# A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypoteneous.Find the volume and the surface area of the double cone so formed. (Take pie = 3.14

Consider the following right angled triangle ABC is rotated through its hypotenuse AC

BD ⊥ AC. In this case BD is the radius of the double cone generated.
Using Pythagoras theorem for ∆ABC it is obtained
AC2 = AB2 + BC2 = (15 cm)2 + (20 cm)2 = 225 cm2 + 400 cm2 = 625 cm2 = (25 cm)2
⇒ AC = 25 cm
∴CD = (25 – x) cm

Using Pythagoras theorem in ∆ABD and ∆CBD
AD2 +BD2 = AB2 and BD2 + CD2 = BC2
⇒ x 2 + BD2 = 152 and BD2 + (25 – x)2 = 202
⇒ BD2 = 152 – x2 and BD2 = 202 – (25 – x)2
⇒ 152 – x 2 = 202 – (25 – x)2
⇒ 225 – x 2 = 400 – (625 + x 2 – 50x)
⇒ 225 – x 2 = – 225 – x 2 + 50x
⇒ 50x = 450
⇒ x = 9
⇒ BD2 = 152 – 92 = 225 – 81 = 144
⇒ BD = 12 cm

Surface area of the double cone formed
= L.S.A of upper cone + L.S.A of the lower cone
= Π (BD) × (AB) + Π (BD) × BC
= Π × 12 cm × 15 cm + Π × 12 cm × 20 cm
= 420 Π cm2

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Thank You.
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The correct ans is -
Volume of double cone - 3768 cube. cm
Surface area of Cone - 1318.8 sq. cm
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multiply it it 3.14
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multiply it it 3.14
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abhinandan joshi pls don't answer without reading the question properly. dash ga
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In right BAC, by Pythagoras theorem BC2 = AB2 + AC2 = 152 + 202 = 225 + 400 = 625 BC = 25cm   Let OA = y cm and OB = x cm x 2 +y2 = 152 [1/2] (25-x)2 + y2 = 202 [1/2] Solving we get x=9 and y=12   OA= 12 cm and OB = 9 cm Volume of double cone =  (OA)2 X OC +  (OA)2 X OB = X 3.14 X (12)2 X (OC + OB) [1/2] = X 3.14 X 144 X 25 = 3768 cm3  Surface area of double cone =  X OA X AC +  X OA X AB =  X 12 X 20 +  X 12 X 15 = 420  cm2 = 1318.8 cm2
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oops!I m sorry i just resend it
=(
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Your answer is wrong 15 square is 225 but u have written as 152 so plz check ur answer it's wrong
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will someone explain me the diagram for thes sum
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square of 15 is 225, not 152
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thanks a lot guys!!!!!
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How do you prove this using similarity??
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FAIZAN COOLBOII 15(square) is 225 not 152!!!

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420 TT cm2
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ek dam bakwaas tha

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It is v difficult question....😣
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420 πcm²
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Correct ans must be multiplid by 3.14
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இது ரோம்பவே கடிணம்
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Ans is 420 pie
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1245
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Consider the following right angled triangle ABC is rotated through its hypotenuse AC

?

?

BD ? AC. In this case BD is the radius of the double cone generated.

Using Pythagoras theorem for ?ABC it is obtained

AC2?= AB2?+ BC2?= (15 cm)2?+ (20 cm)2?= 225 cm2?+ 400 cm2?= 625 cm2?= (25 cm)2

? AC = 25 cm

?CD = (25 ??x) cm

?

Using Pythagoras theorem in ?ABD and ?CBD

??x?2?+ BD2?= 152 and BD2?+ (25 ??x)2?= 202

? BD2?= 152 ? x2?and BD2?= 202?? (25 ? x)2

? 152 ??x?2?= 202 ? (25 ??x)2

? 225 ??x?2?= 400 ? (625 +?x?2?? 50x)

? 225 ??x?2?= ? 225 ??x?2?+ 50x

? 50x?= 450

??x?= 9

? BD2?= 152?? 92?= 225 ? 81 = 144

? BD = 12 cm

?

?
?

?

Surface area of the double cone formed

= L.S.A of upper cone + L.S.A of the lower cone

= ? (BD) ? (AB) + ? (BD) ? BC

= ? ? 12 cm ? 15 cm + ? ? 12 cm ? 20 cm

= 420 ? cm2

?

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602.88cm2
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What type of motion
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Jbfxuncun
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BD ⊥ AC. In this case BD is the radius of the double cone generated.
Using Pythagoras theorem for ∆ABC it is obtained
AC2 = AB2 + BC2 = (15 cm)2 + (20 cm)2 = 225 cm2 + 400 cm2 = 625 cm2 = (25 cm)2
⇒ AC = 25 cm
∴CD = (25 – x) cm

Using Pythagoras theorem in ∆ABD and ∆CBD
AD2 +BD2 = AB2 and BD2 + CD2 = BC2
⇒ x 2 + BD2 = 152 and BD2 + (25 – x)2 = 202
⇒ BD2 = 152 – x2 and BD2 = 202 – (25 – x)2
⇒ 152 – x 2 = 202 – (25 – x)2
⇒ 225 – x 2 = 400 – (625 + x 2 – 50x)
⇒ 225 – x 2 = – 225 – x 2 + 50x
⇒ 50x = 450
⇒ x = 9
⇒ BD2 = 152 – 92 = 225 – 81 = 144
⇒ BD = 12 cm

Surface area of the double cone formed
= L.S.A of upper cone + L.S.A of the lower cone
= Π (BD) × (AB) + Π (BD) × BC
= Π × 12 cm × 15 cm + Π × 12 cm × 20 cm
= 420 Π cm2

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Sample question papersof maths half yearly class 10 solved
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98.56
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