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.
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.