TRUE
5
7
No cheating!
Triangles BDC, CBD, CDB, DBC and DCB.
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.
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.
Triangles BDC, CBD, CDB, DBC and DCB.
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.