OP Malhotra Class 10 Maths Solutions Chapter 12 Similar Triangles Exercise 12 (A)

S Chand Class 10 ICSE Maths Solutions Chapter 12 Similar Triangles Ex 12(a)

 

Question 1. Draw rectangular axes. Scale them from 0 to 18. Plot the points: A (3, 4), B (9, 4), C (9, 6), D (3, 6). Join the points in alphabetical order to form a rectangle. Transform each of the point (x, y) on the points (x1, y1) by the transformation (x, y) → (2x, 2y) Lable image A1B1C1D1 to correspond with ABCD. notice about the length of each side of the image when you compare it with the corresponding side of the original rectangle?
(ii) Are the rectangles similar in shape?
(iii) Write the ratio \( \frac { \text{Area of image } A_1 B_1 C_1 D_1 }{ \text{Area of original } A B C D } \) in the lowest terms.
Answer: First, plot the points A (3, 4), B (9, 4), C (9, 6), and D (3, 6) on a graph and connect them to form rectangle ABCD.
The transformation rule is \( (x, y) \rightarrow (2x, 2y) \). This means each coordinate is multiplied by 2.
So, the new points for the image A'B'C'D' will be:
A' \( (2 \times 3, 2 \times 4) = (6, 8) \)
B' \( (2 \times 9, 2 \times 4) = (18, 8) \)
C' \( (2 \times 9, 2 \times 6) = (18, 12) \)
D' \( (2 \times 3, 2 \times 6) = (6, 12) \)
Now, plot these new points A'(6, 8), B'(18, 8), C'(18, 12), and D'(6, 12) on the same graph and connect them to form the enlarged rectangle A'B'C'D'. The shapes are shown below.
(i) When you compare the length of each side of the image rectangle A'B'C'D' to the original rectangle ABCD, you will find that the sides of the image are twice as long. For example, the length of AB is \( 9 - 3 = 6 \) units, and the length of A'B' is \( 18 - 6 = 12 \) units. So, A'B' is \( 2 \times AB \). Similarly, BC is \( 6 - 4 = 2 \) units, and B'C' is \( 12 - 8 = 4 \) units, so B'C' is \( 2 \times BC \). The ratio of the lengths of corresponding sides is 1:2.
(ii) Yes, the rectangles are similar in shape because their corresponding angles are equal (all 90 degrees) and the ratio of their corresponding side lengths is constant (1:2).
(iii) To find the ratio of the areas:
Area of original rectangle ABCD:
Length \( = 9 - 3 = 6 \) units
Width \( = 6 - 4 = 2 \) units
Area \( = 6 \times 2 = 12 \) square units
Area of image rectangle A'B'C'D':
Length \( = 18 - 6 = 12 \) units
Width \( = 12 - 8 = 4 \) units
Area \( = 12 \times 4 = 48 \) square units
Ratio of areas \( = \frac { \text{Area of image } A_1 B_1 C_1 D_1 }{ \text{Area of original } A B C D } = \frac { 48 }{ 12 } = \frac { 4 }{ 1 } \)
The ratio in lowest terms is 4:1.
In simple words: The new rectangle is twice as big on each side as the original one, but it keeps the same shape. The new area is four times larger than the original area.

🎯 Exam Tip: When a figure is enlarged by a scale factor 'k', its linear dimensions (like side lengths) change by 'k', and its area changes by \( k^2 \). Make sure to show these calculations clearly.

 

Question 2.
(a) What would be the effect (i) on the length of sides, (ii) on the area of the transformation \( (x, y) \rightarrow (3x, 3y) \)?
Label the image A1B1C1D1 to correspond with ABCD.
(b) Which transformation would make the area of the image sixteen times the area of original rectangle?
(c) Which transformation would made the length of each side of the image half the length of the corresponding side of the original rectangle ?
Answer:
(a) For the transformation \( (x, y) \rightarrow (3x, 3y) \):
(i) The length of each side of the image (A'B'C'D') will be three times the length of the corresponding side of the original figure (ABCD). This is because the scale factor is 3.
(ii) The area of the image will be \( (3)^2 = 9 \) times the area of the original rectangle. This happens because the area scales by the square of the scale factor.
(b) If the area of the image is sixteen times the area of the original rectangle, it means the scale factor squared is 16.
So, \( k^2 = 16 \).
\( \implies k = \sqrt{16} = 4 \).
Therefore, the transformation would be \( (x, y) \rightarrow (4x, 4y) \).
(c) If the length of each side of the image is half the length of the corresponding side of the original rectangle, it means the scale factor is \( \frac{1}{2} \).
Therefore, the transformation would be \( (x, y) \rightarrow (\frac{1}{2}x, \frac{1}{2}y) \).
In simple words: When you change a shape by multiplying its coordinates, its side lengths change by that same number, and its area changes by that number squared. To make an area 16 times bigger, you need to multiply coordinates by 4. To make side lengths half, you multiply coordinates by one-half.

