Similarity Class 10 RS Aggarwal Exe-15 Goyal Brothers ICSE Maths Solutions Ch-15. Step by step solutions of exercise-15 questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Similarity Class 10 RS Aggarwal Exe-15 Goyal Brothers ICSE Maths Solutions Ch-15
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-15 | Similarity |
| Writer/ Book | RS Aggarwal |
| Topics | Solution of Exe-15 Questions |
| Academic Session | 2024-2025 |
Similarity
Two figures are similar if they share a common shape, but can be different sizes. The symbol for similar figures is “∼”.
Scale Factor
The common ratio between the dimensions of similar figures is called the scale factor. This number can be used to increase or decrease the size of a figure without changing its shape
Exercise- 15
( Similarity Class 10 RS Aggarwal Exe-15 Goyal Brothers ICSE Maths Solutions Ch-15 )
Que-1: ΔABC with sides AB = 3.6cm, BC = 4.5cm and CA = 6cm is enlarged to ΔA’B’C’ such that the largest side of the enlarged triangle is 10cm. Find the scale factor and use it to find the lengths of the other sides of ΔA’B’C’.
Sol: AB = 3.6cm
BC = 4.5cm
CA = 6cm
The largest side of ΔA’B’C’ = 10cm
Scale factor k = 10/6 = 3/5
A′B′ = k × AB = (5/3) × 3.6cm = 6cm
B′C′ = k × BC = (5/3) × 4.5cm = 7.5cm.
Que-2: A ΔABC with sides AB = 16cm, BC = 12cm, and CA = 18cm is reduced to ΔA’B’C’ such that the smallest side of the image triangle is 4.8cm. Find the scalar factor and use it to find the lengths of the other sides of ΔA’B’C’.
Sol: Given : AB = 16cm
BC = 12cm
CA = 18cm
The smallest side of the image triangle is 4.8cm
Scale factor = 4.8/12
= 2/5
So, A’B’ = 16 × (2/5)
= 32/5 = 6.4 cm
C’A’ = 18 × (2/5)
= 36/5 = 7.2 cm.
Que-3: A ΔPQR is reduced by a scale factor 0.72. If the area of ΔPQR is 62.5cm², find the area of its image.
Sol: Given : Scale factor = 0.72
area of ΔPQR = 62.5cm²
So, area of its image ΔP’Q’R’
= 62.5 × (0.72)²
= 32.4 cm²
Que-4: A rectangle having an area of 60cm² is transformed under enlargement about a point in space. If the area of its image is 135cm², find the scale factor of the enlargement.
Sol: Area of rectangle = 60 cm²
area of its image = 135cm²
Scale factor k = √[area of image/area of rectangle]
k = √(135/60)
k = √(9/4)
k = 3/2 = 1.5.
Que-5: On a map, drawn to a scale of 1:25000, a triangular plot LMN of land has the following measurements : LM = 6cm, MN = 8cm, and ∠LMN = 90°. Calculate :
(i) the actual lengths of MN and LN in kilometres,
(ii) the actual area of the plot in sq. km.
Sol: scale of 1:25000
LM = 6cm
MN = 8cm
∠LMN = 90°
By Pythagoras theorem,
LN² = LM² + MN²
LN² = 6² + 8²
LN² = 36+64
LN = √100 = 10cm
(i) Actual length of MN = 1/k
= 8 × 25000 cm
= (8×25000)/(100×1000) km
= 2 km
Actual length of MN = 1/k
= 10 × 25000 cm
= (10×25000)/(100×1000) km
= 2.5 km.
(ii) Actual length of LM
= (6×25000)/(100×1000) km
= 1.5 km
Actual area of the plot = (1/2)×LM×MN
= (1/2)×1.5×2
= 1.5 km².
Que-6: The scale of a map is 1:200000. A plot of land of area 10km² is to be represented on the map. Find :
(i) the length in km on the ground, represented by 1cm on the map,
(ii) the area in km² that can be represented by 3cm² on the map,
(iii) the area on the map, representing the plot of land.
