Figure it Out
Q1. Identify the missing sidelengths.
Answer:
(i) The given figure has dimensions: 7 in, 4 in, ? in, 3 in, 2 in. Areas given: 14 in², 35 in², 28 in², 21 in². For the missing side, the total area can be found by adding the given partial areas. However, without a clearer diagram, we assume the shape is composed of rectangles. The missing sidelength in the top part: Since area 14 in² has width 7 in, height = 14/7 = 2 in. For 35 in² with width 7 in, total height = 35/7 = 5 in. The missing side between 4 in and 3 in: 5 - 4 = 1 in? Wait, careful: The figure is not fully described, so based on typical problems, the missing side is 5 in? Actually, let's compute: The 28 in² region has width ? in and height 4 in, so ? = 28/4 = 7 in. The 21 in² region has width ? and height 3 in, so ? = 21/3 = 7 in. So missing sidelengths are 7 in for those.
(ii) Area = 50 m², given 29 m² and 11 m². Missing area = 50 - (29+11) = 10 m². The missing side: The figure has width 4 m, so the missing length = 10/4 = 2.5 m. Also, other missing sides: Since total width 4 m and partial widths can be found from areas: 29/4 = 7.25 m, 11/4 = 2.75 m, but that may exceed 4 m? Actually, if the figure is a composite rectangle, the height must be consistent. Assuming the height is something else? Without clear diagram, typical solution: The missing sidelength for the 11 m² part: If width = 4 m, then length = 11/4 = 2.75 m. For 29 m², length = 29/4 = 7.25 m. These are possible lengths if the shape is L-shaped. So missing side = 2.5 m and 2.75 m or similar. For accuracy, refer to standard textbook answer.
Q2. The figure shows a path (the shaded portion) laid around a rectangular park EFGH. (i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area. [Hint: There is a relation between the areas of EFGH, the path, and ABCD.] (ii) If the width of the path along each side is given, can you find its area? If not, what other measurements do you need? Assign values of your choice to these measurements and find the area of the path. Give a formula for the area using these measurements. [Hint: Break the path into rectangles.] (iii) Does the area of the path change when the outer rectangle is moved while keeping the inner rectangular park EFGH inside it, as shown?
Answer:
(i) To find the area of the path, we need the dimensions of the outer rectangle ABCD and the inner rectangle EFGH. Let outer length = L, outer width = W; inner length = l, inner width = w. Area of path = Area of ABCD - Area of EFGH = (L × W) - (l × w). Example: Let L = 10 m, W = 8 m, l = 6 m, w = 4 m. Then area of path = (10×8) - (6×4) = 80 - 24 = 56 m². Formula: Area = (L × W) - (l × w).
(ii) If the width of the path along each side is given as 'd', then the outer dimensions are related to inner: L = l + 2d, W = w + 2d. So area of path = (l+2d)(w+2d) - lw = 4d² + 2d(l+w). Example: l=6 m, w=4 m, d=1 m → area = 4(1) + 2(1)(10) = 4+20=24 m². Formula: Area = 4d² + 2d(l+w).
(iii) No, the area of the path does not change if the outer rectangle is moved as long as the inner rectangle remains inside and the path width is uniform. The area depends only on the dimensions, not the position.
Q3. The figure shows a plot with sides 14m and 12m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose.
Answer:
To find the area of the crosspath, we need the width of the path (say w) and the dimensions of the plot. The plot is 14 m by 12 m. The crosspath consists of two rectangular strips: one horizontal and one vertical, intersecting in a square. Let the width of the path be w = 2 m. Then area of horizontal strip = 14 × w = 14×2 = 28 m². Area of vertical strip = 12 × w = 12×2 = 24 m². Intersection area (square) = w × w = 2×2 = 4 m². Total area of crosspath = 28 + 24 - 4 = 48 m². Formula: Area = (length × w) + (width × w) - w², where length = 14 m, width = 12 m.
Q4. Find the area of the spiral tube shown in the figure. The tube has the same width throughout. [Hint: There are different ways of finding the area.] What should be the length of the straight tube if it is to have the same area as the bent tube on the left?
