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
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