5
7
No cheating!
The triangle is a right angled triangle with sides measuring 3 (and 4) and the hypotenuse of length 5. Note - The length of the third side bc = 4, can be calculated using Pythagoras Theorem. If d is 1 unit of length along the hypotenuse and a perpendicular line is drawn from bc to d (meeting bc at e) then the triangle bde is similar to triangle bac. Then ca/ba = 3/5 = ed/bd = ed/1. Thus ed = 3/5 units in length. The area of a triangle = ½ x base x vertical height. The area of triangle dbc = ½ x bc x ed = ½ x 4 x 3/5 = 1.2 sq units.
Triangles BDC, CBD, CDB, DBC and DCB.
5
Angle ABD = 4x - 4 Angle ABC = twice angle ABD = 7x + 4 So 7x + 4 = 2*(4x - 4) = 8x - 8 So x = 12 Then angle DBC = half of angle ABC = 1/2*(7*12 + 4) = 1/2*88 = 44 degrees.
The measurement of angle ABD is 73 degrees. You find this angle by subtracting angle DBC from angle ABC, or 89-16 is equal to 73 degrees.
The answer is 13. x=13 13*5+13*2=91 Thank you.
7
No cheating!
minus 170 dBc
DBC Pierre was born in 1961.
Danish Bacon Company
The proof is fairly long but relatively straightforward. You may find it easier to follow if you have a diagram: unfortunately, the support for graphics on this browser are hopelessly inadequate.Suppose you have a rhombus ABCD so that AB = BC = CD = DA. Also AB DC and AD BC.Suppose the diagonals of the rhombus meet at P.Now AB DC and BD is an intercept. Then angle ABD = angle BDC.Also, in triangle ABD, AB = AD. therefore angle ABD = angle ADC.while in triangle BCD, BC = CD so that angle DBC = angle BDC.Similarly, it can be shown that angle BAC = angle CAD = angle DCA = angle ACB.Now consider triangles ABP and CBP. angle ABP (ABD) = angle CBP ( CBD or DBC),sides AB = BCand angle BAP (BAC) = angle BCP (BCA = ACB).Therefore, by SAS, the two triangles are congruent.In the same way, triangles BCP and CPD can be shown to congruent as can triangles CPD and DPA. That is, all four triangles are congruent.
The triangle is a right angled triangle with sides measuring 3 (and 4) and the hypotenuse of length 5. Note - The length of the third side bc = 4, can be calculated using Pythagoras Theorem. If d is 1 unit of length along the hypotenuse and a perpendicular line is drawn from bc to d (meeting bc at e) then the triangle bde is similar to triangle bac. Then ca/ba = 3/5 = ed/bd = ed/1. Thus ed = 3/5 units in length. The area of a triangle = ½ x base x vertical height. The area of triangle dbc = ½ x bc x ed = ½ x 4 x 3/5 = 1.2 sq units.
The important property of a rhombus to note is that a rhombus has all sides of equal length. Then: Let the vertices be A, B, C, D; let the diagonal BD be the same length as the sides. Consider the triangle ABD formed by two sides of the rhombus and the diagonal BD. As all sides are equal in length, it must be an equilateral triangle and thus each angle can be calculated, in particular angle BAD of the rhombus. Consider the triangle BCD formed by the other two sides of the rhombus and the diagonal BD. The angle BCD of the rhombus can also be calculated as above. For angles ABC and CDA of the rhombus, note that angle ABC is the same as the sum of angles ABD and DBC. and those angles can be calculated above; similarly for angle CDA. A short cut that can be used is the property of a rhombus that opposite angles of a rhombus are equal which means once angle DAB has been calculate, BCD is then known. Then by subtracting the sum of these from 360o (the sum of the angles of a quadrilateral and a rhombus is an quadrilateral) and dividing the result by 2 will give the other two angles..