Perimeter and area: Questions and Answers

Name: Learnify Academy
Address: Building 27m, Road 19, Block 923, Bukuwara, Riffa, Southern Governorate, BH
Telephone: +97336608476

92 perimeter and area questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

In the figure, you see athletes assembled at the start of a 4 × 100 m relay race. The tracks are laid out, and the athletes are all set to go racing down the tracks. Do you notice that the athletes are not at the same starting line? Those in the outer lanes seem to be starting ahead of those in the inner lanes while the finish line is the same for all of them. What could be the reason for this? The distance between the starting points of adjacent lanes is called the ‘stagger’. Notice that the stagger continues all the way to the outermost lane. Do you think the stagger gives anyone (those in the outer lanes or in the inner lanes) an unfair advantage? Why or why not? On what basis can the organisers work out the length of the stagger between lanes?Answer: The outer lanes go round a bigger curve, so they are longer. The stagger moves each outer runner forward by exactly this extra length, so every runner covers the same distance and no one gets an advantage. The organisers work it out from the circumference of a circle: on a semicircular bend, a lane of width w further out is longer by πw.
In my school, the playground is too small to have a 400 m track, so the school constructed a 200 m track instead. Does this mean that we need a smaller stagger for the race tracks in my school (i.e., smaller than the stagger used in the Olympics), for the same 4 × 100 m relay race?Answer: No. The extra length of an outer lane does not depend on how big the bend is, only on the lane width and how much turning the race has. A 4 × 100 m relay is two laps of a 200 m track, so it has twice as many bends: the stagger is about double (about 15.3 m instead of about 7.7 m for 1.22 m lanes).
Here we see a circle with radius r units. What is its perimeter? How do we find out?Answer: Perimeter (circumference) = 2πr units. We find it from the fact that circumference ÷ diameter is the same number π ≈ 3.14 for every circle, so C = π × 2r.
What is the connection between this question and the one about the 400 m athletics track?Answer: The curved ends of a running track are semicircles. To find how long each lane is, and so how big the stagger must be, we need the perimeter of a circle. Two semicircular ends together make one full circle of length 2πr.
What happens to the perimeter of a square if we double its side?Answer: The perimeter doubles too. Side a gives 4a; side 2a gives 8a = 2 × 4a. Perimeter : side stays 4 : 1 for every square.
What about a circle? What is its perimeter (usually called the circumference) in terms of its diameter? Is the ratio of circumference (C) to diameter (D) the same for circles of all sizes (see the figure)? What do you think?Answer: Yes, C ÷ D is the same for every circle. This constant is π ≈ 3.14, so the circumference is C = πD.
What is the value of the C/D ratio? How would you estimate this ratio?Answer: C/D = π = 3.14159… (about 227 or 3.14). Estimate it by measuring round objects (thread around the edge ÷ diameter), or by geometry: compare the circle with polygons drawn inside and outside it.
You can do a simple measurement at home to estimate the C/D ratio. Take a cotton reel with thin thread around it. Measure the diameter D of the reel as accurately as possible. Unwrap and then tightly wrap the thread around the reel 20 times. Unwrap it again; measure its length L, and calculate L20D. This is the ratio we want. For accuracy, the thread should be very thin. Please do the experiment! Do you get a ratio between 3 and 4? Between 3.1 and 3.2? It is also possible to estimate the C/D ratio using pure geometry, i.e., without any measurements at all! Can you imagine how?Answer: Yes: a careful measurement gives a ratio between 3 and 4, and usually between 3.1 and 3.2 (about 3.14), because L = 20 × circumference. Without measuring, trap the circle between a polygon drawn inside it and one drawn outside it: hexagons give 3 < π < 3.46.
The Mesopotamian Hexagon-to-Circle comparison (see the figure). Can you see why this shows that π > 3?Answer: The hexagon is made of 6 equilateral triangles of side 1, so its perimeter is 6. The circle (radius 1) has circumference 2π, and each arc is longer than the straight side under it. So 2π > 6, i.e. π > 3.