Answer:
The spiral tube consists of various rectangular segments of width 1 unit. The figure shows lengths: 20, 20, 20, 15, 5, 10, 10, 5, 15, 1, 1. The total area is the sum of areas of each rectangular strip (length × width, where width = 1). So areas: 20×1=20, 20×1=20, 20×1=20, 15×1=15, 5×1=5, 10×1=10, 10×1=10, 5×1=5, 15×1=15. Two additional strips of length 1 each: 1×1=1, 1×1=1. Sum = 20+20+20+15+5+10+10+5+15+1+1 = 122. So area = 122 sq. units. For a straight tube of same width (1 unit) to have same area, its length must be 122 units.
Q5. In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.
Answer:
Let the original square have side length s. The regions 1, 2, and 3 likely are parts of the square (e.g., corner squares or triangles). Without the exact figure, assuming region 1, 2, 3 are three of four parts when square is divided. If side is doubled, area becomes 4s². Original area = s². The increase in area for any region depends on its shape. If region is similar to whole square, its area increases by factor of 4. So increase = (4 × original area) - original area = 3 × original area. For specific numbers, need original dimensions. General reason: Area scales with square of side length.
Q6. Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure. Rearrange the pieces to get a larger square, with a hole inside. You can try this activity by constructing the square using cardboard, thick chart paper, or similar materials.
Answer:
Draw a square. Draw two perpendicular lines that intersect at the center (like a plus sign), dividing the square into 4 smaller squares. By cutting along these lines, you get 4 congruent smaller squares. Rearranging them can form a larger square with a hole in the middle if arranged with a gap. This is a classic puzzle: arrange the 4 small squares in a larger square pattern leaving a central square hole. To draw: Draw a square of side length 2 units. Divide into 4 unit squares. Rearrange them to form a larger square of side 2 units with a 1x1 hole in center? Actually, the hole appears if pieces are placed at corners. This is a hands-on activity.
Q1. Find the areas of the following triangles: (i) Triangle with base 3 cm and height 4 cm. (ii) Triangle with base 5 cm and height 3.2 cm. (iii) Triangle with base 3 cm and height 4 cm.
Answer:
(i) Area = 1/2 × base × height = 1/2 × 3 × 4 = 6 cm². (ii) Area = 1/2 × 5 × 3.2 = 1/2 × 16 = 8 cm². (iii) Area = 1/2 × 3 × 4 = 6 cm².
Q2. Find the length of the altitude BY.
Answer:
Given triangle ABC with sides: AB = 6 units, AC = 4 units, BC = 8 units? Actually from figure: Coordinates? The problem references a triangle with points X, Y? Wait, text says: 'Find the length of the altitude BY.' Previous example had triangle with AX = ? From earlier: In triangle ABC, AX is altitude? Actually from the figure: Triangle with sides 5, 3, 4? The specific problem: In triangle ABC, AX = ? Actually the example earlier: Area of triangle ABC = 15/2 sq. units, AC = 4 units, then BY = 15/4 = 3.75 units. So the answer: BY = 3.75 units.
Q3. Find the area of triangle SUB, given that it is isosceles, SE is perpendicular to UB, and the area of triangle SEB is 24 sq. units.
Answer:
Given triangle SUB is isosceles with SE ⟂ UB. Area of triangle SEB = 24 sq. units. Since SE is altitude to base UB, and triangle is isosceles, SE is also median. So E is midpoint of UB. Therefore, triangles SEB and SEU have equal area (same base and height). So area of triangle SUB = area of SEB + area of SEU = 24 + 24 = 48 sq. units.
Q4. [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area.
Answer:
Method: Take a rectangle ABCD. Draw diagonal AC. The diagonal divides the rectangle into two congruent triangles. Each triangle has area exactly half of the rectangle. To get a triangle of equal area to the rectangle, take two such rectangles? Actually, to transform a rectangle into a triangle of equal area: Construct a triangle with base equal to the length of the rectangle and height equal to twice the width, or vice versa. Alternatively, using Śulba method: Cut the rectangle along one diagonal, then rearrange the two triangles to form a larger triangle by joining them along their bases? Standard method: Draw a line from one vertex to the midpoint of the opposite side? Actually, simple method: Extend the base of the rectangle to twice its length, then join the far end to the opposite vertex. The triangle formed has base = 2×length, height = width, area = 1/2 × 2l × w = lw = area of rectangle. So, transform rectangle to triangle: Use base = 2×length, height = width.
Q5. [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.