🎯 Exam Tip: Remember the difference between how linear dimensions (length) and area change with a scale factor. If the scale factor is 'k', length scales by 'k', and area scales by \( k^2 \).

 

Question 3. In the figure, O is the centre of dilatation and the dilatation factor is 3.
(i) Complete
(ii) Determine:
(a) \( \frac{\mathrm{OA}^{\prime}}{\mathrm{OA}} \)
(b) \( \frac{\mathrm{OB}^{\prime}}{\mathrm{OB}} \)
(c) \( \frac{\mathrm{OC}^{\prime}}{\mathrm{OC}} \)
(d) \( \frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}} \)
(e) \( \frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}} \)
(f) \( \frac{C^{\prime} A^{\prime}}{C A} \)
Answer: The centre of dilatation is O, and the dilatation factor (scale factor) is 3. This means the transformation rule is \( (x, y) \rightarrow (3x, 3y) \).
(i) Completed table based on the dilatation factor of 3:

(ii) Determine the ratios, knowing the scale factor is 3:
(a) \( \frac{\mathrm{OA}^{\prime}}{\mathrm{OA}} = \frac{3}{1} \)
(b) \( \frac{\mathrm{OB}^{\prime}}{\mathrm{OB}} = \frac{3}{1} \)
(c) \( \frac{\mathrm{OC}^{\prime}}{\mathrm{OC}} = \frac{3}{1} \)
(d) \( \frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}} = \frac{3}{1} \)
(e) \( \frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}} = \frac{3}{1} \)
(f) \( \frac{C^{\prime} A^{\prime}}{C A} = \frac{3}{1} \)
In simple words: When a shape is enlarged by a scale factor, all lengths from the center of enlargement to any point, and all side lengths within the shape, get multiplied by that scale factor. So, if the factor is 3, everything becomes three times longer.

🎯 Exam Tip: When dealing with dilatation, remember that the ratio of any length in the image to the corresponding length in the preimage is always equal to the dilatation factor. This applies to distances from the center of dilatation and to side lengths within the figure.

 

Question 4. A triangle ABC has its vertices at (1, 1), (1, 3) and (3, 1) respectively. Enlarge this triangle by using a scale factor of 2 with the origin as the centre of enlargement.
Answer: The vertices of the original triangle ABC are A(1, 1), B(1, 3), and C(3, 1).
The centre of enlargement is the origin (0, 0), and the scale factor is 2.
This means the transformation rule is \( (x, y) \rightarrow (2x, 2y) \).
We apply this rule to each vertex to find the coordinates of the enlarged triangle A'B'C':
A' \( (2 \times 1, 2 \times 1) = (2, 2) \)
B' \( (2 \times 1, 2 \times 3) = (2, 6) \)
C' \( (2 \times 3, 2 \times 1) = (6, 2) \)
So, the coordinates of the enlarged triangle are A'(2, 2), B'(2, 6), and C'(6, 2).
In simple words: To enlarge a triangle from the origin by a scale factor of 2, you simply multiply both the x and y coordinates of each point by 2 to get the new, bigger triangle.

🎯 Exam Tip: When enlarging a figure with the origin as the centre, multiply each coordinate \( (x, y) \) by the scale factor 'k' to get the new coordinate \( (kx, ky) \).

 