Archimedes’ method utilising inscribed and circumscribed polygons (see the figure). Can you see why this diagram of an inscribed and circumscribed hexagon tells us that π is between 3 and 2√3? (Hint: Use the Baudhāyana–Pythagoras Theorem.)Answer: Inner hexagon: perimeter 6. Outer hexagon: side a with a² = 1 + (a/2)², so a = 2√3 and perimeter = 6a = 4√3. Since 6 < 2π < 4√3, 3 < π < 2√3 ≈ 3.46.
What will be the length of a semicircle with the same radius r (see the figure)?Answer: Length of a semicircle = πr (half of 2πr), which can also be written 2πr × 180°360°.
What will be the length of a quarter circle with the same radius (see the figure)?Answer: Length of a quarter circle = πr2 (a quarter of 2πr), also written 2πr × 90°360°.
The figure depicts a 400 m athletics track. You can see two straight sections of length 84.39 m each, and two curved portions, which are semicircles with a common centre (points A and B); the innermost semicircle on each side has radius 36.5 m. The width of each lane is 1.22 m. Let an athlete make one complete circuit of the track. What is the total distance she runs?Answer: Taking her path 0.3 m from the inner border: 2 × 84.39 + 2π × 36.8 = 168.78 + 231.22 = 400 m.
What is the difference in radius between the first and second lanes? Use the figure to find the stagger needed by the runner in the second lane. Will an equal stagger be needed between the third and second lanes?Answer: The radius goes up by one lane width, 1.22 m. Over one lap (two semicircles = one full circle) lane 2 is longer by 2π × 1.22 ≈ 7.67 m, so that is the stagger. Yes, lane 3 needs the same 7.67 m stagger ahead of lane 2, because its radius is again 1.22 m more.
Two circles of equal radius are located such that each circle passes through the centre of the other circle (see the figure). Given that the radius of each circle is r units, find the perimeter of the shape formed by the two circles in terms of r units. (Ignore the dotted portions that lie within the circles.)Answer: Each red arc is 23 of its circle, so the perimeter = 2 × 23 × 2πr = 83πr units.
In the figure, we see points P and Q and two paths connecting them. The first path is made up of the semicircle a. The other path is made up of three semicircles (b, c and d). Which path is longer? Choose one: (i) Path a is longer. (ii) Path b + c + d is longer. (iii) The two paths have equal length. (Try to answer this before reading on.)Answer: (iii) The two paths have equal length. With radii a′, b′, c′, d′: path lengths are πa′ and π(b′ + c′ + d′), and since PQ = 2a′ = 2b′ + 2c′ + 2d′, a′ = b′ + c′ + d′.
The perimeter of a circle is 44 cm. What is its radius?Answer: 2πr = 44 ⇒ r = 44 × 744 = 7 cm.
Calculate, correct to 3 significant figures, the circumference of a circle with:Answer: (i) 44.0 cm (ii) 62.9 cm (iii) 75.4 cm
Calculate the length of the arc of a circle if:Answer: (i) 113 cm ≈ 3.67 cm (ii) 13.2 m
Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.Answer: Arc = 2 × 227 × 14 × 75360 = 553 ≈ 18.33 cm; perimeter = 18.33 + 14 + 14 = 1393 ≈ 46.33 cm.
Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (the figure (i) to (ix)):Answer: (i) 348.57 m (ii) 35.43 cm (iii) 62.86 cm (iv) 56.57 cm (v) 176 cm (vi) 88 cm (vii) 37.71 cm (viii) 37.71 cm (ix) 62.86 cm
If the diameter of a car tyre is 56 cm, then:Answer: (i) 176 cm (ii) 1 000 000 ÷ 176 ≈ 5682 revolutions
Find the total perimeter of all the petals in each of the given flowers.Answer: (i) 8 quarter circles of radius 7 cm = 88 cm (ii) 12 arcs of 60° with radius 42 cm = 528 cm
The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?Answer: 2πr₁ : 2πr₂ = r₁ : r₂, so the radii are also in the ratio 5 : 4.
