Similar Triangles Class 10 OP Malhotra Exe-12E ICSE Maths Solutions Ch-12. We Provide Step by Step Solutions / Answer of Exe-12E Questions of S Chand OP Malhotra Maths . In this article you would learn solving problems on Area of Model and Map. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Similar Triangles Class 10 OP Malhotra Exe-12E ICSE Maths Solutions Ch-12
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 10th |
| Chapter-12 | Similar Triangles |
| Writer | OP Malhotra |
| Exe-12E | problems on Area of Model and Map |
| Edition | 2024-2025 |
How to Solve Problems on Area of Model and Map
the “area of the model” or “area of the map” refers to the actual area covered by the representation on the map or model, which either smaller / larger than the real-world area it represents, and the ratio between these areas is equal to the square of the scale factor used to create the model or map.
Resulting Area (Map / Model ) = (scale factor)² x Original Area
Exercise- 12E
Similar Triangles Class 10 OP Malhotra ICSE Maths Solutions Ch-12
Que-1: 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?
Sol: Side of a square = 3 cm
When it is enlarged, the scale factor = 2
i.e. k = 2
Area of the square = (side)² = (3)² = 9 cm²
∴ Area of the image of square = k²
(Area of square) = (2)² x 9 = 4 x 9 = 36 cm²
Que-2: A triangle, whose area is 12 cm, 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.
Sol: Area of the triangle = 12 cm²
Area of enlarged triangle = 108 cm²
Let scale factor = k and then scale factor of area = k²
Then k² x 12 = 108 ⇒ k² = 108/12 = 9
∴ k = √9 = 3
Que-3: ∆ABC with sides AB = 10 cm, BC = 12 cm and AC = 16 cm is enlarged to the triangle A’B’C’ such that the largest side of the triangle A’B’C’ is 24 cm. Find the Scale (enlargement) factor and use it to find the lengths of other sides of the linage triangle A’B’C’.
Sol: Sides of given ∆ABC, AB = 10 cm, BC = 12 cm and AC = 16 cm
Longest side AC = 16 cm
But longest side of enlarged ∆A’B’C’ = 24 cm
i.e. A’C’ = 24 cm
Let k be the scale factor
∴ k = 24/16 = 3/2
Now length of side A’B’ = 10 cm x (3/2) = 15 cm
and length of side B’C’ = 12 cm x (3/2) = 18 cm
Que-4: A ∆PQR right angled at Q has PQ = 8 cm and QR = 15 cm. It is enlarged to a triangle P’Q’R’ such that the image of P is P’, image of Q is Q’, image of R is R’ and length of P’R’ = 42.5 cm. Find :
(i) the scale (enlargement factor);
(ii) the lengths of P’Q’ and Q’R’;
(iii) the area of ∆PQR and hence area of image triangle P’Q’R’.
Sol: In right angled ∆PQR, ∠Q = 90°
PQ = 8 cm and QR = 15 cm
∴ PR = √(PQ²+QR²) = √((8)²+(15)²)
= √(64+225) = √289 = 17 cm
∆PQR is enlarged into AP’Q’R’ in which . image of P is P’ of Q is Q’ and of R is R’
Length of P’R’ = 42.5 cm
(i) Let k be the scale factor of enlargement 42.5 = k x PR = k x 17
∴ k = 42.5/17 = 2.5 = 25/10 = 5/2
(ii) Now length of P’Q’ = (5/2) x PQ = (5/2) x 8 = 20 cm
and length of Q’R’ = (5/2) QR = (5/2) x 15 = 75/2
= 37.5 cm
(iii) Now area of ∆PQR = (1/2) x PQ x QR
∴ (1/2) x 8 x 15 cm² = 60 cm²
∴ area of ∆P’Q’R’ = k² x area of ∆PQR
= (5/2)² x 60 cm² = (25/4) x 60 = 375 cm²
Que-5: On a map scale 1 cm to the km an island has an area of 3.5 cm². Find the actual area.
Sol: Scale of map = 1 cm : 1 km or 1 : 100000
Now area of island on the map = 3.5 cm²
∴ Actual area = k² x area of the map
= (100000)² x 3.5 cm²
= 100000 x 100000 x 3.5 cm²
= {100000×100000×3.5}/{100000×100000} km²
= 3.5 km²
Que-6: The scale of a map is 5 m to 1 cm. (i) What area in m² is represented by a square of side 3 cm on the map ? (ii) What is the length of the side of a square on the map which represents an area of 50625 m²?
