20) D

It is advisable to skip such long and tricky questions and finish the other simpler ones first.

We need to use a lot of Pythagoras in this question.

Side of smaller cube = 1cm

Side of 2nd cube = A

Side of 3rd cube = B

5 faces of the smallest cube can be seen.

**Surface area of the smallest cube = 1 x 1 x 5 = 5 cm²**

The vertices of the smallest cube are at the midpoints of A.

You can see that multiple right-angled triangles are formed.

1² = (½A)² + (½A)²

1 = ¼A² + ¼a²

1 = ½a²

2 = A²

A = √2

4 faces of the 2nd cube can be seen and the top face is partially covered by the smallest cube.

Surface area = 5 x √2 x √2 = 10cm²

Area covered by one of the faces of the smaller cube = 1 x 1 = 1cm²

So total surface area of the 2nd cube = 10 – 1 = 9cm²

Again the vertices of the second cube are at the midpoints of B.

You can see that multiple right-angled triangles are formed.

(√2)² = (½B)² + (½B)²

2 = ¼B² + ¼B²

2 = ½B²

4 = B²

B = √4 = 2cm

5 faces of the largest cube are uncovered and 1 is partially covered.

Surface area = 6 x 2 x 2 = 24cm²

Again the top face is partially covered by the 2nd cube.

Surface area of one of the faces of the 2nd cube = √2 x √2 = 2cm²

Total surface area of largest cube = 24 – 2 = 22cm²

Total surface area of solid = 22 + 9 + 5 = 36 cm²