What happens if the parallelogram is ‘thin’ (see the figure) and the foot of the perpendicular from C to AD does not lie on side AD? The construction then does not seem to work. How do we fix this ‘gap’?Answer: Slide the top side along its own line in small steps: replace ABCD by A′BCD′ with A′A = D′D. A congruent triangle is removed on one side and added on the other, so the area stays the same while the base and height stay the same. Repeat until the parallelogram is ‘fat’, then use the usual cut-and-move. So the area is still base × height.
The area of a rectangle can be found when we know the lengths of its sides. Is the same true for a parallelogram? That is, can we find the area of a parallelogram when we know the lengths of its sides? Why or why not? (Hint: What happens to the area of a parallelogram if we decrease or increase the angle between the adjacent sides while keeping the lengths fixed?)Answer: No. With sides fixed, changing the angle between them changes the height, and so the area. Sides 5 and 3 can give area 15 (a rectangle), 7.5 (angle 30°) or almost 0 (very flat).
You may wonder, like earlier, is there a gap in our argument? What would we do if angle EFG is obtuse and the triangle were shaped like triangle EFG in the figure? Please work out the answer to this question.Answer: Extend GF to H so that EH ⊥ GH (EH = h, HF = x). Then area(EFG) = area(EHG) − area(EHF) = 12(b + x)h − 12xh = 12bh. The formula still holds.
Do you see why the two triangles fit together to make a parallelogram? (If you study the angles in the figure (e.g., ∠B’C’A’ and ∠BCA), you will see why this is so. Keep in mind the criterion by which we check whether two lines are parallel.)Answer: Place the copy so that C′ is at A and A′ at C. Then ∠DAC = ∠B′C′A′ = ∠BCA, and these are alternate angles on AC, so AD ∥ BC. In the same way ∠DCA = ∠B′A′C′ = ∠BAC gives AB ∥ DC. Both pairs of opposite sides are parallel, so ABCD is a parallelogram.
Since ΔABD and ΔACD have equal area, you may wonder — Can we divide ΔABD using straight cuts into two or more pieces that we can then rearrange to exactly cover ΔACD? What do you think? Is it possible?Answer: Yes, with just one cut (2 pieces). Let E be the midpoint of AB and cut ΔABD along DE. Slide piece EBD along BC by the length BD: it becomes NDC (N = midpoint of AC). Turn piece AED half round the midpoint M of AD: it becomes DNA. Together NDC and DNA make ΔACD exactly.
Suppose we are given two polygons P and Q with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other polygon? Try this out for familiar shapes, e.g.,Answer: Yes, every time. (1) 9 × 4 rectangle → 6 × 6 square with one staircase cut. (2) Triangles: each → parallelogram → the same rectangle. (3) Triangle → rectangle → square. Conjecture: any two polygons with equal area can be cut into pieces that rearrange from one into the other (this is a true theorem).
Think of various rectangles with perimeter 40 units (the sides do not have to be integers).Answer: 1. Infinitely many (length + breadth = 20). 2. Yes: the 10 × 10 square, area 100. 3. No smallest: thin rectangles have areas as close to 0 as we like, but never 0.
Let us test Heron’s formula against some known cases: an equilateral triangle with side a units.Answer: s = 3a2, so area = √(3a2 × a2 × a2 × a2) = √34a² sq. units, the same as 12 × base × height.
Let us test Heron’s formula against some known cases: an isosceles triangle with equal sides a units and base 2b units.Answer: s = a + b, so area = √((a + b)(b)(b)(a − b)) = b√(a² − b²) sq. units, the same as 12 × base × height.
Let us test Heron’s formula against some known cases: a triangle with sides 3 units, 4 units and 5 units.Answer: s = 6, area = √(6 × 3 × 2 × 1) = √36 = 6 sq. units. Since 3² + 4² = 5², it is right-angled, and 12 × 3 × 4 = 6 too.
