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Maths geometry
i just need 4)
i’ve worked everything else out

Maths geometry i just need 4) i’ve worked everything else out-example-1
User Flux
by
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1 Answer

3 votes

Explanation:

good job then !

the first thing I see in the graphic is that

BG is half of BD (as a major definition of diagonals in a parallelogram).

BD on the other hand is the Hypotenuse (the side opposite of the 90° angle at A) of the triangle ABD.

to be able to use Pythagoras to get BD, I need to find the length of AB.

this I get from the triangle ABE. this is also a right-angled triangle. AE = 4cm. the angle B1 = 50°. the angle A = 90°.

so, the angle E1 = 180 - 90 - 50 = 40° (remember, the sum of all angles in a triangle is always 180°).

now we use the law of sine :

a/sin(A) = b/sin(B) = c/sin(C)

with a, b, c being the sides of the triangle, and A, B, C being the corresponding opposite angles.

so, we have here

AE / sin(50) = AB / sin(40)

4 / sin(50) = AB / sin(40)

AB = 4×sin(40) / sin(50) = 3.356398525... cm

so (Pythagoras),

BD² = AB² + AD² = 3.356398525...² + 9² =

= 11.26541106... + 81 = 92.26541106...

BD = 9.60548859... cm

BG = BD/2 = 4.802744295... cm

and now look at that.

because of the fact that I did not have to go through the other points, I chose the first approach that went through my mind.

and it showed that your teacher created an impossible (contradicting) scenario :

the angle of B1 does not fit to the ratios of the various line segments.

as you correctly calculated CD as 3 cm, that must mean that AB is also 3 cm.

that makes

BD² = 9² + 3² = 81 + 9 = 90

BD = 9.486832981... cm

BG = BD/2 = 4.74341649... cm

PLEASE, USE THIS VALUE.

there is a difference to my previous calculation. and the reason for this is : the 50° for B1 are not correct.

with AE = 4, AB = 3 and then BE = 5, we get

AE / sin(B1) = BE / sin(A)

4 / sin(B1) = 5 / sin(90) = 5

sin(B1) = 4/5 = 0.8

B1 = 53.13010235...°

and not 50° !!!!

User Gaurawerma
by
7.5k points

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