Question #8c23f
2 Answers
As below
Explanation:
Given
ABCD is an Isosceles trapezium
Length of its minor base
The diagonal DB bisects
ABCD being an Isosceles trapezium
If we draw a line BE parallel to AD and it intersects DC at E, then quadrilateral ABED will have
and
being two parallel sides of a trapezium.
So ABDE is a parallelogram.
Now by given condition diagonal
DB bisects
Hence
Again
This implies AD=AB=10cm
So ABED must be a rhombus
So BE=AB =BC=10cm=AD=DE
Now
so
ABQP being a rectangle
Area of
Similarly Area of
Area of rectangle
So
Area of the Trapezium
Perimeter of the Trapezium
Angles
Perimeter
Area
Explanation:
Angle
Now
Similarly due to symmetry property of Isosceles trapezium side
Draw
Given
Using Pythagoras theorem
-
angle D= sin^-1 (3/4)=48.6^@
AsDC and AB are|| andDA is transversal,=>angles A and D are supplementary angles.
:. angle A=180-48.6=131.4^@ ,
As the base angles of an isosceles trapezium are equal,=>angleC=angle D=48.6^@ and angle B=angle A=131.4^@ , all rounded to one decimal place. -
Perimeter of the Isosceles trapezium
=AB+BC+CF+FE+ED+AD
=4xx10+2xx6.61approx53.2cm - Area of the Isosceles trapezium
=
Area of rectangleAEFB+2xx Area ofDeltaADE
(Deltas ADE and BFC are congruent)
or Area of the Isosceles trapezium=10xx7.5+2xx1/2xx6.61xx7.5approx124.6cm^2