In the same way we ask: can we find the area of a 4-gon if we only know the lengths of its sides? The figures below reveal the answer to this question. The problem (see the figure) is about a 4-gon whose sides are known to be 3, 3, 3, 3 (it is a ‘rhombus’). As you can see, the areas of the three figures are different. (We drew the figures using GeoGebra and found the areas using the ‘Area’ tool. Please try this exercise yourself, or by using four rods joined together at their ends.)Answer: No. Four rods of length 3 can be pushed into many shapes. Area = base × height = 3 × height, and the height changes as the rhombus leans: 3 (square, area 9), about 2.67 (area 8.01), about 1.80 (area 5.41). The sides alone do not fix the area.
Verify Brahmagupta’s formula for the case of a rectangle.Answer: Sides a, b, a, b: s = a + b, so area = √(b × a × b × a) = √(ab · ab) = ab, which is correct.
The formula states that if the sides of the cyclic 4-gon have lengths a, b, c, d, and the semi-perimeter s is s = 12(a + b + c + d), then: Area of 4-gon = √((s − a)(s − b)(s − c)(s − d)). Please check out the formula against other special cases.Answer: Square (side a): s = 2a, √(a · a · a · a) = a² ✓. Kite with two right angles, sides 3, 4, 4, 3: s = 7, √(4 × 3 × 3 × 4) = 12 = 2 × 12 × 3 × 4 ✓. Triangle (d = 0): it becomes Heron’s formula ✓.
Verify Brahmagupta’s formula for the case of an isosceles trapezium.Answer: With AB = 2a, DC = 2b, slant sides c: s = a + b + c, and the formula gives (a + b)√(c² − (b − a)²) = (a + b)h, exactly 12 × (sum of parallel sides) × height.
Here is another example. Compare the following identities from algebra: (a + b)² = a² + b² + 2ab, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. Can you see that the first identity is a special case of the second one (put c = 0 in the second identity), and the second identity is a generalisation of the first one?Answer: Put c = 0 in (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca: every term with c vanishes, leaving (a + b)² = a² + b² + 2ab. So the first is a special case; the second, which works for any c, generalises it.
Try to work out why this method works. You will find that it is a geometrical translation of the formula (a + b2)² − (a − b2)² = ab.Answer: AF = a + b2, so HK = HG = a + b2, and KP = BH = AF − AB = a − b2. In right triangle HPK: HP² = HK² − KP² = (a + b2)² − (a − b2)² = ab. So square HPQS has the area of the rectangle.
What procedure would you use to square a given triangle? Here, the task is to construct a square whose area is equal to the area of some given triangle. Think carefully. How would you proceed?Answer: Two stages: (1) turn the triangle into a rectangle of equal area: base b and height h2 (cut along the line joining the midpoints of two sides); (2) square that rectangle with Baudhāyana’s construction. The square has side √(bh2).
Find the area of triangle ADE in the figure.Answer: Take AD = 8 cm as the base; the height from E to AD is the length of the rectangle, 10 cm. Area = 12 × 8 × 10 = 40 cm² (half the rectangle), wherever E is on BC.
The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.Answer: Each end overhangs (40 − 20) ÷ 2 = 10 cm, so height = √(26² − 10²) = 24 cm. Area = 12(40 + 20) × 24 = 720 cm².
Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.Answer: Third side = 32 − 8 − 11 = 13 cm; s = 16. Area = √(16 × 8 × 5 × 3) = √1920 = 8√30 ≈ 43.82 cm².
The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.Answer: Sides 60 m, 100 m, 140 m; s = 150. Area = √(150 × 90 × 50 × 10) = √6 750 000 = 1500√3 ≈ 2598 m².
One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.Answer: Area = 12 × d × 2d = d² = 128, so d = √128 = 8√2 ≈ 11.31 cm.
ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (∆PCD): area (∆QCD)?Answer: Both triangles have base CD and the same height (the distance between AB and CD), so their areas are equal: ratio 1 : 1 (each is half the parallelogram).
