Finding of the area in itself of a rectangle is quite simple type of tasks. But very often this type of exercises is complicated by introduction of additional unknown. For their decision you will need the broadest knowledge of various sections of geometry.

## It is required to you

- - Notebook;
- - ruler;
- - pencil;
- - handle;
- - calculator.

## Instruction

1. The rectangle is a quadrangle which has all right angles. A special case of a rectangle is the square. The area of a rectangle is the size equal to the work of its length and width. And the area of a square is equal to its length of its party built in the second degree. If only width is known, then you have to find at first length, and then calculate the area.

2. For example, the rectangle of ABCD (Fig. 1) where AV = 5 cm, IN = 6.5 cm is given. Find the area of a rectangle of ABCD.

3. Since ABCD – a rectangle, the joint-stock company = OS, IN = OD (as rectangle diagonals). Consider AVS triangle. AV = 5 (on a condition), the EXPERT = 2 joint-stock companies = 13 cm, AVS corner = 90 (since ABCD – a rectangle). Therefore AVS – a rectangular triangle., in which AB and BC – legs, and the EXPERT – a hypotenuse (since it is opposite to a right angle).

4. Pythagorean theorem says: the square of a hypotenuse is equal to the sum of squares of legs. On Pythagorean theorem you find leg BC.BC ^2 = the EXPERT ^2 – AV ^2ВС ^2 = 13 ^2 – 5 ^2ВС ^2 = 169 - 25BC ^2 = 144BC = 144BC = 12

5. Now you can find the area of a rectangle of ABCD.S = AV * BCS = 12 * 5S = 60.

6. Also the option where width will be known partially is possible. For example, ABCD rectangle where AB=1/4AD, OHM – a median of a triangle of AOD, OM=3, AO=5 is given. Find the area of a rectangle of ABCD.

7. Consider AOD triangle. The corner of OAD is equal to ODA corner (since. The expert and BD – rectangle diagonals). Therefore, AOD triangle – isosceles. And in an isosceles triangle the median of OHMS is at the same time a bisector and height. Means, AOM triangle – rectangular.

8. In AOM triangle where OHM and AM – legs, find what it is equal OHM (hypotenuse) to. On Pythagorean theorem АМ^2 = АО^2 - ОМ^2АМ = 25-9AM = 16AM = 4

9. Now calculate the area of a rectangle of ABCD. AM = 1/2AD (since. Ohm, being a median, halves AD). Therefore AD = 8.AB=1/4AD (on a condition). From here AV = 2.S = AB*ADS = 2*8S = 16