NCERT Solutions Class 8 Maths Chapter 4 2026-27

Let the midpoint joins split each side into two equal parts. Each corner triangle has two equal sides and a 60^ degrees included angle, so it is equilateral. The middle triangle has sides joining midpoints, so each side is parallel to a side of the original triangle and has the same length as the other two midpoint segments.

Answer: The four small triangles are congruent equilateral triangles.

Start with an equilateral triangle. Join the midpoints of its sides and remove the central triangle. For Step 2, repeat the same midpoint construction inside each of the three remaining corner triangles.

Answer: Step 0 is one equilateral triangle. Step 1 removes the middle triangle. Step 2 repeats the same removal in each remaining triangle.

Each remaining triangle gives three smaller remaining triangles at the next step, so R_n=3^n. One new hole is created in every triangle that remained at the previous step, so H_n=1+3+...+3^n-1=(3^n-1)/2.

Answer: At Step n, remaining triangles are 3^n and holes are 1+3+...+3^n-1=(3^n-1)/2.

For the carpet, each step keeps 8 of 9 equal parts, so the remaining area is (8/9)^n. For the triangle, each step keeps 3 of 4 equal parts, so the remaining area is (3/4)^n.

Answer: Carpet: (8/9)^n. Triangle: (3/4)^n.

At Step 0 there are 3 sides. At each next step, every side is replaced by 4 smaller sides. Hence the count is 3, 12, 48, ..., or 34^n.

Answer: The number of sides is 3 4^n.

At each step, every side becomes 4 sides, each one-third as long. So the perimeter is multiplied by 4/3 every step. Starting from 3, we get 3(4/3)^n.

Answer: The perimeter is 3(4/3)^n units.

The four cuts join consecutive side midpoints. These four midpoint-join segments form a diamond-shaped square inside the original square. Its sides are equal and adjacent sides are perpendicular.

Answer: A smaller square is left in the middle.

Each corner cut removes a small equilateral triangle. The boundary left has six equal short sides, so the remaining central shape is a regular hexagon.

Answer: A regular hexagon is left.

Cutting the four corners replaces each corner by one new slant side. The four middle parts of the original sides remain. Thus the boundary has 4+4=8 sides, so it is an octagon.

Answer: An octagon is left.

When we look straight at one face of a cube, the visible outline is exactly that square face.

Answer: A cube viewed directly from one face has a square profile.

A sphere looks circular from every direction. A cylinder also looks circular when viewed from the top or bottom along its axis.

Answer: A sphere, or a cylinder viewed along its axis, has a circular profile.

The end face of a triangular prism is a triangle. Looking straight at that end hides the length and shows the triangular outline.

Answer: A triangular prism viewed from one end has a triangular profile.

A cylinder viewed from the side gives a rectangle. The same cylinder viewed along its axis gives a circle.

Answer: A cylinder works.

A cone viewed from the top gives a circle. The same cone viewed from the side gives a triangle.

Answer: A cone works.

Viewed along its length, a triangular prism gives a triangle. Viewed from a rectangular side, it gives a rectangle.

Answer: A triangular prism works.

There are two n-sided bases and n side faces, so faces =n+2. Each base contributes n edges and there are n joining edges, so edges =3n. There are n vertices on each base, so vertices =2n.

Answer: For a 10-sided prism: 12 faces, 30 edges, 20 vertices. For an n-sided prism: n+2 faces, 3n edges, 2n vertices.

The pyramid has one base and n triangular side faces. Hence faces =n+1. The base has n edges and the apex joins to n base vertices, so edges =2n. The base has n vertices plus the apex, so vertices =n+1.

Answer: For a 10-sided pyramid: 11 faces, 20 edges, 11 vertices. For an n-sided pyramid: n+1 faces, 2n edges, n+1 vertices.

Test a net by choosing one square as the base and folding neighbouring squares around its four edges. A valid net must produce six distinct faces: top, bottom, left, right, front, and back.

Answer: Any arrangement of 6 connected squares that folds so no two squares overlap on the same face is a cube net.

The four rectangles in a strip make the side belt. The two remaining rectangles are the top and bottom faces. Their dimensions must match the unused pair 5 cm by 1 cm.

Answer: One valid net has four side rectangles in a row: 5 by 3, 1 by 3, 5 by 3, 1 by 3, with two 5 by 1 rectangles attached to one 5 by 3 rectangle.

A cuboid has opposite equal faces. Use a strip for the four side faces, then attach the two congruent top and bottom faces of size 6 cm by 2 cm.

Answer: One valid net has side rectangles 6 by 3, 2 by 3, 6 by 3, 2 by 3 in a row, with two 6 by 2 rectangles attached.

Unrolling the curved surface keeps the cylinder height as one side. The other side equals the circumference of the circular base, which is 2pi r.

Answer: The rectangle is h by 2pi r.

Every point on the base circle is at the same slant distance from the vertex. After unrolling, that slant distance becomes the radius of a circular sector.

Answer: It is a sector of a circle, together with the circular base.

A path on the cuboid surface unfolds into a path of the same length on a net. In the plane, the shortest path between two points is the straight line. We compare suitable unfoldings and pick the shortest straight segment.

Answer: Unfold the relevant faces into a plane and draw the straight line between the two points.

Using the Baudhayana theorem, d^2=24^2+32^2=576+1024=1600. Hence d=40 cm.

Answer: The distance is 40 cm.

If the actual length is l and the projected length is p, the projection forms a right triangle with hypotenuse l and leg p. Hence p<= l. Equality occurs when there is no tilt away from the plane.

Answer: The projected length is never greater than the actual length. It is equal only when the line is parallel to the plane.

Projection keeps parallel lines parallel. Since opposite sides of a parallelogram are parallel, their projections stay parallel in pairs.

Answer: No, its projection remains a parallelogram, unless it collapses into a line segment.

A single view may match many different solids. Front, top, and side views record height, length, and depth information from three directions.

Answer: One projection can hide depth, so three perpendicular views give more complete information.

When the light source is very far away, rays reaching the object are almost parallel. That makes the shadow behave like a perpendicular projection.

Answer: Sunlight is nearly parallel, so a shadow made on a perpendicular plane is close to a projection.

Start with the bottom layer and count every occupied position. Then count the next layer above it, and so on. Any cube in an upper layer must have a position below it in a stable stack.

Answer: Count layer by layer, including hidden cubes needed to support upper cubes.

When a cube is oriented so its three edge directions make equal projected scales, the drawing shows equal unit lengths along height, length, and depth axes.

Answer: It is a projection in which the projected lengths of all cube edges are equal.

The grid has three families of parallel lines. These match the three main directions of a solid. Equal grid steps let us count units along each direction.

Answer: It gives three equal-length drawing directions for height, length, and depth.

Place the first cube, then attach each new cube along a full face. If rotations and flips are not counted again, the familiar five flat tetromino arrangements appear. Three-dimensional arrangements add more possibilities.

Answer: There are several tetromino shapes: straight, square, T, L, and skew forms, with rotations and flips treated as the same shape.

The eye reads each corner as a valid cubical edge arrangement. Globally, the near and far relationships contradict each other, so an actual cube model cannot be built.

Answer: Each small part follows local isometric directions, but the parts cannot join consistently in three-dimensional space.