Answer:
Question 7:
∠L = 124°
∠M = 124°
∠J = 118°
Question 8:
∠Q = 98°
∠T = 98°
∠R = 82°
Question 15:
m∠G = 110°
Question 16:
∠G = 60°
Question 17:
∠G = 80°
Question 18:
∠G = 70°
Step-by-step explanation:
The angles can be solving using Symmetry.
Question 7.
The sum of interior angles in an isosceles trapezoid is 360°, and because it is an isosceles trapezoid
∠K = ∠J = 118°
∠L = ∠M
∠K+∠J+∠L +∠M = 360°
236° + 2 ∠L = 360°
Therefore,
∠L = 124°
∠M = 124°
∠J = 118°
Question 8.
In a similar fashion,
∠Q+∠T+∠S +∠R = 360°
and
∠R = ∠S = 82°
∠Q = ∠T
∠Q+∠T + 164° = 360°
2∠Q + 164° = 360°
2∠Q = 196°
∠Q = ∠T =98°.
Therefore,
∠Q = 98°
∠T = 98°
∠R = 82°
Question 15.
The sum of interior angles of a kite is 360°.
∠E + ∠G + ∠H + ∠F = 360°
Because the kite is symmetrical
∠E = ∠G.
And since all the angles sum to 360°, we have
∠E +∠G + 100° +40° = 360°
2∠E = 140° = 360°
∠E = 110° = ∠G.
Therefore,
m∠G = 110°
Question 16.
The angles
∠E = ∠G,
and since all the interior angles sum to 360°,
∠E + ∠G + ∠F +∠H = 360°
∠E + ∠G + 150 + 90 = 360°
∠E + ∠G = 120 °
∠E = 60° = ∠G
therefore,
∠G = 60°
Question 17.
The shape is a kite; therefore,
∠H = ∠F = 110°
and
∠H + ∠F + ∠E +∠G = 360°
220° + 60° + ∠G = 360°,
therefore,
∠G = 80°
Question 18.
The shape is a kite; therefore,
∠F = ∠H = 90°
and
∠F +∠H + ∠E + ∠G = 360°
180° + 110° + ∠G = 360°
therefore,
∠G = 70°.
A circle has infinit lines of symmetry. They pass through the centre of the circle and are the circle's diameter.
Hope this helps! (:
Well, we dont know because we cant see the barmodels
