Area Class 8 Solutions

There are infinitely many ways. Simple examples are two perpendicular midlines, both diagonals, four equal parallel strips, or four regions obtained by making equal-area inward and outward adjustments along shared boundaries. Equal area, not equal shape or perimeter, is the requirement.

Their areas are 7\times4=28\text{ cm}^2 and 8\times3=24\text{ cm}^2. With even colouring, the 7 cm × 4 cm rectangle needs more powder.

A diagonal divides a rectangle into two congruent triangles. Each area is ½ of 7\times4, so each triangle has area 14\text{ cm}^2.

Perimeter measures boundary length while area measures covered surface, so neither determines the other. Rectangles 1×10 and 4×4 have perimeters 22 and 16 but areas 10 and 16: the larger perimeter has smaller area. Also, a thin jagged region can have a long boundary while enclosing less area than a compact circle or square.

In (i), use area ÷ known side: 28/4=7 in, 21/7=3 in, the central height is 35/7=5 in, and the vertical 14 in² rectangle has height 5+2=7 in, so its width is 2 in. In (ii), the 29 m² rectangle has width 29/4=7.25 m; the 11 m² rectangle has width 11/4=2.75 m; the lower area is 50-29=21 m², so its height is 21/7.25=84/29\approx2.90 m.

If outer rectangle dimensions are L,W and inner park dimensions are l,w, path area is LW-lw. If the uniform path width is t, then L=l+2t,W=w+2t, so area =(l+2t)(w+2t)-lw=2t(l+w)+4t^2. Moving the inner park inside a fixed outer rectangle does not change the area difference, provided it remains inside.

If the two perpendicular path widths are p and q, their areas are 14p and 12q. Their overlap (pq) is counted twice, so crosspath area is 14p+12q-pq. For equal width t, it is 26t-t^2.

Add the nine rectangular runs: 20+20+20+15+15+10+10+5+5=120. Eight 1×1 corner overlaps were counted twice, so area =120-8=112 square units. A straight tube of width 1 therefore needs length 112 units.

Every length is multiplied by 2, so every area is multiplied by 2² = 4. Each region's new area is four times its old area, an increase of three times its original area, regardless of the internal division.

Cut along the two perpendicular lines and rotate/rearrange the four pieces so their suitable outer edges form a larger square. Area is preserved by dissection; therefore the central hole's area equals the new outer square's area minus the original square's area.

All displayed triangles occupy half of an identical rectangle: each uses a full rectangle side as base and has the same corresponding height. Therefore ΔXDC, ΔYDC, and ΔYBC have equal area even though their shapes differ.

Taking DC=5 as base and the rectangle's height 4, area =½\times5\times4=10 square units.

Place or extend the triangle inside a rectangle, or subtract two right triangles for an obtuse case. In every case, the result is ½ of base × perpendicular height: A=½ bh. The height may fall outside the triangle.

Using base (BC), area =½\times5\times3=15/2. Using base (AC), area =½\times4\times BY=2BY. Thus 2BY=15/2, so BY=15/4=3.75 units.

Yes. Diagonals bisect each other, so adjacent triangles have equal bases along one diagonal and the same altitude from the opposite vertex. Repeating this argument around the intersection shows each triangle has one-fourth of the rectangle's area.

All such triangles have the same base and the same perpendicular height, so all have equal area; none has a maximum or minimum area. Their perimeters vary. Reflecting one endpoint across the parallel line turns the two variable sides into a broken path; the minimum occurs where that path becomes straight.

This point need not lie on the perpendicular bisector unless the geometry is symmetric.

(i) ½×4×3 = 6\text{ cm}^2. (ii) use side EF = 5 cm and perpendicular DN = 3.2 cm: ½×5×3.2 = 8\text{ cm}^2. (iii) ½×3×4 = 6\text{ cm}^2.

Area using (BC) is ½×6×4 = 12. Using (AC), it is ½×8×BY = 4BY. Hence BY=3 units.

In an isosceles triangle, the altitude from the apex also bisects the base. Thus triangles SEU and SEB have equal bases and the same height, so equal areas. Total area of ΔSUB is 24 + 24 = 48 square units.

Rectangle to triangle: keep the same base and choose a triangle whose perpendicular height is twice the rectangle's height. Triangle to rectangle: keep the triangle's base and use half its height, or keep its height and use half its base. These dimensions give equal areas by ½ bh.

Let each square have area s^2. The blue triangle has area s^2/4; the full red region has area s^2. Therefore red : blue = 4 : 1. If red is 49, blue is 49/4=12.25 sq units. If red + blue = 180, then 5s^2/4=180, so each square has area 144 sq units.

XM and XN are each half of the corresponding sides from X, so ΔXMN is similar to ΔXYZ with scale factor ½. Areas scale by the square of the factor, giving ¼ of the area of ΔXYZ.

Reflect the water tank across the river line. Join the house to the reflected tank with a straight segment; its intersection with the river is the optimal water-collection point. Reflecting back gives the shortest broken path house → river → tank.

Draw non-crossing diagonals from one vertex to divide a convex polygon into triangles, or triangulate a concave polygon carefully. Measure a base and perpendicular height for each triangle and add their ½ bh areas.

Split along AC. Area =½\times22\times3+½\times22\times3=66\text{ cm}^2. Equivalently, ½×diagonal×sum of the two perpendicular distances.

Rectangle area is 180 cm². The unshaded left triangle has area ½×10×6 = 30 cm²; the right triangle has area ½×8×10 = 40 cm². Shaded area =180-30-40=110\text{ cm}^2.