O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.Answer: The diagonal PR halves the parallelogram, so S and Q are at equal distances from PR. △PSO and △PQO share the base PO, so they have equal heights and hence equal areas.
If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered later, when quadrilaterals are studied.)Answer: Each corner triangle is 14 of a triangle cut off by a diagonal: ar(APS) = 14ar(ABD), ar(CQR) = 14ar(CBD), etc. The four corners add up to 14 × 2 × ar(ABCD) = 12ar(ABCD), so PQRS = 12 ABCD.
In ∆ABC, the midpoint of BC is D (see the figure). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).Answer: ar(ABD) = ar(ACD) (AD is a median of △ABC) and ar(PBD) = ar(PCD) (PD is a median of △PBC). Subtracting: ar(ABP) = ar(ACP).
Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (see the figure). What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?Answer: With side a, the heights of △PAB and △PCD from P add up to a, so red = 12a × a = 12a². Then green = 12a² too. Ratio 1 : 1, wherever P is.
In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (see the figure). Prove that Area (∆BPQ) = 12 Area (∆ABC).Answer: Join CD. △DPQ and △DPC have the same base DP and lie between the parallels DP and QC, so they have equal area. Then ar(BPQ) = ar(BPD) + ar(DPQ) = ar(BPD) + ar(DPC) = ar(BDC) = 12ar(ABC) (CD is a median).
Why were human beings so fond of using circular shapes? Was this only for practical reasons, or could there have been other reasons too? What kinds of uses have human beings found for the circular shape?Answer: Practical: a circle is easy to draw with a rope and a peg, it rolls (wheels), it has no weak corners (pots, wells, domes), and it encloses the most area for a given boundary, saving wall or fence material. Other reasons: its perfect symmetry, the round sun and moon, beauty and religious or cultural meaning. Uses: wheels, pots, wells, round huts and granaries, coins, plates, bangles, clocks, gears, stupas and stadiums.
Such reasoning suggests that for a circle too, if C = circumference and A = area, the ratio C² : A must be some fixed constant. But what is this constant?Answer: C = 2πr and A = πr², so C² : A = 4π²r² : πr² = 4π ≈ 12.57. (The Babylonians measured it as about 12.)
As the slices become smaller and smaller, the arcs in the figure (B) become more and more closer to a line. This makes the figure more and more closer to a parallelogram with base = half the circumference (why?) = πr, height = radius r.Answer: The slices alternate: half of them point up and half point down. The arcs of the up-pointing slices make the bottom edge and the arcs of the down-pointing slices make the top edge. So each long edge gets half of all the arcs, i.e. half of 2πr = πr.
Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.Answer: πr² × θ360 = 227 × 49 × 60360 = 773 ≈ 25.67 cm².
Find the area of a quadrant of a circle whose circumference is 44 cm.Answer: 2πr = 44 ⇒ r = 7 cm. Quadrant = 14πr² = 14 × 227 × 49 = 38.5 cm².
The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.Answer: In 60 min the hand turns 360°, so in 10 min it turns 60°. Area = 227 × 7² × 60360 = 773 ≈ 25.67 cm².
A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (Use π ≈ 3.14.)Answer: (i) 78.5 cm² (ii) 235.5 cm²
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73.)Answer: Sector = 3.14 × 225 × 60360 = 117.75 cm²; △OAB is equilateral, area = 1.734 × 225 = 97.3125 cm². Minor segment = 20.4375 ≈ 20.44 cm²; major segment = 706.5 − 20.4375 = 686.0625 ≈ 686.06 cm².
A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.Answer: One wiper: 227 × 28 × 28 × 120360 = 24643 ≈ 821.33 cm². Two wipers (no overlap): 49283 ≈ 1642.67 cm².
A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr²(16 − √34).Answer: Minor segment = sector − equilateral △OAB = πr²6 − √34r² = r²(π6 − √34). Note: the expression as printed, πr²(16 − √34), is a misprint: it is negative (√34 ≈ 0.433 > 16), so it cannot be an area. The π belongs only to the first term.
