Step-by-step answer:
Please refer to attached image.
1. Quad PQRS is cyclic (all vertices on the same circle), so opposite angles are supplementary, meaning
that
angles QPS and QRS are supplementary =>
QPS+QRS=180 =>
QRS = 180 - 74 = 106
2. Triangle RSQ is isosceles with RS=RQ =>
angles RSQ and RQS are congruent.
3. Angle RSQ = (180 - 106) /2 = 74 / 2 = 37
4. QP is a diameter => angle QSP is a right-angle.
5. From (3) and (4) above,
angle RSP = 37+90 = 127
6. Since PQRS is cyclic, angles RQP and RSP are supplementary =>
RQP+RSP = 180 =>
x + 127 = 180 =>
x = 180 - 127 = 53 degrees.
The area and perimeter of the triangle is 2/5 square units and (2√10 + 4√5) / 5 units
<h3>Determining the perimeter and area of the triangle giving line equation</h3>
In order to determine the area and perimeter of the lines, we will plot the giving lines, determine the point of intersection and then use the Pythagoras theorem to determine the dimension of the right triangle.
The points of intersection of the line are;
(x₁, y₁) = (- 0.4, 5.2),
(x₂, y₂) = (-0.8, 4.4),
(x₃, y₃) = (0, 4)
Determine the base
b² = c² -a²
b = √(-0.8)² + (4 - 4.4)²
b = 2√5 / 5
Determine the height
h = √((- 0.4) - (- 0.8))² + (5.2 - 4.4)²
height = 2√5 / 5
For the hypotenuse
r = √2 · b
r = 2√10 / 5
<h3>Determine the Perimeter and area</h3>
Perimeter = s1+s2+s3
Perimeter = 2√5 / 5 + 2√5 / 5 + 2√10 / 5
Perimeter = (2√10 + 4√5) / 5 units
<u>For the area</u>
area = 1/2* base * height
A = 0.5 · (2√5 / 5) · (2√5 / 5)
A = 2/5 square units
Hence the area and perimeter of the triangle is 2/5 square units and (2√10 + 4√5) / 5 units
Learn more on area and perimeter of triangles here: brainly.com/question/12010318
#SPJ1
Answer:
y = 1/2x - 1
Step-by-step explanation:
When you put all three equations from the question into a graph and then input the equation i just gave you it will be perpendicular to line B & C, also it will be parallel to line A. Finally you can check yourself that it also has a y-intercept of (0,-1) & an x-intercept of (2,0). Hopefully this helps out & gets brainliest. xD
From the given figure ,
RECA is a quadrilateral
RC divides it into two parts
From the triangles , ∆REC and ∆RAC
RE = RA (Given)
angle CRE = angle CRA (Given)
RC = RC (Common side)
Therefore, ∆REC is Congruent to ∆RAC
∆REC =~ ∆RAC by SAS Property
⇛CE = CA (Congruent parts in a congruent triangles)
Hence , Proved
<em>Additional</em><em> comment</em><em>:</em><em>-</em>
SAS property:-
"The two sides and included angle of one triangle are equal to the two sides and included angle then the two triangles are Congruent and this property is called SAS Property (Side -Angle-Side)
<u>also</u><u> </u><u>read</u><u> </u><u>similar</u><u> questions</u><u>:</u> Complete this proof. Given: EC AC, DB AC, ∠A = ∠F Prove: ΔMDF ∼ ΔNCA..
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Consider the proof. Given: Segment AB is parallel to line DE. Prove: AD/DC = BE/EC What is the missing statement in Step 5? A.) AC = BC B.) AC/DC = BC/EC C.) AD...
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Answer:
any of the below:
Step-by-step explanation:
12/30
6/15
2/5