Question 5. The vertices being given, enlarge the given figure by using the given scale factor. For each question you may use a grid.
Answer: We will find the new coordinates for each figure based on the given vertices and scale factor.
(a) For vertices P(1, 2), Q(2, 2), R(2, 1), scale factor 4, and centre of enlargement at origin (0, 0):
The transformation is \( (x, y) \rightarrow (4x, 4y) \).
P' \( (4 \times 1, 4 \times 2) = (4, 8) \)
Q' \( (4 \times 2, 4 \times 2) = (8, 8) \)
R' \( (4 \times 2, 4 \times 1) = (8, 4) \)
So, the enlarged figure's vertices are P'(4, 8), Q'(8, 8), and R'(8, 4).
(b) For vertices A(0, 0), B(3, 2), C(4, 0), scale factor 2, and centre of enlargement at origin (0, 0):
The transformation is \( (x, y) \rightarrow (2x, 2y) \).
A' \( (2 \times 0, 2 \times 0) = (0, 0) \) (Since it's at the origin, it stays there)
B' \( (2 \times 3, 2 \times 2) = (6, 4) \)
C' \( (2 \times 4, 2 \times 0) = (8, 0) \)
So, the enlarged figure's vertices are A'(0, 0), B'(6, 4), and C'(8, 0).
(c) For vertices A(5, 5), B(5, 7), C(7, 7), D(7, 5), scale factor 4, and centre of enlargement at (6, 6):
First, find the vector from the centre of enlargement (CE) to each point: \( \vec{CE \rightarrow P} = P - CE \).
A_vec \( = (5-6, 5-6) = (-1, -1) \)
B_vec \( = (5-6, 7-6) = (-1, 1) \)
C_vec \( = (7-6, 7-6) = (1, 1) \)
D_vec \( = (7-6, 5-6) = (1, -1) \)
Now, multiply these vectors by the scale factor (4):
A_scaled \( = (4 \times -1, 4 \times -1) = (-4, -4) \)
B_scaled \( = (4 \times -1, 4 \times 1) = (-4, 4) \)
C_scaled \( = (4 \times 1, 4 \times 1) = (4, 4) \)
D_scaled \( = (4 \times 1, 4 \times -1) = (4, -4) \)
Finally, add the centre of enlargement coordinates back to find the new points: \( P' = CE + P_{scaled} \).
A' \( = (6 + (-4), 6 + (-4)) = (2, 2) \)
B' \( = (6 + (-4), 6 + 4) = (2, 10) \)
C' \( = (6 + 4, 6 + 4) = (10, 10) \)
D' \( = (6 + 4, 6 + (-4)) = (10, 2) \)
The enlarged figure's vertices are A'(2, 2), B'(2, 10), C'(10, 10), and D'(10, 2).
(d) For vertices A(5, 4), X(5, 8), Y(7, 8), Z(7, 4), scale factor 2, and centre of enlargement at (6, 6):
First, find the vector from the centre of enlargement (CE) to each point: \( \vec{CE \rightarrow P} = P - CE \).
A_vec \( = (5-6, 4-6) = (-1, -2) \)
X_vec \( = (5-6, 8-6) = (-1, 2) \)
Y_vec \( = (7-6, 8-6) = (1, 2) \)
Z_vec \( = (7-6, 4-6) = (1, -2) \)
Now, multiply these vectors by the scale factor (2):
A_scaled \( = (2 \times -1, 2 \times -2) = (-2, -4) \)
X_scaled \( = (2 \times -1, 2 \times 2) = (-2, 4) \)
Y_scaled \( = (2 \times 1, 2 \times 2) = (2, 4) \)
Z_scaled \( = (2 \times 1, 2 \times -2) = (2, -4) \)
Finally, add the centre of enlargement coordinates back to find the new points: \( P' = CE + P_{scaled} \).
A' \( = (6 + (-2), 6 + (-4)) = (4, 2) \)
X' \( = (6 + (-2), 6 + 4) = (4, 10) \)
Y' \( = (6 + 2, 6 + 4) = (8, 10) \)
Z' \( = (6 + 2, 6 + (-4)) = (8, 2) \)
The enlarged figure's vertices are A'(4, 2), X'(4, 10), Y'(8, 10), and Z'(8, 2).
In simple words: When you enlarge a shape, you multiply the distance of each point from the center of enlargement by the scale factor. If the center is the origin, you just multiply the coordinates directly. If the center is elsewhere, you first find the vector from the center to the point, scale that vector, and then add the center's coordinates back.

🎯 Exam Tip: For enlargement with a centre that is not the origin, remember the formula: \( P' = C + k(P - C) \), where \( P \) is the original point, \( C \) is the centre of enlargement, \( k \) is the scale factor, and \( P' \) is the image point.

 

Question 6. Draw the image of the shape KLMN after an enlargement by scale factor \( \frac{1}{2} \) with centre C. Label the image K'L'M'N'.
Answer: Let's assume C is a given point, and KLMN is a quadrilateral. We are enlarging it by a scale factor of \( \frac{1}{2} \), which means we are reducing its size.
1. Locate point C, the centre of enlargement.
2. Draw lines from C to each vertex of KLMN (CK, CL, CM, CN).
3. Since the scale factor is \( \frac{1}{2} \), find the midpoint of each of these lines. These midpoints will be K', L', M', N'. For example, K' is the midpoint of CK, so CK' will be half of CK.
4. Connect K'L'M'N' to form the image.
The figure K'L'M'N' is the image of figure KLMN with a scale factor of \( \frac{1}{2} \) and centre C. Each point in the image is halfway between the centre C and the original point.
In simple words: To make a shape half its size around a point C, draw lines from C to each corner of the shape. Then, mark new corners exactly halfway along these lines. Connect these new halfway points to create the smaller, new shape.

