Exploring Some Geometric Themes Class 8 Solutions

Step 0 is one filled square. For Step 1, divide it into a 3×3 grid and remove the centre. For Step 2, repeat that operation independently in each of the eight remaining squares.

Each remaining square creates eight at the next step, so R_n=8^n. Also H_{n+1}=H_n+R_n, with H_0=0. Hence H_n=1+8+\cdots+8^{n-1}=(8^n-1)/7.

Each midpoint segment is parallel to the third side and half its length. Every small triangle therefore has three sides equal to half the original side, so all four are equilateral and congruent by SSS.

Step 0 is one filled equilateral triangle. At Step 1 join the side midpoints and remove the central inverted triangle. At Step 2 repeat this in each of the three remaining corner triangles.

At Step n, R_n=3^n. Each remaining triangle creates one new hole, so H_n=1+3+\cdots+3^{n-1}=(3^n-1)/2. Thus (R,H) begins (1,0),(3,1),(9,4),(27,13).

The Carpet retains 8/9 of each surviving area per step, so A_n=(8/9)^n. The Triangle retains 3/4, so A_n=(3/4)^n.

Start with an equilateral triangle. Replace the middle third of every side by the other two sides of an outward equilateral triangle. Repeat the same four-segment replacement on every new side for Step 2.

Step 0 has 3 sides and every side becomes 4 sides. Therefore N_n=3\cdot4^n.

Each new side is one-third as long while the count is multiplied by 4. Hence P_n=3(4/3)^n. The perimeter grows without bound.

Reverse the visible letter order, not the sounds. For example, MATHWISE becomes ESIWHTAM. The result for the learner depends on the chosen name.

The remaining figure is a smaller square rotated 45 degrees. The four removed congruent isosceles right triangles can be arranged with their right-angle vertices together; their hypotenuses form the boundary of another square.

A hexagon remains. For an equilateral starting triangle, all six new boundary segments have equal length and the angles are 120 degrees, so it is a regular hexagon.

An octagon remains. It is not regular: the four horizontal/vertical sides have length one-third of the original side, while the four sloping cuts have length \sqrt2/3 of it.

Look straight at a face of a cube. Its outline is a square. A square prism viewed along its axis is another example.

A sphere has a circular profile from every direction. A cylinder viewed along its axis also has a circular outline.

A cone viewed from the side has a triangular outline. A triangular prism viewed along its length is another example.

A right circular cylinder: viewed from the side its profile is a rectangle, and viewed along its axis it is a circle.

A right circular cone: the base view is a circle and a side view through its axis is a triangle.

A triangular prism: viewed along its length it is a triangle; viewed perpendicular to a rectangular lateral face it is a rectangle.

A frustum of a right circular cone: viewed along its axis it is circular, while an axial side view is an isosceles trapezium.

A pentagonal prism: its end view is a pentagon and a side view is a rectangle.

No. Profiles lose depth information. Dimensions, slants, hollows, and added features hidden behind the same outline can vary, so multiple solids can share the stated profiles.

An n-gonal prism has n+2 faces, 3n edges, and 2n vertices. For n=10, these are 12 faces, 30 edges, and 20 vertices.

An n-gonal pyramid has n+1 faces, 2n edges, and n+1 vertices. For n=10, these are 11 faces, 20 edges, and 11 vertices.

Arrangements (ii), (iii), (iv), and (vi) fold into a cube. In (i) and (v), two squares try to occupy the same cube face.

Enumerate connected arrangements of six unit squares, discard any arrangement whose folded faces overlap, then identify rotations and reflections as the same. The 11 survivors are the complete set; the crop provides the square-grid format for drawing them.

For each, use a strip of the four lateral rectangles in alternating dimensions, then attach the two congruent base rectangles on opposite usable edges. The first needs pairs of 5×3, 5×1, and 3×1 faces; the second needs pairs of 6×3, 6×2, and 3×2 faces.

The first and third displayed arrangements are the two net types: one central equilateral triangle with three attached, and a chain of four alternating equilateral triangles. The other displayed arrangements create overlap or invalid face adjacency.

Choose a side length s. Draw four congruent equilateral triangles of side s, either one with three attached to its sides or a four-triangle alternating chain. Fold along the shared edges.

Draw a square of side s. Attach one congruent isosceles triangle to each side; all four equal triangle sides must equal the desired slant edges. Fold the triangles upward so their apex vertices meet.

For radius r and height h, the curved surface unfolds to a rectangle of height h and width equal to the base circumference 2\pi r. Add two circles of radius r.

For base radius r and slant height l, the curved surface is a sector of radius l whose arc length is 2\pi r, together with the base circle. Its sector angle is 360^\circ r/l.

Joining the straight edges produces an oblique cone-like surface: the apex is not vertically above the centre of the circular base. A physical paper model makes the unequal slant directions visible.

Draw three rectangles in a strip, with widths equal to the three triangle side lengths and common height equal to the prism length. Attach two congruent triangles to suitable opposite edges of the strip.

The eight equilateral triangles form two groups of four meeting at opposite apexes. Fold the triangles around the central chain so corresponding free edges meet; this makes two square pyramids joined base-to-base.

