Question

The angles A, B, C and D of a quadrilateral ABCD are in the ratio 2:4:5:7. Find the measures of these angles. What type of quadrilateral is it? Give reasons. (2) Prove that the opposite angles of an isosceles trapezium are supplementary.(3)Show that the bisectors of angles of a parallelogram form a rectangle.

Answer 1

ANSWER1--

Let the angles of ABCD be 2x, 4x, 5x and 7x

2x + 4x + 5x + 7x = 360o

18x = 360o

x = 20o

So, the angles are A = 40o, B = 80o, C = 100oand D = 140o

A + D = 40o + 140o = 180o

B + C = 80o + 100o = 180o

So, ABCD is a trapezium with AB parallel to CD.

ANSWER 2--

In an isosceles trapezium AB is parallel to DC and the alternate interior angles are supplementary. So, the angles ∠A and ∠D are supplementary.which means 
∠A + ∠D = 180

Similarly, ∠B and ∠C are supplementary.
so,  ∠B + ∠C = 180
Since the trapezium is the isosceles trapezium, the base angles are equal. ∠C = ∠DSo by subsituting the anles we get
∠A + ∠C = 180
∠B + ∠D = 180 

so the opposite angles are also supplementary 

ANSWE 3--

Given:- ABCD is a parellelogram, AP, BR, CP and BR are angle bisectors.
To prove:- PQRS is a rectangle.
Proof:- In Parellelogram ABCD, ADllCB
Therefore Angle A + Angle B=180°. [ Sum of Angles on the same the same side of transversal is 180°]
1/2 Angle A+ 1/2 Angle B=180°/2
1/2( Angle A + Angle B)=90° .....(1)
By Angle Sum Property in Triangle ASB,
Angle ABS + Angle BSA + Angle SAB= 180°
1/2 Angle B + Angle BSA + 1/2 Angle A= 180°( given that AP and BR are angle bisectors)
1/2( A+B) + Angle BSA=180°
90°+ Angle BSA =180° (From (1))
Angle BSA = 90°
Similarly , Angle BRC= CQD= APD =90°
Therefore a quadrilateral PQRS in which all angles are right angles is a rectangle. 'Proved'

Answer 2

2x +4x+5x+7x =360 18x=360 x=360/18 x=20 2(20)=40 4(20)=80 5(20)=100 7(20)=140
So sides are 40,80,100,140