🎯 Exam Tip: When the scale factor is a fraction less than 1, the image will be smaller than the original (a reduction). When drawing, make sure all new points are on the lines connecting the original points to the centre of enlargement, and at the correct proportional distance.

 

Question 7. Figure shows the enlargement transformation of shape ABCD to PQRS by a scale factor of 2. Find the centre of enlargement and state its coordinates.
Answer: To find the centre of enlargement, we need to draw lines connecting corresponding points of the original shape ABCD and its image PQRS.
1. Draw a line from A to P.
2. Draw a line from B to Q.
3. Draw a line from C to R.
4. Draw a line from D to S.
The point where all these lines intersect is the centre of enlargement.
Based on the typical geometric construction shown in the figure:
After joining the corresponding points P to A, Q to B, R to C, and S to D and extending these lines, they all meet at a single point L.
From the provided solution, the centre of enlargement L has coordinates (-2, 3).
In simple words: To find where an enlargement started from, draw lines connecting the matching corners of the small shape and the big shape. Where all these lines cross is the center of the enlargement. In this case, it's at the point (-2, 3).

🎯 Exam Tip: Always use a ruler to draw precise lines when finding the centre of enlargement. Accuracy in drawing will ensure you find the correct intersection point and its coordinates.

 

Question 8. A square with side 3 cm is drawn any where in a plane which is then enlarged about any point in the plane with a scale factor 2. What is the area of the image of the square?
Answer: The side of the original square is 3 cm.
The area of the original square is \( (\text{side})^2 = (3)^2 = 9 \) square cm.
The scale factor for the enlargement is 2. Let's call the scale factor 'k', so \( k = 2 \).
When a figure is enlarged by a scale factor 'k', its area is enlarged by \( k^2 \).
So, the area of the image square will be \( k^2 \times \) (Area of original square).
Area of image \( = (2)^2 \times 9 \)
Area of image \( = 4 \times 9 \)
Area of image \( = 36 \) square cm.
In simple words: If you have a square with a side of 3 cm, its area is 9 square cm. When you make it twice as big (scale factor 2), its area becomes four times larger, so it will be 36 square cm.

🎯 Exam Tip: Remember the relationship between linear scale factor and area scale factor: if the linear scale factor is 'k', the area scale factor is \( k^2 \). This is a common point of confusion.

 

Question 9. Construct a square ABCD of side 2 cm. Enlarge ABCD with enlargement factor m = 2 and label its image A'B'C'D'. What is the relation between the area ABCD and its image A'B'C'D'.
Answer: First, draw a square ABCD with each side measuring 2 cm.
The area of the original square ABCD is \( 2 \times 2 = 4 \) square cm.
The enlargement factor (scale factor) is \( m = 2 \).
This means each side of the image square A'B'C'D' will be \( 2 \times 2 = 4 \) cm long.
Now, construct the image square A'B'C'D' with each side measuring 4 cm.
The area of the image square A'B'C'D' is \( 4 \times 4 = 16 \) square cm.
The relation between the area of ABCD and its image A'B'C'D' is that the area of A'B'C'D' (16 square cm) is four times the area of ABCD (4 square cm).
In simple words: When you make a square twice as big (each side is doubled), its total area becomes four times larger.

🎯 Exam Tip: Always clarify if the enlargement factor applies to length or area. If it's for length, square it to find the area scale factor. Visualizing the squares on a grid can help confirm this relationship.

 

Question 10. A triangle, whose area is 12 cm², is transformed under enlargement about a point in space. If the area of its image is 108 cm², find the scale factor of the enlargement.
Answer: The area of the original triangle is 12 square cm.
The area of the enlarged triangle (image) is 108 square cm.
Let the scale factor of the enlargement be 'k'.
We know that the ratio of the areas of similar figures is equal to the square of their linear scale factor.
So, \( \frac { \text{Area of image} }{ \text{Area of original} } = k^2 \)
\( \implies \frac { 108 }{ 12 } = k^2 \)
\( \implies 9 = k^2 \)
Now, we find the square root of 9 to get 'k':
\( \implies k = \sqrt{9} \)
\( \implies k = 3 \)
The scale factor of the enlargement is 3.
In simple words: The new triangle's area is 108 square cm, and the old one's area was 12 square cm. Since 108 is 9 times 12, the new area is 9 times bigger. This means the scale factor for the sides must be the square root of 9, which is 3. So, each side of the triangle became 3 times longer.

