# three cubes of a metal whose edges are in ratio 3:4:5 are melted and converted into a single cube whose diagonal is 12root 3 cm find the edges of the three cubes . a right triangle ,whose sides are 15cm nd 20cm,is made to revlve about its hypotenuse. find the surface area of the double cone so formed.

SOLUTION 1)

Let the edges of the three cubes = 3x, 4x and 5x

Then,

Their volume  Given : The length of the diagonal of the single cube = As we know that,

the length of the diagonal of a cube of side ⇒ edge of the single cube = 12 cm

Now,

Volume of the single cube = (12 cm)3 = 1728, cm3

Since the three cubes are melted and converted into single cube

⇒ Sum of volumes of three cubes = volume of single cube ∴ The edges of the three cubes are 6cm, 8cm and 10 cm.

• 165

1. New side of cube after melting = cube root of ((3x)3 + (4x) 3 + (5x)3 ))= 6x

Also given , diagonal of cube = 12 * root 3

=> a * root 3 = 12 * root 3

=> a = 12 cm

6x = 12

x = 2

=> Side of cubes were 3*2 , 4*2,5*2 = 6 cm, 8 cm , 10 cm

Cheers

Best of Luck

• 25

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
• 15
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