7. Area

There are infinitely many ways.

For example, draw both diagonals, or draw the two lines joining the midpoints of opposite sides, or draw three lines parallel to one side at equal gaps.

We can also bend the dividing lines. If one part is pushed in along an edge and pushed out by the same amount along another edge, the 4 areas stay equal.

Area of first rectangle = 7 × 4 = 28 cm².

Area of second rectangle = 8 × 3 = 24 cm².

So the 7 cm × 4 cm rectangle needs more powder.

The diagonal divides the rectangle into two congruent triangles.

Area of each triangle = 12 × 7 × 4 = 14 cm²

Region 1: a rectangle of 12 cm × 1 cm. Perimeter = 2(12 + 1) = 26 cm. Area = 12 cm².

Region 2: a square of 5 cm × 5 cm. Perimeter = 20 cm. Area = 25 cm².

Region 1 has the larger perimeter but the smaller area.

Take a circle and a thin star (or a zig-zag shape) drawn inside it.

The star has many sharp points. So its boundary is very long. But it covers only a small part of the circle.

So the star has a larger perimeter but a smaller area than the circle.

We need the width of each path.

Let the path parallel to the 14 m side have width a. Let the path parallel to the 12 m side have width b.

Area of the two paths = 14a + 12b. The square in the middle (a × b) is counted twice. So we subtract it once.

Area of crosspath = 14a + 12b − ab

Example: a = b = 1 m. Area = 14 + 12 − 1 = 25 m²

Use the hint. A bent tube with arms 5 and 5 has the same area as a straight tube of length 5 + 4 = 9.

At every bend, the corner square is shared by both arms. So each new arm adds (its length − 1).

The outer lengths of the arms are 20, 20, 20, 15, 15, 10, 10, 5 and 5.

Area of the spiral tube = 112 square units.

Straight tube of the same area (width 1) should have length 9 for the bent tube 5 + 5.

Region 3 is half the square. Regions 1 and 2 are each one-fourth of the square.

Let the side be s. When the side becomes 2s, the area of the square becomes 4s². It is 4 times the old area.

Each region also becomes 4 times. So each area increases by 3 times its old area.

Region 1: from s²4 to s², increase = 3s²4

Region 2: increase = 3s²4

Region 3: from s²2 to 2s², increase = 3s²2

The two lines cross inside the square at right angles. So each piece has one right angle at the crossing point and one corner of the square.

Turn each piece so that its square corner points inwards and its cut edges lie outside.

Place the 4 pieces around a centre, each turned by 90° from the next.

Their cut edges form the sides of a bigger square. A small square hole is left in the middle.

Both triangles have the same base DC. Their height is the width of the rectangle.

So each triangle is half of the rectangle. They have equal areas.

∆XDC has base DC and height AD. ∆YBC has base BC and height AB (= DC).

Each is half of the rectangle ABCD. So they have equal areas.

Area of ∆XDC = 12 × DC × AD = 12 × 5 × 4 = 10 sq. units

Yes. Area (∆ABC) = Area (∆ADC) − Area (∆ADB) = 12 × h × DC − 12 × h × DB = 12 × h × (DC − DB) = 12 × h × BC.

(i) All the triangles have the same base BC. Their height is the distance between l and BC. So all have equal areas. No triangle has more or less area.

(ii) The perimeter is least when A lies on the perpendicular bisector of BC (the isosceles triangle). As A moves farther along l, AB + AC keeps increasing. So there is no maximum perimeter.

BCC′B′ is a rectangle. BC′ and B′C are its diagonals. They bisect each other at the centre of the rectangle.

This centre lies on l. It also lies on the perpendicular bisector of BC.

The point A is where BC′ meets l. So A is this centre. Hence yes, A lies on the perpendicular bisector of BC.

Area (∆ABC) = 12 × BC × AX = 12 × 6 × 4 = 12 sq. units.

Also, Area (∆ABC) = 12 × AC × BY = 12 × 8 × BY = 4 BY.

So 4 BY = 12. BY = 3 units

In an isosceles triangle, the perpendicular SE bisects the base. So UE = EB.

∆SUE and ∆SEB have equal bases and the same height SE. So they have equal areas.

Area (∆SUB) = 24 + 24 = 48 sq. units

Let ABCD be the rectangle. Mark M, the midpoint of BC.

Join AM and extend it to meet DC extended at E.

∆ABM ≅ ∆ECM (BM = CM, vertically opposite angles are equal, ∠B = ∠C = 90°).

So cut off ∆ABM and place it at ∆ECM. The rectangle ABCD becomes ∆ADE of the same area.