🎯 Exam Tip: When given areas and asked for the scale factor, always remember to take the square root of the ratio of the areas to find the linear scale factor. Ensure units are consistent.

 

Question 11. The parallelogram ABCD has vertices (6, 3), (9, 3), (12, 9), (9, 9) respectively. Copy the grid and the parallelogram. An enlargement scale factor \( \frac{1}{3} \) and centre (0, 0) transforms parallelogram ABCD onto parallelogram A'B'C'D'.
(a) (i) Draw the parallelogram A'B'C'D'.
(ii) Calculate the area of parallelogram A'B'C'D'.
(b) The side AB has length 3 cm. The original shape ABCD is now enlarged with a scale factor of \( \frac{2}{5} \) to give A”B”C”D”. Calculate the length of the side A”B”.
Answer: The original parallelogram ABCD has vertices A(6, 3), B(9, 3), C(12, 9), D(9, 9).
(a) (i) The scale factor is \( \frac{1}{3} \) and the centre of enlargement is the origin (0, 0).
To find the new vertices A'B'C'D', multiply each coordinate by the scale factor \( \frac{1}{3} \).
A' \( (6 \times \frac{1}{3}, 3 \times \frac{1}{3}) = (2, 1) \)
B' \( (9 \times \frac{1}{3}, 3 \times \frac{1}{3}) = (3, 1) \)
C' \( (12 \times \frac{1}{3}, 9 \times \frac{1}{3}) = (4, 3) \)
D' \( (9 \times \frac{1}{3}, 9 \times \frac{1}{3}) = (3, 3) \)
So, the vertices of the image parallelogram A'B'C'D' are A'(2, 1), B'(3, 1), C'(4, 3), and D'(3, 3).
(ii) To calculate the area of parallelogram A'B'C'D'.
First, find the base and height of the original parallelogram ABCD.
Base AB = \( 9 - 6 = 3 \) cm.
The perpendicular distance between AB (y=3) and CD (y=9) is \( 9 - 3 = 6 \) cm.
Area of parallelogram ABCD = base \( \times \) height \( = 3 \times 6 = 18 \) square cm.
Since the scale factor is \( k = \frac{1}{3} \), the area scale factor is \( k^2 = (\frac{1}{3})^2 = \frac{1}{9} \).
Area of A'B'C'D' = Area of ABCD \( \times k^2 = 18 \times \frac{1}{9} = 2 \) square cm.
(b) The original side AB has length 3 cm.
The new enlargement scale factor is \( \frac{2}{5} \).
The length of the side A''B'' in the new enlarged parallelogram will be the original length multiplied by the new scale factor.
Length of A''B'' \( = 3 \times \frac{2}{5} = \frac{6}{5} = 1.2 \) cm.
In simple words: When you shrink a shape by a scale factor of 1/3, its new sides are 1/3 as long, and its new area is \( (1/3)^2 = 1/9 \) of the original area. If you then enlarge a side by a different scale factor, you just multiply its current length by that new factor.

🎯 Exam Tip: Remember to apply the scale factor correctly: directly for lengths, and squared for areas. When the centre of enlargement is the origin, simply multiply the coordinates by the scale factor.

 

Question 12. A model of a van is constructed using a scale of 1 : 40. If the length of the real van is 8 metres, what is the length of the model?
Answer: The scale of the model to the real van is 1 : 40. This means that 1 unit on the model represents 40 units on the real van.
The length of the real van is 8 metres.
To find the length of the model, we use the scale ratio:
\( \frac { \text{Length of model} }{ \text{Length of real van} } = \frac { 1 }{ 40 } \)
Length of model \( = \frac { 1 }{ 40 } \times \text{Length of real van} \)
Length of model \( = \frac { 1 }{ 40 } \times 8 \) metres
Length of model \( = \frac { 8 }{ 40 } = \frac { 1 }{ 5 } \) metres
To express this in centimetres, multiply by 100 (since 1 metre = 100 cm):
Length of model \( = \frac { 1 }{ 5 } \times 100 \) cm
Length of model \( = 20 \) cm.
The length of the model van is 20 cm.
In simple words: If a model is 1/40th the size of a real van, and the real van is 8 meters long, the model will be 1/40th of 8 meters. This works out to be 20 cm.

🎯 Exam Tip: Ensure that the units are consistent throughout the calculation. Convert all measurements to a common unit (e.g., cm or m) before applying the scale factor to avoid errors.

