Advertisements
Advertisements
प्रश्न
An aeroplane is 30m long and its model is l5 cm long. If the total outer surface area of the model is 150 cm2 , find the cost of painting the outer surface of the aeroplane at Rs. 120 per m2, if 5O m2 is left out for windows.
उत्तर
15 cm represents = 30m
1 cm represents `30/15` = 2 m
1 cm2 represents 2m x 2m = 4 m2
Surface area of the model = 150cm2
Actual surface area of aeroplane = 150 x 2 x 2 m2 = 600 m2
50 m2 is left out for windows
Area to be painted = 600 - 50 = 50 m2
Cost of painting per m2 = Rs. 120
Cost of painting 550 m2 = 120 x 550 = Rs 66000
APPEARS IN
संबंधित प्रश्न
In the following figure, DE || OQ and DF || OR, show that EF || QR.
In the given figure, QR is parallel to AB and DR is parallel to AB and DR is parallel to QB.
Prove that: PQ2 = PD × PA.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5,
Find:
- ∆APB : ∆CPB
- ∆DPC : ∆APB
- ∆ADP : ∆APB
- ∆APB : ∆ADB
The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5cm, find the length of QR.
In the following figure, seg BE ⊥ seg AB and seg BA ⊥ seg AD. If BE = 6 and \[\text{AD} = 9 \text{ find} \frac{A\left( \Delta ABE \right)}{A\left( \Delta BAD \right)} \cdot\]

Prove that the area of the triangle BCE described on one side BC of a square ABCD as base is one half of the area of similar triangle ACF described on the diagonal AC as base.
The areas of two similar triangles are 16cm2 and 9cm2 respectively. If the altitude of the smaller triangle is 1.8cm, find the length of the altitude corresponding to the larger triangle.
A model of a ship is made to a scale of 1:500. Find: The area other deck o the ship, if the area of the deck of its model is m2
In ∆LMN, ∠L = 60°, ∠M = 50°. If ∆LMN ~ ∆PQR then the value of ∠R is
Are triangles in figure similar? If yes, then write the test of similarity.