Answer:
Method: Given triangle ABC with base BC and height h. Find the midpoint of the altitude from A to BC. Draw a line through this midpoint parallel to BC. This line cuts the triangle into a smaller triangle at top and a trapezoid. The trapezoid can be cut and rearranged into a rectangle. Alternatively, construct a rectangle with base = BC/2 and height = h. Area of rectangle = (BC/2) × h = 1/2 × BC × h = area of triangle. So, transform triangle to rectangle: Use base = half of triangle's base, height = same as triangle's height.
Q6. ABCD, BCEF, and BFGH are identical squares. (i) If the area of the red region is 49 sq. units, then what is the area of the blue region? (ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square?
Answer:
(i) The figure shows three identical squares arranged. Without the figure, assume the red and blue regions are triangles or parts. Typically, the red region is a triangle formed by connecting vertices. Since the squares are identical, the area of blue region likely equals the area of red region if symmetrical. So if red = 49, blue = 49 sq. units. (ii) If total area of blue and red = 180, and they are equal, each = 90. But careful: The figure may have more regions. Assuming red and blue together cover part of the squares. If each square area = s², and the colored regions cover half of each? Not enough info. Standard solution: The three squares form a shape where red and blue are triangles with equal area, each being half of a square? Then area of each square = 2×49 = 98? Or total 180 means each region 90, so square area = 180? Actually, from known problem: Area of each square = 49? Wait, no. Re-solving: If red area = 49 and it's a triangle with base = side of square, height = side, then area = 1/2 s² = 49, so s² = 98. So area of each square = 98 sq. units. Blue region would also be 49. (ii) If total of blue and red = 180, and they are equal, each = 90, so 1/2 s² = 90, s² = 180. Area of each square = 180 sq. units.
Q7. If M and N are the midpoints of XY and XZ, what fraction of the area of triangle XYZ is the area of triangle XMN? [Hint: Join NY]
Answer:
Join NY. Since M and N are midpoints, MN is midsegment parallel to YZ. In triangle XYZ, M is midpoint of XY, N is midpoint of XZ. By midsegment theorem, MN = 1/2 YZ and MN ∥ YZ. Triangles XMN and XYZ are similar with ratio 1:2. So area of triangle XMN = (1/2)² = 1/4 of area of triangle XYZ. Answer: 1/4.
Q8. Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.
Answer:
This is a classic reflection problem. To find the shortest path from house to river to tank, reflect the tank across the river line (or house across the river), then draw a straight line from the house to the reflected point. The point where this line meets the river is the optimal point to reach. Draw the map: Place house on one side, river as a line, tank on the other side. Reflect tank across river to get tank'. Draw straight line from house to tank'. The intersection with river is the point. The path is house → river point → tank. This minimizes total distance.
Q1. Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.
Answer:
Quadrilateral ABCD is divided by diagonal AC into two triangles: triangle ABC and triangle ADC. Area of triangle ABC = 1/2 × AC × BM = 1/2 × 22 × 3 = 33 cm². Area of triangle ADC = 1/2 × AC × DN = 1/2 × 22 × 3 = 33 cm². Total area = 33 + 33 = 66 cm².
Q2. Find the area of the shaded region given that ABCD is a rectangle.
Answer:
Given rectangle ABCD with dimensions: AB = 18 cm, AD = 10 cm? Actually figure shows: A, B, D, C with points E, F. Dimensions: 18 cm, 10 cm, 10 cm, 8 cm, 6 cm, 4 cm. Without the diagram, we assume the shaded region is the area of the rectangle minus areas of unshaded triangles or shapes. For example, total area of rectangle = 18 × 10 = 180 cm². Subtract areas of triangles or other shapes. Typical: The shaded region might be a central shape. Accurate solution requires knowing which parts are shaded. From typical problems, the area of shaded region = 180 - (1/2×8×6 + 1/2×4×10?) Not sure. Suggest using grid and counting.
Q3. What measurements would you need to find the area of a regular hexagon?
Answer:
For a regular hexagon, you need the side length. The area can be calculated as (3√3/2) × side². Alternatively, you can find the apothem (distance from center to side) or the circumradius. Minimum measurement: side length.
Q4. What fraction of the total area of the rectangle is the area of the blue region?
Answer:
Without the figure, assume the blue region is a triangle or shape inside the rectangle. Typically, if the blue region is a triangle formed by joining midpoints or vertices, the fraction is 1/2, 1/4, etc. For example, if the blue region is a triangle with base = length of rectangle and height = width, then area = 1/2 lw, so fraction = 1/2. If it is a smaller triangle, need dimensions. Cannot determine exactly without the figure.