Sol: Scale of map = 5 m to 1 cm
or 500 : 1 ⇒ k = 500/1
Side of a square on the map = 3 cm
∴ Its area = (side)² = (3 cm)² = 9 cm²
(i) ∴ Actual area of the square = k² x area of square on the map = (500)² x 9 cm²
= 500 x 500 x 9 cm²
= {500×500×9}/{100×100} = 225 m²
(ii) Actual area of a square = 50625 m²
Area on the map = (1/k²) x actual area
= {1/(500)²} x 50625
= {50625×100×100}/{500×500} cm² = 2025 cm²
∴ Side = √2025 = 45 cm
Que-7: A ground plan of a house is made on the scale 1 cm to 15 m. Find the scale factor of the plan. Find the length and breadth on the plan of a room 18 m by 12 m. What area on the ground is represented by 1 cm² on the plan ?
Sol: Scale of a ground plan of a house = 1 cm : 15 m
Scale factor (k) = 1 cm : 1500 cm
= 1 : 1500 = 1/1500
Now length of a room = 18 m
and breadth = 12 m
∴ Length on plan = 18 x (1/1500) m
= (18×100)/1500 cm = 6/5 cm = 1.2 cm
and breadth = 12 x (1/1500) m = 1200/1500 m = 1200/1500 cm
= 4/5 cm = 0.8 cm
Area on the plan = 1 cm³
Area on the ground = 1 x (1500)² cm²
= {1×1500×1500}/{100×100} = 225 m²
Que-8: On a map of 1 cm to the 4 kni, an estate occupies an area of 9.37 cm². Find the actual area to the nearest km².
Sol: Scale on the map = 1 cm to 4 km = 1 cm : 400000 cm
= 1 : 400000
⇒ k = 1/400000
Area of an estate on the map = 9.37 cm²
Actual area = k² (area on the map)
= (400000)² x 9.37 cm²
= {937×(400000)²}/100
= {937×400000×400000}/{100×100000×100000}
= 14992/100 = 149.92 km²
= 150 km² (nearest km²)
Que-9: The map of a rectangular field drawn to a scale of 1 : 20000 measures 20 cm by 15 cm. Calculate the actual area of the Field in sq. km (km²).
Sol: Scale of a map = 1 : 20000
Measure of field on the map = 20 cm x 15 cm = 300 cm²
Actual area = k² (Area of the field on the map)
= (20000)² x 300 cm²
= 20000 x 20000 x 300 cm²
= {20000×20000×300}/{100000×100000}
= 12 km²
Que-10: The scale of a map is 1 : 50,000. In the map, a triangular plot ABC of land has the following dimensions : AB = 2 cm, BC = 3.5 cm and ∠ABC = 90°. Calculate: (i) the actual length of BC,in km of the land (ii) the area of the plot in sq. km.
Sol: Scale on the map = 1 : 50000
On the map in right angled ∆ABC, AB = 2
cm, BC = 3.5 cm and ∠ABC = 90°
∴ Area = (1/2) x AB x BC = (1/2) x 2 x 3.5 cm²
= 3.5 cm²
(i) Now actual length of BC = 3.5 cm x 50000
= (3.5×50000)/100000
= (35×50000)/(10×100000) = (35×5)/100
= 175/100
and actual area = 3.5 cm² x (50000)²
= (3.5×50000×50000)/(100000×100000) km²
= (3.5×25)/100 = 3.5/4 = 0.875 k²
Que-11: The dimensions of the model of a multistorey building are 1 m by 60 cm by 1.20 m. If the scale factor is 1 : 50, find the actual dimensions of the building. Also, find :
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq. cm;
(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90 m³.
Sol: Scale factor of the model = 1 : 50
Dimensions of a model of a multistorey building are 1 m x 60 cm x 1.20 m
or 100 cm x 60 cm x 120 cm
∴ Actual length = 100 x 50 cm = (100×50)/100 m
= 50 m
Breadth = 60 x 50 cm
= (60×50)/100 = 30 m
and height = 120 x 50 cm
= (120×50)/100 = 60 m
∴ Actual dimensions = 50 m x 30 m x 60 m
(i) Floor area of a room on the model = 50 sq. cm
∴ Actual area = 50 x k² = 50 x (50 x 50) cm²
= (50×2500)/(100×100) m² = 12.5 m²
(ii) Actual volume of the room space = 90 m³
∴ Volume on the model = 90/(50)³ m³
= 90/(50×50×50) m³
= (90×100×100×100)/(50×50×50) cm³
= 90000/125 cm² = 720 cm³
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