The geometry studies properties and characteristics of two-dimensional and spatial figures. The numerical sizes characterizing such designs are the area and perimeter which calculation is made on the known formulas or one is expressed through another.
Instruction
1. Rectangle. Task: calculate area rectangle if it is known that its perimeter is equal 40, and length of b is 1.5 times more than a width.
2. Decision. Use the known formula of perimeter, it is equal to the sum of all parties of a figure. In this case P = 2•a + 2•b. From initial data of a task you know that b = 1.5•a, therefore, P = 2•a + 2•1,5•a = 5 · a from where a = 8. Find b length = 1.5•8 = 12.
3. Write down a formula for the area of a rectangle: S = a • b, Substitute the known sizes: S = 8 · *12 = 96.
4. Square. Task: find the area of a square if the perimeter is equal to 36.
5. Decision. The square – a special case of a rectangle where all parties are equal, therefore, its perimeter is equal 4•a from where a = 8. The area of a square determine by formula S = a² = 64.
6. Triangle. Task: let any triangle of ABC which perimeter is equal to 29 be given. Learn the size of its area if it is known that BH height lowered on the party of AC divides it into pieces with lengths of 3 and 4 cm.
7. Decision. For a start remember an area formula for a triangle: S = 1/2 · with • h, where with – the basis and h – figure height. In our case the party of AC which is known for a statement of the problem will be the basis: AC = 3+4 = 7, it was necessary to find BH height.
8. Height is the perpendicular which is carried out to the party from opposite top, therefore, it to divide a triangle of ABC into two rectangular triangles. Knowing this property, consider ABH triangle. Remember Pythagoras's formula according to which: AB² = BH² + AH² = BH² + 9 → AB = √ (h² + 9). In BHC triangle by the same principle write down: BC² = BH² + HC² = BH² + 16 → BC = √ (h² + 16).
9. Apply a perimeter formula: P = AB + BC + ASPodstavte the sizes expressed through height: P = 29 = √ (h² + 9) + √ (h² + 16) + 7.
10. Solve the equation: √ (h² + 9) + √ (h² + 16) = 22 → [replacement of t² = h² + 9]: √ (t² + 7) = 22 - t, square both parties of equality: t² + 7 = 484 - 44 ·t + t² → t≈10.84h² + 9 = 117.5 → h ≈ 10.42
11. Find the area of a triangle of ABC: S = 1/2·7 · 10.42 = 36.47.