Let ABC be the triangle with base BC. Mark P and Q, the midpoints of AB and AC. Join PQ.

Draw AD ⊥ PQ. Cut along PQ and AD. We get two small right triangles APD and AQD.

Turn ∆APD down to fill the gap near B. Turn ∆AQD down to fill the gap near C.

We get a rectangle with base BC and height half the height of the triangle. Its area = BC × h2 = area of the triangle.

Let each square have side s. Then DH goes from D to H, and CH = 2s.

Red region = ∆DCH. Area = 12 × DC × CH = 12 × s × 2s = s². So the red region equals one square.

DH cuts AB at its midpoint. Blue region = right triangle with legs AD = s and s2. Area = 12 × s × s2 = s²4.

Join NY.

YN joins Y to the midpoint of XZ. So Area (∆XNY) = 12 Area (∆XYZ).

In ∆XNY, NM joins N to the midpoint of XY. So Area (∆XMN) = 12 Area (∆XNY).

Hence, Area (∆XMN) = 12 × 12 Area (∆XYZ) = 14 of Area (∆XYZ)

Treat the river bank (near side) as a mirror. Mark the image of the water tank on the other side of the bank, at the same distance.

Join the house to this image by a straight line. Let it meet the bank at P.

The shortest path is House → P → Water tank. The path P → tank is as long as P → image. A straight line is the shortest.

Diagonal AC divides ABCD into ∆ABC and ∆ADC.

Area (∆ABC) = 12 × 22 × 3 = 33 cm².

Area (∆ADC) = 12 × 22 × 3 = 33 cm².

Area of ABCD = 33 + 33 = 66 cm²

AB = 10 + 8 = 18 cm. AD = 6 + 4 = 10 cm. Area of ABCD = 18 × 10 = 180 cm².

Area (∆AEF) = 12 × 10 × 6 = 30 cm².

Area (∆EBC) = 12 × 8 × 10 = 40 cm².

Shaded area = 180 − 30 − 40 = 110 cm²

Join the centre to all six vertices. We get 6 identical equilateral triangles.

We need the sidelength and the distance from the centre to a side (the height of each triangle).

Area of hexagon = 6 × 12 × side × height.

A point inside is joined to all four corners. The blue triangles stand on the top and bottom sides.

Both have base = length of the rectangle. Their heights add up to the width of the rectangle.

So blue area = 12 × length × width. Blue region = 12 of the rectangle

Mark the midpoints of the four sides of the quadrilateral. Join them in order.

The new quadrilateral has half the area.

Reason: Each corner triangle cut off has 14 the area of a triangle formed by a diagonal. These 4 corner triangles together make half the quadrilateral.

Draw AX ⊥ CD. Cut off ∆AXD.

Slide it to the other side so that AD falls on BC. It fits exactly as ∆BYC (RHS congruence).

We get rectangle ABYX. Area of parallelogram = base × height.

(i) Each parallelogram has base 5 units and height 3 units. So all have equal areas = 15 sq. units.

(ii) Their perimeters are different. The more slanted the side, the longer it is.

(g) appears to have the maximum perimeter. (a) (nearly a rectangle) has the minimum perimeter.

(i) 7 × 4 = 28 cm²

(ii) 5 × 3 = 15 cm²

(iii) Side 5 cm, height 4.8 cm. Area = 5 × 4.8 = 24 cm²

(iv) Side 2 cm, height 4.4 cm. Area = 2 × 4.4 = 8.8 cm²

Area of PQRS = SR × QM = 12 × 6 = 72 cm².

Also, Area = PS × QN = 7.6 × QN.

QN = 72 ÷ 7.6 = 18019 cm ≈ 9.47 cm

Put both on the same base of 5 cm.

Height of rectangle = 4 cm. Height of parallelogram is less than its slant side 4 cm.

So the rectangle has the greater area (20 cm²).

Method 1: Through A draw a line l ∥ BC. At B and C draw perpendiculars to BC meeting l. This rectangle has base BC and height h. Its area = BC × h = 2 × 12 BC × h.

Method 2: Take two copies of the triangle. Join them to make a parallelogram. Then convert the parallelogram into a rectangle by dissection.

Draw the rectangle with the same base BC and half the height of the triangle.

Area = BC × h2 = 12 × BC × h = area of the triangle.

By dissection: cut along the line joining the midpoints of AB and AC. Cut the top part along the altitude. Turn the two small triangles down to fill the corners.

Draw the altitude AD. In an isosceles triangle, it bisects BC. So ∆ADB ≅ ∆ADC.

Each is a right triangle, i.e., half of a rectangle with sides BD and AD.