An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to 3√34π ≈ 0.413.Answer: Joining O to the vertices makes 3 triangles with two sides r and a 120° angle; each has height r2 and base √3r, area √34r². Triangle = 3√34r²; ratio = 3√34π ≈ 0.413.
A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2π ≈ 0.637.Answer: The diagonals of the square are diameters, 2r each. Area of square = 12 × 2r × 2r = 2r². Ratio = 2r² : πr² = 2π ≈ 0.637.
A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to 3√32π ≈ 0.827. Can you see why the answer is exactly twice the answer for the inscribed equilateral triangle?Answer: The (regular) hexagon splits into 6 equilateral triangles of side r: area = 6 × √34r² = 3√32r²; ratio = 3√32π ≈ 0.827. It is twice 3√34π because the inscribed equilateral triangle (joining alternate corners of the hexagon) is exactly half the hexagon.
Identities in algebra can sometimes be shown as area relationships. For example: The figure shown corresponds to the identity (a + b)² = a² + 2ab + b². Do you see how? Draw figures corresponding to the identities (a + b)(a − b) = a² − b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.Answer: The big square has side a + b, so area (a + b)²; its four pieces have areas a², ab, ab, b². For a² − b²: cut a b × b corner from an a × a square and rearrange the L-shape into an (a + b) × (a − b) rectangle. For (a + b + c)²: split a square of side a + b + c into a 3 × 3 grid.
An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.Answer: Base = 40 − 30 = 10 cm; height = √(15² − 5²) = √200 = 10√2 cm. Area = 12 × 10 × 10√2 = 50√2 ≈ 70.71 cm².
An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?Answer: 12 × 10 × h = 60 ⇒ h = 12 cm. Equal side = √(5² + 12²) = 13 cm.
The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.Answer: 12 × 12 × x = 54 ⇒ x = 9 cm; hypotenuse = √(144 + 81) = 15 cm. Perimeter = 12 + 9 + 15 = 36 cm.
The sides of a triangle are in the ratio 2: 3: 4, and its perimeter is 45 cm. Find its area.Answer: Sides 10, 15, 20 cm; s = 22.5. Area = √(22.5 × 12.5 × 7.5 × 2.5) = 75√154 ≈ 72.62 cm².
The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.Answer: Way 1 (Heron): s = 28, √(28 × 21 × 4 × 3) = √7056 = 84. Way 2: 7² + 24² = 25², so it is right-angled: 12 × 7 × 24 = 84. Area = 84 cm².
If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.Answer: One turn = πd = 227 × 60 = 13207 ≈ 188.57 cm. 100 turns ≈ 18 857 cm ≈ 188.57 m.
Find the area of a quadrant of a circle whose circumference is 66 cm.Answer: 2πr = 66 ⇒ r = 10.5 cm. Quadrant = 14 × 227 × 10.5² = 6938 = 86.625 cm².
The wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km.Answer: One turn = 2 × 227 × 28 = 176 cm. In 1 km (100 000 cm): 100 000 ÷ 176 ≈ 568.2 turns (568 complete turns).
Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?Answer: Yes. If the perimeter is 2p and the area is A, the sides satisfy l + b = p and lb = A. Then (l − b)² = p² − 4A is fixed, so l − b is fixed, and l, b are fixed. Same sides ⇒ congruent rectangles.
You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., 12(a + b)h.Answer: The line through the top-right corner parallel to the left side splits the trapezium into a parallelogram (base a, height h) and a triangle (base b − a, height h). Area = ah + 12(b − a)h = 12(a + b)h.
By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).Answer: Diagonal AC splits trapezium ABCD (AB = a ∥ DC = b, height h) into △ABC (base a, height h) and △ACD (base b, height h). Area = 12ah + 12bh = 12(a + b)h.
Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?Answer: Turn a copy upside down (half-turn) and place it against a slant side of the first: the bottom becomes b + a and the top a + b, giving a parallelogram with base a + b and height h. Area = (a + b)h, so one trapezium = 12(a + b)h.
