Advice 1: How to find the sides of the rectangle

A special case of a parallelogram - rectangle – only known in Euclidean geometry. Have a rectangle all the angles are equal, and each of them individually is 90 degrees. On the basis of private properties of the rectangleand from the properties of the parallelogram of the parallel opposite sides can find a hand figure by the given diagonals and angle from their intersection. Calculating sides of a rectangle based on additional constructions and apply the properties of the resultant shapes.
Instruction
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Draw a rectangle EFGH. Record the known data: the diagonal of the rectangle EG, and the angle α, obtained from the intersection of the two equal diagonals FH and EG. Build the figure and mark the diagonal between them the angle α.
How to find <b>hand</b> <strong>rectangle</strong>
2
The letter And mark the point of intersection of the diagonals. Consider the constructions formed by the triangle EFА. According to the property of the rectangle, its diagonals are equal and are bisected by the intersection point A. Calculate the values of FA and EA. Since the triangle is isosceles EFА and his side of EA and FA are equal and accordingly equal to half the diagonal EG.
3
Next, calculate the first side EF of the rectangle. This is the third unknown side of the considered triangle EFА. According to the theorem of the cosines of the appropriate formula, find the side EF. To do this, substitute in the formula for cosines of the previously obtained values of FA equal to the sides EA and the cosine of the known angle α between them. Calculate and record the obtained value of EF.
How to find <b>hand</b> <strong>rectangle</strong>
4
Find the second side of the rectangle FG. For this we consider another triangle EFG. It is rectangular, where the hypotenuse BC and the side EF. According to the Pythagorean theorem, find the second leg FG with the applicable formula.
How to find <b>hand</b> <strong>rectangle</strong>
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In accordance with the properties of the rectangle, its opposite edges are equal. Thus, the side GH is equal to the side EF, and = FG. Write in the answer all the computed sides of the rectangle.

Advice 2: How to calculate the area perimeter

Geometry studies the properties and characteristics of planar and spatial figures. Numeric quantities characterizing such structures are area and perimeter, the calculation of which is produced by known formulae or expressed one through another.
Instruction
1
Rectangle.Task: compute the area of a rectangleif you know its perimeter is 40 and the length b is 1.5 times the width a.
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Solution.Use the well-known formula of the perimeter, it is equal to the sum of all sides of the figure. In this case, P = 2•a + 2•b. From the initial data of the problem, you know that b = 1,5•a, therefore P = 2•a + 2•1,5•a = 5•a where a = 8. Find the length of b = 1.5 x 8 = 12.
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Write down the formula for the area of the rectangle:S = a•b,Substitute the known values:S = 8•*12 = 96.
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Square.Task: find the area of a square if the perimeter is 36.
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Solution.A square is a special case of rectangle where all sides are equal, so the perimeter is 4•a where a = 8. The area of a square define by the formula S = a2 = 64.
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Triangle.Problem: suppose that we are given an arbitrary triangle ABC with a perimeter of 29. Find out the magnitude of its area, if it is known that the height BH, descended on the AC side, and divides it into segments of lengths 3 and 4, see
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Solution.First recall the formula for area of a triangle:S = 1/2•c•h, where c is the base and h is the height of the figure. In our case, the basis is the AC side, which is known according to the problem: AC = 3+4 = 7, it remains to find the height BH.
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The height is a perpendicular drawn to the side from the opposite vertex, therefore, it is to divide the triangle ABC into two right-angled triangles. Knowing this property, consider the triangle ABH. Remember the formula of Pythagoras that:AB2 = BH2 + AH2 = BH2 + 9 → AB = √(h2 + 9).In the triangle BHC on the same principle record:BC2 = BH2 + HC2 = BH2 + 16 → BC = √(h2 + 16).
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Apply the formula for perimeter:P = AB + BC + Aspositive values expressed through the height:P = 29 = √(h2 + 9) + √(h2 + 16) + 7.
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Solve the equation:√(h2 + 9) + √(h2 + 16) = 22 → [replace t2 = h2 + 9]:√(t2 + 7) = 22 - t, lift both sides of the equality to the square:t2 + 7 = 484 – 44•t + t2 → t≈10,84h2 + 9 = 117,5 → h ≈ 10,42
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Find the area of triangle ABC:S = 1/2•7•10,42 = 36,47.
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