General Idea:
When we are given a point P(x, y) centered at origin with a scale factor of k, then the dilated point will be given by P' (kx, ky)
Applying the concept:
In the diagram given, the coordinate of C is (6, 3). Triangle ABC is being dilated with the center of dilation at the origin. The image of C, point C' has coordinates of (7.2, 3.6).
If C (x, y) centered at origin with a scale factor of k, then the dilated point will be given by C' (kx, ky)
We know 
& 
Substituting 6 for x in the equation , we get 
.
Dividing 6 on both sides
Simplifying fractions on both sides
Point A from the diagram is (-3, 3)
x-coordinate of A' = 
y-coordinate of A' = 
Point A' is given by (-3.6, 3.6)
Conclusion:
<u>Scale factor of dilation is 1.2</u>
<u>The coordinates of point A' is (-3.6, 3.6)</u>
3/5*3/3=9/15
1/3*5/5=5/15
15 is a common denominator.
Answer:
Yes, we can conclude that Triangle ABC is similar to triangle DEF because the measures of the 3 angles of both triangles are congruent.
Step-by-step explanation:
We have the measure of 2 angles from both triangles, and we know that triangles have 180°, so we can solve for the measure of the third angle for both triangles.
Triangle ABC:
Measure of angle A= 60°
Measure of angle C= 40°
Measure of angle B = 180°- (measure of angle A + measure of angle C) = 180° - (60° + 40°) = 80°
Triangle DEF
Measure of angle E= 80°
Measure of angle F= 40°
Measure of angle D= 180° - (measure of angle E + measure of angle F) = 180° - (80° + 40°) = 60°
The measures of the angles in Triangle ABC are: 60°, 40°, and 80°.
The measures of the angles in Triangle DEF are: 60°, 40°, and 80°.
Since the measure of 3 angles of the two triangles are the same, we know that the two triangles are similar.
Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS = 
= 57.5°
Now, tan(57.5°) = 
⇒ 1.5697 = 
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) = 
⇒ tan65° =
⇒ QM = 
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = (
) × (ST)
= (
) × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²