Sol: Scale of a map is 1:200000
area = 10km²
(i) Length of the map is 1cm
Actual length = (1/k) ×1
= 200000 × 1
= (200000)/(100×1000)
= 2km
(ii) Area of the map is 3 cm²
Actual length = 3 × (1/k)²
= 3 × (200000)² cm²
= [3×200000×200000]/[100×100×1000×1000] km²
= 12 km²
(iii) Actual are of plot = 10 km²
Area on the map = k² × (10 km²)
= (1/200000)² × (10 km²)
= [10×(100)²×(1000)²]/(200000)²
= [10×100×100×1000×1000]/[200000×200000]
= 2.5 cm²
Que-7: On a map drawn to a scale of 1:20000, a rectangular plot of land ABCD has AB = 32cm and BC = 24cm, Calculate :
(i) the diagonal distance of the plot in km,
(ii) the area of the plot in sq. km.
Sol: Scale :- 1 : 20000
1 cm represents 20000 cm = (200000)/(100×1000) = 0.2 km
(i) AC2 = AB2 + BC2
= 242 + 322
= 576 + 1024
= 1600
AC = 40 cm
Actual length of diagonal = 40 × 0.2 km = 8 km
(ii) 1 cm represents 0.2 km
1 cm2 represents 0.2 × 0.2 km2
The area of the rectangle ABCD = AB × BC
= 24 × 32
= 768 cm2
Actual area of the plot = 0.2 × 0.2 × 768 km2 = 30.72 km2
Que-8: The dimensions of the model of a multistorey building are 1m×60cm×1.25m. If the model is drawn to a scale 1:60, find the actual dimensions of the building in metres. Find :
(i) the floor area of a room of the building, whose area in the model is 250cm².
(ii) the volume of the room in the model, whose actual volume is 648 cubic metres.
Sol: Length of the model = 1 m = 100 cm
Width of the model = 60 cm
Height of the model = 1.25 m = 125 cm
The scale is 1:60,
Now, calculate the actual dimensions:
Actual length = 1×60 = 60 m
Actual width = 0.60×60 = 36 m
Actual height = 1.25×60 = 75 m
(i) The scale for area is (1:60)² = 1:3600, meaning the actual area is 3600 times larger than the model.
Actual area = 250 cm²×3600
Actual area = 900000 cm² = 90 m²
(ii) The scale for volume is (1:60)³ = 1:216000, meaning the model volume is 216,000 times smaller than the actual volume.
Model volume = 648/216000 m³
Model volume = 0.003 m³
Since 1 cubic meter = 1,000,000 cm³:
Model volume = 0.003 m³×1,000,000 = 3000 cm³.
Que-9: A model of a ship is made to a scale of 1:250. Find :
(i) the length of the ship, if the length of its model is 1.2m,
(ii) the area of the deck of the ship, if the area of the deck of its model is 1.6m².
(iii) the volume of its model, when the volume of the ship is 1 cubic kilometres.
Sol: Scale of 1:250
(i) length of model = 1.2m
Actual length = 1.2×250
= 300m.
(ii) Area of deck of its model = 1.6 m²
Actual area = 1.6×62500 = 100000 m²
(iii) The actual volume of the ship is given as 1 km³
Since 1 km³ = 1,000,000,000 m³, we will calculate the volume of the model using the volume scale, which is (1:250)³ = 1:15625000. The volume of the model is:
Model volume = 1,000,000,000/15625000
= 64m³.
Que-10: A model of a ship is made to a scale of 1:200.
(i) If the length of the model is 4m, calculate the length of ship
(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model.
(iii) If the volume of the model is 200 litres, calculate the volume of the ship in cubic metres.
Sol: Scale factor = k = 1/200
(i) Length of model = k × actual length of the ship
⟹ Actual length of the ship = 4 × 200 = 800 m
(ii) Area of the deck of the model = k2 × area of the deck of the ship
= (1/200)² × 160000 m² = 4m²
(iii) Volume of the model = k3 × volume of the ship
Volume of the ship
k = (1/k)³×200 liters
=(200)3 x 200 liters
= 1600000000 liters
= 1600000 m3
— : End of Similarity Class 10 RS Aggarwal Exe-15 Goyal Brothers ICSE Maths Solutions :–
Return to:- ICSE Class 10 Maths RS Aggarwal Solutions
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