 

Question 13. Akhil is drawing a plan of his classroom using a scale of 1 : 50. If the length and width of the classroom are 8 metres and 6 metres, how long will they each be on his drawing?
Answer: The scale of the plan is 1 : 50. This means 1 unit on the drawing represents 50 units in the actual classroom.
The actual length of the classroom is 8 metres.
The actual width of the classroom is 6 metres.
To find the length on the drawing:
Length on drawing \( = \frac { 1 }{ 50 } \times \text{Actual length} \)
Length on drawing \( = \frac { 1 }{ 50 } \times 8 \) metres
Convert to centimetres (1 metre = 100 cm):
Length on drawing \( = \frac { 8 }{ 50 } \times 100 \) cm \( = 8 \times 2 = 16 \) cm.
To find the width on the drawing:
Width on drawing \( = \frac { 1 }{ 50 } \times \text{Actual width} \)
Width on drawing \( = \frac { 1 }{ 50 } \times 6 \) metres
Convert to centimetres:
Width on drawing \( = \frac { 6 }{ 50 } \times 100 \) cm \( = 6 \times 2 = 12 \) cm.
On Akhil's drawing, the length will be 16 cm and the width will be 12 cm. This makes the drawing a smaller, accurate representation.
In simple words: Akhil's drawing is 50 times smaller than the real classroom. So, for every 8 meters of real length, it's 16 cm on the drawing, and for every 6 meters of real width, it's 12 cm on the drawing.

🎯 Exam Tip: Always make sure to convert all measurements to a consistent unit (like centimetres for drawings) before applying the scale factor to avoid calculation errors.

 

Question 14. Nikhil is drawing a plan of the gymnasium at his school using a scale of 1 : 75. If the dimensions of the gymnasium on his plan are 28 cm and 16 cm, what are the dimensions of the real gymnasium?
Answer: The scale of the plan is 1 : 75. This means 1 unit on the plan represents 75 units in the real gymnasium.
The length on the plan is 28 cm.
The width (breadth) on the plan is 16 cm.
To find the actual length of the gymnasium:
Actual length \( = \text{Length on plan} \times 75 \)
Actual length \( = 28 \times 75 \) cm \( = 2100 \) cm.
Convert to metres (100 cm = 1 metre):
Actual length \( = \frac { 2100 }{ 100 } \) m \( = 21 \) m.
To find the actual width (breadth) of the gymnasium:
Actual width \( = \text{Width on plan} \times 75 \)
Actual width \( = 16 \times 75 \) cm \( = 1200 \) cm.
Convert to metres:
Actual width \( = \frac { 1200 }{ 100 } \) m \( = 12 \) m.
The dimensions of the real gymnasium are 21 m by 12 m. It's a large space suitable for sports.
In simple words: The plan is 75 times smaller than the real gymnasium. So, to find the real size, multiply the plan's length (28 cm) and width (16 cm) by 75, then convert those answers from centimeters to meters. The real gym is 21 meters long and 12 meters wide.

🎯 Exam Tip: When converting from a smaller scale (plan) to a larger scale (real object), you multiply by the scale factor. Always remember to convert between cm and m (1 m = 100 cm) if the question asks for different units.

 

Question 15. Krishna has a model of the Eiffel Tower which is 16 cm high. If the height of the real tower is 320 m, what is the scale of the model?
Answer: The height of the model of the Eiffel Tower is 16 cm.
The height of the real Eiffel Tower is 320 m.
To find the scale, we need to compare the model's height to the real tower's height in the same units. Let's convert the real height to centimetres:
1 metre = 100 cm
320 m \( = 320 \times 100 \) cm \( = 32000 \) cm.
Now, the scale of the model is Model height : Real height.
Scale \( = 16 \text{ cm} : 32000 \text{ cm} \)
To simplify the ratio, divide both sides by 16:
Scale \( = \frac{16}{16} : \frac{32000}{16} \)
Scale \( = 1 : 2000 \)
The scale of the model is 1 : 2000. This means 1 cm on the model represents 2000 cm (or 20 meters) in real life.
In simple words: To find the scale, you compare the size of the model to the size of the real thing. Make sure both are in the same units (like centimeters). The model Eiffel Tower is 16 cm tall, and the real one is 32,000 cm tall. So, 1 cm on the model stands for 2000 cm in real life.

🎯 Exam Tip: The most crucial step when finding a scale is to ensure both measurements are in the same units before forming the ratio. A scale ratio must always be simplified to the form 1:N.

