Construct a quadrilateral ABCD , Where AB- 6cm , CD- 6.5cm , LL- 70°, LA- 95°​

Solution:

Given : A quadrilateral A B CD, in which, AB=5  cm, BC = 6 cm, CD =6.5 cm , angle B= 105° and angle C = 75

Steps of construction

(1) Draw a line segment B C of length 6 cm.

(2) At Point B and C , with the help of Compass draw angle of  105° and 75° respectively.Name the rays as C M and B Z respectively.

105°→Draw an angle of 120° and 90° and draw angle bisector between 90° and 120°.

75°→Draw angle bisector of 60° and 90°.

(3) From C , mark an arc at a distance of 6.5 cm on ray CM.You will get point D.

(4) From B , mark an arc at a distance of 5 cm on ray B Z.You will get point A.

(5) Join A and D.

This is required trapezoid A B CD.

here, AB ║CD→→∠B+∠C=180°[If sum of interior angles on the same side of transversal is supplementary then lines are parallel.]

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

the answer is two. hope this !