Measure side s and apothem a. Divide the hexagon into six triangles to get A=6(½ sa)=3sa. With only side length, a=\sqrt3s/2, so A=3\sqrt3s^2/2.

The two blue triangles use the rectangle's full top and bottom as their bases. Their perpendicular heights add to the rectangle's height. Their combined area is therefore ½ × width × height, exactly ½ of the rectangle.

Join the midpoints of the four sides. Varignon's theorem gives a parallelogram whose area is exactly half the original quadrilateral's area. The construction works for any simple convex quadrilateral.

Drop a perpendicular height, cut off the resulting end triangle, and move it to the opposite side. The pieces form a rectangle with the same base and height. Hence parallelogram area = base × perpendicular height; any side can serve as base with its matching height.

They share the same base and perpendicular height, so their areas are equal. Their sloping sides, and hence perimeters, differ. Figures c and g appear most slanted and have the maximum perimeter; a and d appear least slanted and have the minimum.

Use base × corresponding perpendicular height: (i) 7\times4=28\text{ cm}^2; (ii) 5\times3=15\text{ cm}^2; (iii) 5\times4.8=24\text{ cm}^2; (iv) 2\times4.4=8.8\text{ cm}^2.

Using base SR = 12 cm and height QM = 6 cm, area is 72 cm². Using base PS = 7.6 cm and height QN, 7.6QN=72. Thus QN=180/19\approx9.47 cm.

The rectangle has area 5×4 = 20 cm². A non-rectangular parallelogram with base 5 and sloping side 4 has perpendicular height less than 4, so its area is less than 20 cm². The rectangle has the greater area; equality occurs only when the parallelogram is also a rectangle.

Twice the triangle's area: use a rectangle with the same base and full perpendicular height. Equal area: use the same base and half the height, or half the base and the full height. A dissection can cut and translate triangle pieces to realize these rectangles.

For a triangle, cut along an altitude, then rearrange the two right triangles to make a rectangle of the same area with half-height or half-base. For an isosceles triangle, split at the apex-to-base midpoint and fit the congruent halves into a rectangle. Reverse these cuts to convert a rectangle into an isosceles triangle.

For side s, one equilateral triangle has area \sqrt3s^2/4\approx0.433s^2, two have \sqrt3s^2/2\approx0.866s^2, and the square has s^2. The square is larger than either one or two of the triangles.

The diagonals are perpendicular bisectors. Splitting along one diagonal gives two triangles whose combined area is ½×BD×AO+CO = ½×BD×AC. Thus A=½ d_1d_2. The dissection also forms a rectangle with sides one diagonal and half the other.

Drop perpendiculars to split the trapezium into a rectangle and two triangles, or join two congruent copies into a parallelogram. If parallel sides are a,b and height is h, two copies have area h(a+b); one has A=½ h(a+b). The result also holds for an overhanging trapezium.

A=½\times20\times15=150\text{ cm}^2.

Reverse the rhombus-to-rectangle construction: treat the rectangle's length as one rhombus diagonal and twice its width as the other. Split the rectangle appropriately into the four right-triangle pieces and rearrange them around the diagonal intersection.

(i) Parallelogram: 16\times10=160\text{ ft}^2. (ii) Trapezium: ½×14×24+36 = 420\text{ m}^2. (iii) Trapezium with parallel sides 14 in and 6 in, distance 10 in: ½×10×20 = 100\text{ in}^2. (iv) ½×8×12+18 = 120\text{ ft}^2.

For an isosceles trapezium, cut the two congruent end triangles and move half of each to the opposite end, forming a rectangle of height h and width a+b/2. In the general figure, mark I and J as midpoints of AD and BC, draw perpendiculars through them to the parallel bases, and extend AB/DC as needed to meet those perpendiculars at H,G,E,F.

The paired corner triangles are congruent, so area is preserved.

Choose height 8 cm and parallel sides whose sum is 36 cm, for example 16 cm and 20 cm. Then A=½\times8\times(16+20)=144\text{ cm}^2. Draw the parallel segments 8 cm apart and join their endpoints.

A regular hexagon consists of six congruent equilateral triangles. The top trapezium contains 3 such units, the lower-left equilateral triangle contains 1, and the lower-right rhombus contains 2. The ratio is 3:1:2.

Let the parallel sides be ZY=a, WX=b, with height h. Because A is the midpoint of XY and Z,A,B are collinear, the extension gives triangle ZWB base WB=a+b with the same height h. Thus both areas equal ½h(a+b).

A4 area =21\times29.7=623.7\text{ cm}^2. Using 1 in = 2.54 cm: 5 in = 12.7 cm; 7.4 in = 18.796 cm; 5.08 cm = 2 in; 11.43 cm = 4.5 in. Also 1\text{ in}^2=2.54^2=6.4516\text{ cm}^2.

10\text{ in}^2=64.516\text{ cm}^2. 161.29\text{ cm}^2=161.29/6.4516=25\text{ in}^2. 1\text{ ft}^2=12^2=144\text{ in}^2. 1\text{ km}^2=1000^2=1,000,000\text{ m}^2.

Measure or obtain consistent boundary data, convert every quantity to one unit, and state whether the figure is floor area, campus area, or administrative area. For irregular regions, triangulate a scaled map or use an official GIS/municipal source. City rankings depend on how 'city' and its boundary are defined, so cite the source and date rather than treating one ranking as universal.