 

Question 16. The dimensions of the model of a multi-storey building are 1.2 m x 75 cm x 2 m. If the scale factor is 1 : 30, find the actual dimensions of the building.
Answer: The dimensions of the model building are given as 1.2 m x 75 cm x 2 m.
Let's convert all model dimensions to centimetres for consistency.
Length of model \( = 1.2 \) m \( = 1.2 \times 100 \) cm \( = 120 \) cm.
Width of model \( = 75 \) cm.
Height of model \( = 2 \) m \( = 2 \times 100 \) cm \( = 200 \) cm.
So the model dimensions are 120 cm x 75 cm x 200 cm.
The scale factor is 1 : 30. This means 1 unit on the model represents 30 units on the actual building.
To find the actual dimensions, multiply each model dimension by 30.
Actual Length \( = 120 \text{ cm} \times 30 = 3600 \) cm.
Convert to metres: \( \frac { 3600 }{ 100 } \) m \( = 36 \) m.
Actual Width \( = 75 \text{ cm} \times 30 = 2250 \) cm.
Convert to metres: \( \frac { 2250 }{ 100 } \) m \( = 22.5 \) m.
Actual Height \( = 200 \text{ cm} \times 30 = 6000 \) cm.
Convert to metres: \( \frac { 6000 }{ 100 } \) m \( = 60 \) m.
Hence, the actual dimensions of the building are 36 m x 22.5 m x 60 m.
In simple words: The model is 30 times smaller than the real building. So, we change all the model's measurements to centimeters, then multiply each by 30 to get the real size in centimeters. Finally, we convert these back to meters. The real building is 36 meters long, 22.5 meters wide, and 60 meters high.

🎯 Exam Tip: Always make sure to convert all given dimensions to a single, consistent unit (e.g., cm) before applying the scale factor to avoid errors, and then convert back to the desired final unit.

 

Question 17. On a map drawn to a scale of 1 : 2500, a triangular plot of land has the following measurements: AB = 3 cm, BC = 4 cm and \( \angle ABC = 90^\circ \).
(i) Calculate the actual lengths of AB and AC in km;
(ii) The actual area of the plot in km².
Answer: The map scale is 1 : 2500. This means 1 cm on the map represents 2500 cm in real life.
The measurements of the triangular plot on the map are AB = 3 cm, BC = 4 cm, and \( \angle ABC = 90^\circ \).
(i) Calculate the actual lengths of AB and AC in km:
Actual length of AB:
\( = 3 \text{ cm} \times 2500 = 7500 \) cm.
Convert to metres (100 cm = 1 m): \( = \frac { 7500 }{ 100 } \) m \( = 75 \) m.
Convert to kilometres (1000 m = 1 km): \( = \frac { 75 }{ 1000 } \) km \( = 0.075 \) km.
Actual length of BC:
\( = 4 \text{ cm} \times 2500 = 10000 \) cm.
Convert to metres: \( = \frac { 10000 }{ 100 } \) m \( = 100 \) m.
Convert to kilometres: \( = \frac { 100 }{ 1000 } \) km \( = 0.1 \) km.
To find the actual length of AC, we first find AC on the map using Pythagoras theorem, since \( \triangle ABC \) is a right-angled triangle:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = 3^2 + 4^2 \)
\( AC^2 = 9 + 16 \)
\( AC^2 = 25 \)
\( AC = \sqrt{25} = 5 \) cm (on the map).
Actual length of AC:
\( = 5 \text{ cm} \times 2500 = 12500 \) cm.
Convert to metres: \( = \frac { 12500 }{ 100 } \) m \( = 125 \) m.
Convert to kilometres: \( = \frac { 125 }{ 1000 } \) km \( = 0.125 \) km.
(ii) The actual area of the plot in km²:
Area of the triangle on the map \( = \frac { 1 }{ 2 } \times \text{base} \times \text{height} = \frac { 1 }{ 2 } \times AB \times BC \)
Area on map \( = \frac { 1 }{ 2 } \times 3 \text{ cm} \times 4 \text{ cm} = 6 \) square cm.
The scale factor for area is \( (2500)^2 \).
Actual Area \( = \text{Area on map} \times (2500)^2 \)
Actual Area \( = 6 \text{ cm}^2 \times (2500)^2 \)
Actual Area \( = 6 \times 6250000 \) square cm \( = 37500000 \) square cm.
Convert to square metres (1 \( \text{m}^2 = 100 \times 100 = 10000 \text{ cm}^2 \)):
Actual Area \( = \frac { 37500000 }{ 10000 } \) \( \text{m}^2 = 3750 \) \( \text{m}^2 \).
Convert to square kilometres (1 \( \text{km}^2 = 1000 \times 1000 = 1000000 \text{ m}^2 \)):
Actual Area \( = \frac { 3750 }{ 1000000 } \) \( \text{km}^2 = 0.003750 \) \( \text{km}^2 \).
Alternatively, using actual lengths in metres first:
Actual Area \( = \frac { 1 }{ 2 } \times \text{Actual AB} \times \text{Actual BC} \)
Actual Area \( = \frac { 1 }{ 2 } \times 75 \text{ m} \times 100 \text{ m} = 3750 \) \( \text{m}^2 \).
Convert to square kilometres: \( = \frac { 3750 }{ 1000000 } \) \( \text{km}^2 = 0.00375 \) \( \text{km}^2 \).
In simple words: The map is 2500 times smaller than reality. So, to find real lengths, multiply map lengths by 2500 and convert to kilometers. For area, multiply the map area by \( 2500^2 \) and convert to square kilometers. The real plot's sides AB and AC are 0.075 km and 0.125 km respectively, and its area is 0.00375 square kilometers.