Cut along AD. Turn ∆ADC over and join it to ∆ADB along the hypotenuse (AC on AB).

We get a rectangle with sides BD and AD. Its area = BD × AD = 12 × BC × AD.

Cut the rectangle along a diagonal. We get two identical right triangles.

Place them back to back so that the two shorter legs (width) coincide.

The two longer legs (length) now lie on one straight line. The base becomes 2 × length and the height is the width.

The new figure is an isosceles triangle of the same area.

Let the side be a. The height h of the triangle is less than a, because a slant side is longer than the perpendicular.

Triangle: 12 × a × h < 12 × a × a. Square: a × a.

So the square has greater area than one triangle.

Two triangles: 2 × 12 × a × h = a × h < a × a. So the square has greater area in this case also.

Area = 12 × AO × BD + 12 × CO × BD = 12 × BD × (AO + CO) = 12 × BD × AC.

So, Area of a rhombus = 12 × product of diagonals.

Let FD = x. Then DE = a − x and EC = b − (a − x) = b − a + x.

Area ABED = Area ABEF − Area ∆AFD = ah − 12xh.

Area ∆BEC = 12h(b − a + x).

Area ABCD = ah − 12xh + 12hb − 12ha + 12hx = 12 h(a + b)

Draw BG ∥ AD. ABGD is a parallelogram with base DG = a. Its area = ah.

∆BGC has base GC = b − a. Its area = 12h(b − a).

Total = ah + 12h(b − a) = 12h(a + b).

Yes. It works for any trapezium, if we draw BG from the end of the shorter parallel side.

Area = 12 × 20 × 15 = 150 cm²

Let the rectangle have length L and width W. Cut it along its middle into two halves of L/2 × W.

Cut each half along a diagonal. We get 4 identical right triangles with legs L/2 and W.

Arrange them around a point O, with the right angles at O. The legs L/2 lie along one line and the legs W along the perpendicular line.

We get a rhombus with diagonals L and 2W. Area = 12 × L × 2W = L × W.

Let AB be the shorter parallel side. Draw AF ⊥ DC and BE ⊥ DC.

∆AFD ≅ ∆BEC (RHS). Cut off ∆BEC.

Turn it over and join it to ∆AFD along the slant side AD. Two equal right triangles joined along the hypotenuse make a rectangle.

This rectangle and ABEF together form one rectangle of height h and length AB + FD.

Mark I and J, the midpoints of AD and BC.

Through I, draw a line perpendicular to DC. It meets DC at G and AB extended at H.

Through J, draw a line perpendicular to DC. It meets DC at F and AB extended at E.

∆AHI ≅ ∆DGI and ∆BEJ ≅ ∆CFJ (ASA). So EFGH has the same area as ABCD.

We need 12 × h × (a + b) = 144.

Take h = 12 cm. Then a + b = 24 cm. Take a = 10 cm and b = 14 cm.

Draw a trapezium with parallel sides 10 cm and 14 cm, at a distance of 12 cm.

Check: 12 × 12 × 24 = 144 cm².

A regular hexagon is made of 6 identical equilateral triangles.

Trapezium (half the hexagon) = 3 triangles. Equilateral triangle = 1 triangle. Rhombus = 2 triangles.

Trapezium : triangle : rhombus = 3 : 1 : 2

ZA extended meets WX extended at B.

In ∆ZYA and ∆BXA: YA = XA (A is the midpoint), ∠ZAY = ∠BAX (vertically opposite), ∠ZYA = ∠BXA (alternate angles, ZY ∥ WB).

So ∆ZYA ≅ ∆BXA (ASA). They have equal areas.

Area of ZYXW = Area ZAXW + Area ∆ZYA = Area ZAXW + Area ∆BXA = Area ∆ZWB.

Area = 21 × 29.7 = 623.7 cm²

(i) 5 × 2.54 = 12.7 cm

(ii) 7.4 × 2.54 = 18.796 cm

(i) 5.08 ÷ 2.54 = 2 in

(ii) 11.43 ÷ 2.54 = 4.5 in

161.29 ÷ 6.4516 = 25 in²

1 ft = 12 in. So 1 ft² = 12 × 12 = 144 in²

1 km = 1000 m. So 1 km² = 1000 × 1000 = 10,00,000 m²

These are activities. Measure or estimate the length and width, and multiply.

Example: a classroom of 30 ft × 24 ft has area 720 ft².

Local units of area include bigha, gaj, katha, guntha, cent and ankanam. Find the one used in your region.

The smallest city in the world is Vatican City (about 0.44 km²). For the others, look up an atlas or a reliable website.