No exact finite polygonal net can cover a sphere without stretching, wrinkling, gaps, or overlap because a sphere has non-zero Gaussian curvature while paper is flat. Narrow gores only approximate it.

Unfold every crossed face into one plane. A surface path keeps the same length on the net, so the shortest valid path is a straight segment between the corresponding points. A bent segment on a chosen net cannot be shortest if a valid straight segment exists.

Use the valid unfolding shown. The two points are separated by 4 cm across the joined faces and 2 cm vertically, so the straight net distance is \sqrt{4^2+2^2}=2\sqrt5 cm. The earlier blue segment leaves its net and is not a valid surface route.

Compare the distinct valid unfoldings. One route is 42 cm. The displayed diagonal unfolding gives d=\sqrt{24^2+32^2}=\sqrt{1600}=40 cm, which is the minimum among the distinct cases.

In the right triangle from the construction, l is the hypotenuse and p is a leg, so 0\le p\le l. Equality holds when the line is parallel to the projection plane; p=0 when it is perpendicular.

A face-on square projects to a square. Tilting it generally gives a parallelogram, including rectangles and rhombi as special cases; an edge-on orientation degenerates to a line segment.

No, under parallel orthogonal projection. Each pair of parallel opposite sides remains parallel, so any non-degenerate quadrilateral projection is still a parallelogram. It may degenerate to a segment.

It is formed by projecting its n sides. It generally becomes an n-sided, not necessarily regular, polygon; some consecutive projected sides may become collinear or vanish in special orientations, reducing the visible side count.

A cube generally projects to a centrally symmetric hexagon, with squares or other parallelograms in special orientations. A right cone gives a circle from the base direction and a triangle from the side, with other orientations giving curved oval-like outlines.

A triangular lamina can match a cone's triangular side projection, and a circular disc or cylinder can match its circular top projection. A single projection therefore does not identify the original solid uniquely.

If the line vector is (x,y,z), then p_F^2=x^2+z^2, p_T^2=x^2+y^2, and p_S^2=y^2+z^2. Hence p_F^2+p_T^2+p_S^2=2(x^2+y^2+z^2)=2l^2, and each view is at most l.

Cube: squares in aligned views. Cuboid: rectangles. Vertical cylinder: rectangle, circle, rectangle. Vertical cone: triangle, circle, triangle. A prism shows its base polygon along its axis and parallelogram/rectangular lateral views. A pyramid shows its base polygon from above and triangular side views.

A parallelepiped gives parallelogram or rectangular outlines depending on alignment.

Numbering the projection rows from top to bottom: mug → 8, funnel → 6, hammer → 7, car → 1, slide → 3, chair → 4, ceiling fan → 5, and the lidded container → 2.

Light rays diverge from a nearby torch, so the object intercepts a wider cone of rays and magnifies the shadow. Increasing torch distance makes the rays more nearly parallel; with perpendicular parallel rays, the shadow approaches the orthogonal projection.

For the top view, mark every occupied floor position. For the front view, use the maximum stack height in each left-right column. For the side view, use the maximum height in each front-back row. Applying these three scans to each cropped model gives the required outlines without showing hidden internal edges.

Treat each view as a binary silhouette. Start with one cube for every occupied cell in the front view, assign depths so the top silhouette is met, then add cubes only where they do not change an already-correct outline. Multiple solutions are possible; verify each candidate by rescanning from all three arrows.

Candidate (vi) matches all three silhouettes. Its occupied floor cells reproduce the top view, while the maximum heights along the two horizontal directions reproduce the stepped front and side outlines.

Each row of three drawings is a constraint set. Use the top view as the occupied-cell footprint, then assign positive integer heights so column maxima equal the front view and row maxima equal the side view. Hidden cubes are allowed, so more than one solid may satisfy a set; the three checks must all agree.

The visible rows contain 4+3+2+1=10 cubes. No extra hidden depth layer is indicated, so the stack contains 10 cubes.

A generic orientation gives a centrally symmetric hexagon. Special orientations give a parallelogram, including a rectangle, rhombus, or square. Edge coincidences can create degenerate limiting cases, but a true triangular or pentagonal orthogonal outline is impossible.

Balancing the cube on a body-diagonal vertex makes the three edge directions symmetric relative to the projection plane. They meet the plane at equal angles, so equal spatial edge lengths receive the same projection scale factor.

Yes. The five shown are planar tetracubes. There are three additional free non-planar tetracubes, obtained by folding cubes out of the plane in the distinct non-equivalent ways. Thus there are 8 free tetracubes in total.

Choose the three grid directions as length, depth, and height. Transfer every edge by counting its units along the matching direction, draw nearer visible edges first or lightly draw all edges, then remove hidden segments. Equal spatial unit edges must occupy equal grid steps.

The drawing joins locally consistent segments that use the three isometric directions, but their depth ordering changes around the loop. Trace one physically consistent portion at a time; the apparent closed path cannot exist as one ordinary 3D track.

No. Following the beams around forces contradictory near/far and height relationships at the final corner. Its front, top, and side silhouettes can each look locally plausible, but they cannot come from one globally consistent cube model. Isometric projection hides enough depth information for the incompatible joins to appear continuous.