🎯 Exam Tip: Pay close attention to unit conversions for both lengths and areas. When converting square units, remember that \( 1 \text{ m}^2 = 100^2 \text{ cm}^2 \) and \( 1 \text{ km}^2 = 1000^2 \text{ m}^2 \). For right-angled triangles, use Pythagoras theorem to find unknown side lengths.

 

Question 18. The scale of a map is 1 : 200000. A plot of land of area 20 km² is to be represented on the map; find :
(i) the number of kilometres on the ground which is represented by 1 centimetre on the map;
(ii) the area in km² that can be represented by 1 cm²;
(iii) the area on the map that represents the plot of land.
Answer:
(i) The map scale 1:200000 means that 1 cm on the map shows 200000 cm on the actual ground. To change centimetres to kilometres, we divide by 100 to get meters, then by 1000 to get kilometres. So, 200000 cm is equal to 2 km on the ground. This helps us understand real distances from map measurements.
(ii) From part (i), we know that 1 cm on the map represents 2 km on the ground. So, to find the area represented by 1 cm² on the map, we multiply the ground distance by itself: \( 2 \text{ km} \times 2 \text{ km} = 4 \text{ km}^2 \). This means a small square of 1 cm by 1 cm on the map covers a very large area of 4 km² in reality.
(iii) The actual plot of land has an area of 20 km². We found in part (ii) that 1 cm² on the map represents 4 km² on the ground. To find how much area this 20 km² plot takes up on the map, we divide its actual area by the area represented by 1 cm² on the map: \( \frac{20 \text{ km}^2}{4 \text{ km}^2/\text{cm}^2} = 5 \text{ cm}^2 \). So, a 20 km² plot of land would be shown as a 5 cm² area on the map.
In simple words: A map scale tells us how much real distance 1 unit on the map shows. For lengths, you multiply by the scale factor. For areas, you multiply or divide by the square of the scale factor.

🎯 Exam Tip: Remember that when dealing with scale factors, linear measurements (like length) are directly proportional, but area measurements involve the square of the scale factor.

 

Question 19. A model of a ship is made to a scale of 1 : 200.
(i) The length of the model is 4 m. Calculate the length of the ship.
(ii) The area of the deck of the ship is 160000 m². Find the area of the deck of the-model.
(iii) The volume of the model is 200 litres. Calculate the volume of the ship in m³.
Answer:
(i) The model has a scale of 1:200, which means every 1 unit on the model represents 200 units on the real ship. If the model's length is 4 m, we multiply this by the scale factor to find the real length. So, \( 4 \text{ m} \times 200 = 800 \text{ m} \). The real ship is very long, 800 metres.
(ii) The area of the ship's actual deck is 160000 m². For areas, the scale factor is squared. So, if the linear scale is 1:200, the area scale is \( 1^2 : 200^2 \), which is \( 1 : 40000 \). To find the model's deck area, we divide the actual area by the square of the scale factor: \( \frac{160000 \text{ m}^2}{200^2} = \frac{160000}{40000} = 4 \text{ m}^2 \). This is how large the deck area would be on the scaled model.
(iii) The volume of the model is 200 litres. First, we convert this to cubic metres, knowing that 1000 litres equals 1 m³. So, \( \frac{200}{1000} = 0.2 \text{ m}^3 \). For volumes, the scale factor is cubed. The scale factor is 200, so we multiply the model's volume by \( 200^3 \). The actual volume of the ship is \( 0.2 \text{ m}^3 \times (200)^3 = 0.2 \text{ m}^3 \times 8,000,000 = 1,600,000 \text{ m}^3 \). This shows how much bigger the real ship is in terms of space it occupies.
In simple words: When a model is made with a scale, lengths are multiplied by the scale, areas by the scale squared, and volumes by the scale cubed. Remember to convert units like litres to cubic meters before calculating.

🎯 Exam Tip: Remember that for scale factors, length changes by k, area by \( k^2 \), and volume by \( k^3 \). Always check units and convert them properly before calculations.