Q5. Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral.
Answer:
Method: Draw one diagonal of the given quadrilateral. Find the midpoint of that diagonal. Join this midpoint to the two vertices not on the diagonal. This forms a new quadrilateral (a triangle? Actually, joining the midpoint to opposite vertices creates a quadrilateral? Better: Construct a triangle with base equal to the diagonal and height half, but for quadrilateral: The line joining the midpoints of two opposite sides divides the quadrilateral into two equal areas? Another method: Draw both diagonals. The triangle formed by the intersection and two vertices? Standard: Join the midpoints of all four sides to get a parallelogram whose area is half the original quadrilateral. So, the quadrilateral formed by joining midpoints of sides has area half of original. Method: Mark midpoints of all sides, then connect them in order.
Q1. Observe the parallelograms in the figure below. (i) What can we say about the areas of all these parallelograms? (ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter?
Answer:
(i) All these parallelograms have the same base and same height (since they are between the same parallel lines), so their areas are equal. (ii) The perimeters are different. The parallelogram that is most 'slanted' (i.e., has the greatest tilt) will have the maximum perimeter, while the rectangle (least slanted) will have the minimum perimeter.
Q2. Find the areas of the following parallelograms: (i) base 7 cm, height 4 cm; (ii) base 5 cm, height 3 cm; (iii) base 4.8 cm, height 5 cm; (iv) base 4.4 cm, height 2 cm.
Answer:
Area of parallelogram = base × height. (i) 7 × 4 = 28 sq. cm (ii) 5 × 3 = 15 sq. cm (iii) 4.8 × 5 = 24 sq. cm (iv) 4.4 × 2 = 8.8 sq. cm
Q3. Find QN.
Answer:
From the diagram, PQRS is a parallelogram with PQ = 7.6 cm, PS = 6 cm, and the height from P to SR (QM) = 12 cm. Area of parallelogram = base × height = SR × QM = 7.6 × 12 = 91.2 sq. cm. Also area = PS × QN = 6 × QN. So 6 × QN = 91.2, giving QN = 91.2/6 = 15.2 cm.
Q4. Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm. Which has the greater area? [Hint: Imagine constructing them on the same base.]
Answer:
The rectangle has the greater area. For a parallelogram with sides 5 cm and 4 cm, the height is less than 4 cm (the side length) because the height is perpendicular distance, and the slant side is longer than the height. So rectangle area = 5 × 4 = 20 sq. cm, while parallelogram area = 5 × height (where height < 4 cm), so area < 20 sq. cm.
Q5. Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?
Answer:
Method 1: Draw a rectangle with the same base and height as the triangle. The rectangle area = base × height, while triangle area = (1/2) × base × height, so rectangle area is twice the triangle area. Method 2: Take two identical copies of the triangle and join them along their longest side to form a parallelogram, then convert that parallelogram into a rectangle of equal area by dissection.
Q6. [?ulba-S?tras] Give a method to obtain a rectangle of the same area as a given triangle.
Answer:
Take the triangle ABC. Find the midpoint D of side BC. Draw a line through D parallel to AB. Draw a line through A parallel to BC, intersecting the parallel line at E and F. The rectangle formed (by cutting and rearranging) will have the same area as the triangle. Specifically, cut along the altitude from A to BC, and rearrange the two pieces to form a rectangle.
Q7. [?ulba-S?tras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it? [Hint: Show that triangles ?ADB and ?ADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cut-outs if necessary.]
Answer:
In an isosceles triangle ABC with AB = AC and D as the midpoint of BC, draw AD perpendicular to BC. Cut along AD to get two right triangles ?ADB and ?ADC. These two triangles can be rearranged to form a rectangle by placing them so that their hypotenuses form opposite sides and the legs align to form right angles. The resulting rectangle has dimensions AD (height) and BD (half the base).
Q8. [?ulba-S?tras] Give a method to convert a rectangle into an isosceles triangle by dissection.
Answer:
Take a rectangle ABCD. Find the midpoint E of side AB. Draw line segments from E to C and E to D. Cut along these segments. The three pieces can be rearranged to form an isosceles triangle with base equal to CD and vertex at E (or a point where the two cut edges meet). The triangle will have the same area as the rectangle.
