Answer:
Both are similar by SAS similarity.
This SAS similarity is equivalent to the congruence.
Step-by-step explanation:
Step 1:
To prove that ACB and HIG as similar triangles.
We have to look upon the corresponding sides.
SAS= Side angle sides , there the angle must be in between two sides.
ACB = HIG
Lets work on the corresponding sides.
IG/AC = IH/AC
= 
Reducing each to lowest form, we divide numerator and denominator by 3 for the 1st fraction and by 4 for the 2nd fraction.
We have
=
Both sides are equal.
So its proved that both are similar with SAS similarity theorem.
B they only have one pair of parallel sides. The top and bottom
Answer:
approximately 21, when rounded to nearest tenth.
Step-by-step explanation:
First, let's find the measure of angle BAD. Angles BAD and DAC add up to 180 degrees, so 180-57= 123 degrees. Now, let's find the circumference of the circle. The circumference is diameter * π, which is 19.6(π). Then, the ratio is 123/360, so multiply 19.6(π) by 123/360, and you get 21.0381988...... which simplifies to 21 (rounded to nearest tenth)
I use the sin rule to find the area
A=(1/2)a*b*sin(∡ab)
1) A=(1/2)*(AB)*(BC)*sin(∡B)
sin(∡B)=[2*A]/[(AB)*(BC)]
we know that
A=5√3
BC=4
AB=5
then
sin(∡B)=[2*5√3]/[(5)*(4)]=10√3/20=√3/2
(∡B)=arc sin (√3/2)= 60°
now i use the the Law of Cosines
c2 = a2 + b2 − 2ab cos(C)
AC²=AB²+BC²-2AB*BC*cos (∡B)
AC²=5²+4²-2*(5)*(4)*cos (60)----------- > 25+16-40*(1/2)=21
AC=√21= 4.58 cms
the answer part 1) is 4.58 cms
2) we know that
a/sinA=b/sin B=c/sinC
and
∡K=α
∡M=β
ME=b
then
b/sin(α)=KE/sin(β)=KM/sin(180-(α+β))
KE=b*sin(β)/sin(α)
A=(1/2)*(ME)*(KE)*sin(180-(α+β))
sin(180-(α+β))=sin(α+β)
A=(1/2)*(b)*(b*sin(β)/sin(α))*sin(α+β)=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
KE/sin(β)=KM/sin(180-(α+β))
KM=(KE/sin(β))*sin(180-(α+β))--------- > KM=(KE/sin(β))*sin(α+β)
the answers part 2) areside KE=b*sin(β)/sin(α)side KM=(KE/sin(β))*sin(α+β)Area A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
