl) =2(4×4 + 4×8 +4×8)cm² = 2(16+32+32)cm² = 2(80) cm² = 160 cm² Hence, the surface area of the resulting cubRead more
Given that : Volume of cube = 64 cm³
See less⇒ (Side)³ = 64 cm³ ⇒ Side = 4 cm
The sides of cuboids formed by joining the cubes are 4 cm, 4 cm and 8 cm.
Surface area of resulting cuboid
= 2 (Ib + bh +hl)
=2(4×4 + 4×8 +4×8)cm²
= 2(16+32+32)cm²
= 2(80) cm²
= 160 cm²
Hence, the surface area of the resulting cuboid 160 cm².
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Radius of cylinder = 7 cm Height of cylinder = 13 - 7 = 6 cm Radius of hemi- sphere = 7 cm Inner surface area of the vessel = CSA of Cylinder + CSA of hemisphere = 2πrh +2πr² = 2 × 22/7 ×7×6 + 2 ×22/7 ×7² = 44(6 + 7) = 44 × 13 = 572cm² Hence, the inner surface area of the vessel is 572 cm²
Radius of cylinder = 7 cm
See lessHeight of cylinder = 13 – 7 = 6 cm
Radius of hemi- sphere = 7 cm
Inner surface area of the vessel = CSA of Cylinder + CSA of hemisphere
= 2πrh +2πr²
= 2 × 22/7 ×7×6 + 2 ×22/7 ×7²
= 44(6 + 7)
= 44 × 13
= 572cm²
Hence, the inner surface area of the vessel is 572 cm²
Draw a line segment AB of length 8 cm. Taking A as Centre, draw a circle of radius 4 cm and taking B as Centre, draw another circle of radius 3 cm. Construct tangents to each circle from the Centre of the other circle.
The tangents can be constructed on the given circles as follows. Step 1 Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius. Step 2 Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect theRead more
The tangents can be constructed on the given circles as follows.
See lessStep 1
Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius.
Step 2
Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AS, And AR. These are the required tangents.
Justification
The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.
∠ASB is an angle in the semi-circle. We Know that an angle in a semi-circle is a right angle.
∴ ∠ASB = 90° ⇒ BS ⊥AS
Since BS is the radius of the circle, As has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
The tangents can be constructed in the following manner: Step 1 Draw a circle of radius 5 cm and with centre as O. Step 2 Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A. Step 3 Draw a radius OB, making an angle of 120° (180° - 60°) with OA. StepRead more
The tangents can be constructed in the following manner:
See lessStep 1
Draw a circle of radius 5 cm and with centre as O.
Step 2
Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.
Step 3
Draw a radius OB, making an angle of 120° (180° – 60°) with OA.
Step 4
Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°.
Justification
The construction can be justified by proving that ∠APB = 60°
By our construction
∠OAP = 90° ∠OBP = 90° and ∠AOB = 120°
We know that the sum of all interior angles of a quadrilateral = 360°
∠OAP + ∠AOB + ∠OBP + ∠ABP = 360° ⇒ 90° + 120° + 90° + ∠APB = 360°
∠APB = 60°
This justifies the construction.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its Centre. Draw tangents to the circle from these two points P and Q.
The tangent can be constructed on the given circle as follow. Step 1 Taking any point O on the given plane as centre, draw a circle of 3 cm radius. Step 2 Take one of its diameters, PQ and extent it on both sides. Locate two points on this diameter surch that OR = OS = 7 cm Step 3 Bisect OR and OS.Read more
The tangent can be constructed on the given circle as follow.
See lessStep 1
Taking any point O on the given plane as centre, draw a circle of 3 cm radius.
Step 2
Take one of its diameters, PQ and extent it on both sides. Locate two points on this diameter surch that OR = OS = 7 cm
Step 3
Bisect OR and OS. Let T and U be the mid-points of OR and OS respectively.
Step 4
Taking T and U as its centre and with TO and UO as radius, draw two circles. These two circles will intersect the circle at point V, W, X, Y respectively. join RV, RW, SX, and SY. These are the required tangents.
Justification
The construction can be justified by proving that RV, RW, SY, and SX are the tangents to the circle (whose centre is O and radius 3 cm). For this, join OV, OW, OX, and OY.
∠RVO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.
∴ ∠RVO = 90° ⇒ OV⊥RV
Since OV is the radius of the circle, RV has to be a tangent of the circle. Similarly, OW, OX, and OY are the tangents of the circle.