Q9. Which has greater area?—an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area?—two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons.
Answer:
Let the side length be 'a'. Area of square = a². Area of equilateral triangle = (√3/4)a² ≈ 0.433a². Since a² > 0.433a², the square has greater area. Two identical equilateral triangles together have area = 2 × (√3/4)a² = (√3/2)a² ≈ 0.866a², which is still less than a². So the square has greater area in both cases.
Q1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm.
Answer:
Area of rhombus = (1/2) × product of diagonals = (1/2) × 20 × 15 = 150 sq. cm.
Q2. Give a method to convert a rectangle into a rhombus of equal area using dissection.
Answer:
Take a rectangle. Draw its diagonals to find the center. From the center, draw lines to the midpoints of each side. Cut along these lines to get 4 congruent triangles. Rearrange these triangles to form a rhombus by placing them around a central point such that the sides of the rhombus are formed by the hypotenuses of the triangles. The resulting rhombus has the same area as the original rectangle.
Q3. Find the areas of the following figures:
Answer:
(i) The figure with dimensions 16 ft: Assuming it's a trapezium or parallelogram, area = base × height = 16 × (given height). (ii) 24 m, 36 m, 14 m: This appears to be a trapezium with parallel sides 24 m and 36 m, height 14 m. Area = (1/2) × (24 + 36) × 14 = (1/2) × 60 × 14 = 420 sq. m. (iii) 10 in, 6 in, 14 in: Trapezium with parallel sides 10 in and 14 in, height 6 in. Area = (1/2) × (10 + 14) × 6 = (1/2) × 24 × 6 = 72 sq. in. (iv) 12 ft, 18 ft, 8 ft, 7 ft, 10 ft: This is a composite figure. Break it into rectangles and triangles, find areas, and sum them.
Q4. [?ulba-S?tras] Give a method to convert an isosceles trapezium to a rectangle using dissection.
Answer:
In an isosceles trapezium ABCD with AB || CD and AD = BC, draw perpendiculars from A and B to CD meeting at E and F. This divides the trapezium into a rectangle ABFE and two right triangles ?AED and ?BFC. Since the trapezium is isosceles, triangles ?AED and ?BFC are congruent. Cut off triangle ?BFC and attach it to the other side (by rotating 180°) to form a rectangle with the remaining pieces.
Q5. Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area. Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH? [Hint: If ?AHI ? ?DGI and ?BEJ ? ?CFJ, then the trapezium and rectangle have equal areas.]
Answer:
To find vertices of rectangle EFGH: Draw perpendiculars from A and B to the base DC (extended if needed). Let these perpendiculars meet at points on the base. The rectangle EFGH is formed such that its top side is AB (or parallel to AB), its bottom side is a line through points where the perpendiculars meet, and its height is the perpendicular distance between AB and DC. The hint shows that by cutting triangles ?AHI and ?DGI (congruent) and ?BEJ and ?CFJ (congruent), the trapezium can be rearranged into rectangle EFGH.
Q6. Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm².
Answer:
First, draw a rectangle of area 144 cm², e.g., 12 cm by 12 cm (square) or 18 cm by 8 cm. Then convert this rectangle into a trapezium by cutting along a diagonal line from one corner to a point on the opposite side and rearranging. For example, take a 12 cm × 12 cm square. Mark a point on the top side 4 cm from the left corner. Cut from this point to the bottom right corner. Rearrange the two pieces to form a trapezium with parallel sides of different lengths. The area remains 144 cm².
Q7. A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.
Answer:
In a regular hexagon, connections from the center to vertices divide it into 6 equilateral triangles. If the hexagon is divided as shown, the trapezium covers 3 of these triangles, the equilateral triangle covers 1, and the rhombus covers 2. So the ratio of areas (trapezium : triangle : rhombus) = 3 : 1 : 2.
Q8. ZYXW is a trapezium with ZY?WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of ?ZWB.
Answer:
Join ZA and extend to meet WX at B. Since A is the midpoint of XY, by properties of trapezium and midpoints, triangles ?ZYA and ?ABW (or similar) are equal in area. Trapezium area = area of ?ZYW + area of ?ZWA. Triangle ?ZWB = area of ?ZWA + area of ?ABW. Since area of ?ABW = area of ?ZYA, we get trapezium area = area of ?ZWB.