Show that the area of a kite is half the product of its diagonals. Show this:Answer: (i) Diagonal AC (= d₁) splits the kite into two triangles of height d₂2: 2 × 12d₁ × d₂2 = 12d₁d₂. (ii) The kite is half of the d₁ × d₂ rectangle around it.
Three problems about fitting congruent shapes together:Answer: (i) 2a × 2b = 4ab; yes, 4 copies fit (a 2 × 2 grid). (ii) Heron with s doubled gives √16 = 4 times; yes, joining midpoints gives 4 copies. (iii) √81 = 9 times; yes, trisecting the sides and drawing parallels gives 9 copies.
What fraction of the triangle is shaded? What fraction of the square is shaded?Answer: (a) 12 of the triangle (b) 15 of the square
What fraction of the rectangle is covered by the circles?Answer: (a) π4 ≈ 0.785 (b) π4 ≈ 0.785: the same fraction, 1114 with π = 227.
Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!Answer: Conjecture: however many equal circles are fitted in a row, they cover π4 ≈ 78.5% of the rectangle. Tests: 10 circles: 10πr²40r² = π4; 20: 20πr²80r² = π4; 50: 50πr²200r² = π4. Proof: n circles cover nπr² of a 2nr × 2r rectangle: nπr²4nr² = π4.
The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle.Answer: Let each small rectangle be L × W. Top row: 4L; bottom row: 5W; so 4L = 5W. 9LW = 72 ⇒ LW = 8 ⇒ W² = 325, W = 4√105 ≈ 2.53 cm, L = √10 ≈ 3.16 cm. Perimeter = 2(L + W) = 18√105 ≈ 11.38 cm.
Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.Answer: Both triangles have equal bases (one-third of the base each) and the same height (from the top vertex), so their areas are equal. 4 pieces are enough: cut the blue triangle with three straight cuts; slide two pieces to the right and turn the other two half round (see the picture).
The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B . Show that A and B have equal area.Answer: Side s: quarter disc = πs²4; each semicircle (radius s2) = πs²8, so the two together = πs²4 too. The quarter disc = (semicircle 1 + semicircle 2 − A) + B. So πs²4 = πs²4 − A + B, giving A = B.
In the figure, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.Answer: Each petal is bounded by two quarter circles of radius 1: perimeter of flower = 8 × π2 = 4π ≈ 12.57 units (887). Each petal has area 2(π4 − 12) = π2 − 1, so the flower = 2π − 4 ≈ 2.29 sq. units (167 with π = 227).
In the figure we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is l. Show that the area of the green region enclosed between the two circles is 14πl².Answer: OA ⊥ BC (radius to the point of contact) and A is the midpoint of BC, so R² = r² + (l2)². Ring = πR² − πr² = π(l2)² = 14πl².
In the figure, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).Answer: Legs a, b, hypotenuse c. Semicircle areas πa²8 + πb²8 = πc²8 (since a² + b² = c²). The big semicircle = C + two segments; the small ones = A + segment and B + segment. Subtract: A + B = C.
The figure shows two circles passing through each other’s centres. Find the area of the region enclosed by the two circles in terms of the common radius r.Answer: The shaded common region = 2 segments, each with central angle 120°: 2(πr²3 − √34r²) = (2π3 − √32)r² ≈ 1.23r². (The whole region covered by the two circles is 2πr² minus this, (4π3 + √32)r² ≈ 5.05r².)
In the figure, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is 2(A + C)(B + C)C.Answer: Let the rectangle be w × h with the corner O at the bottom left, P = (p, h) on the top, Q = (w, q) on the right and R = (p, q). Then A + C = 12w(h − q), B + C = 12h(w − p), C = 12(w − p)(h − q), so 2(A + C)(B + C)C = wh.
In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.Answer: Let OA = OB = R. Triangle AOB = R²2. AB = R√2, so the semicircle on AB has area πR²4. Segment AFB = quarter disc − triangle = πR²4 − R²2. Crescent AEBF = semicircle − segment = R²2 = area of △AOB.

All maths questions