Question #8c23f

2 Answers
Jun 4, 2016

As below

Explanation:

self drawn
Given
ABCD is an Isosceles trapezium
Length of its minor base AB=10 cmAB=10cm
sinD=3/4sinD=34
The diagonal DB bisects /_ADCADC

ABCD being an Isosceles trapezium
/_ADC=/_BCDADC=BCD

If we draw a line BE parallel to AD and it intersects DC at E, then quadrilateral ABED will have DA||BE" ""by construction"DABE by construction
and
AB||EDABED
being two parallel sides of a trapezium.
So ABDE is a parallelogram.
Now by given condition diagonal
DB bisects/_ADEADE,
Hence /_ADB=/_BDEADB=BDE
Again /_BDE="alternate"/_ABDBDE=alternateABD
:.In DeltaABD,/_ABD=/_ADB
This implies AD=AB=10cm

So ABED must be a rhombus

So BE=AB =BC=10cm=AD=DE

Now
Sin /_ADF=h/"AD"=>3/4=h/10=>h=7.5cm

:. **Height** (h)=7.5cm

DP=sqrt(AD^2-AF^2)=sqrt(10^2-7.5^2)~~6.6cm

so CQ=DP =6.6cm

ABQP being a rectangle

PQ =AB =10 cm

Area of Delta ADP=1/2xxAPxxDP=1/2xx6.6xx7.5=24.75 "cm"^2

Similarly Area of Delta BQC=24.75 "cm"^2

Area of rectangle ABQP=10xx7.5cm^2=75cm^2

So
Area of the Trapezium

2xxDelta ABP+ "rectangle"ABQP=2xx24.75+75=124.5cm^2

Perimeter of the Trapezium

DA + AB + BC+2xxQC+PQ=(10+10+10+2xx6.6+10)cm=53.2cm

Angles
/_ADC=/_BCD=sin^-1 (3/4)=48.6^@

/_DAB=/_CBA=(180-48.6)=131.4^@

Jun 6, 2016

angle A=angleB=131.4^@and angleC=angle D=48.6^@ , all rounded to one decimal place
Perimeter approx53.2cm
Area approx124.6cm^2

Explanation:

my computer (Not to Scale)
ABCD is given Isosceles trapezium with short base AB=10 cm.
Angle D is the vertex angle and diagonal DB bisects it.
=> angleADB=angleBDC
Now DC and AB are || and DB is transversal.
=> angle BDC=angleABD, alternate interior angles
=>Delta ADB is isosceles triangle with angle A as vertex
=>AB=AD=10cm
Similarly due to symmetry property of Isosceles trapezium side BC=10cm

Draw AE ⊥ from A on side DC
AE=BF=h
Given sin D=3/4, in DeltaDEA
=>sin D=h/10=3/4, solving for h
h=3/4xx10=7.5cm
Using Pythagoras theorem
DE=sqrt(10^2-7.5^2)approx6.61cm

  1. angle D= sin^-1 (3/4)=48.6^@
    As DC and AB are || and DA 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.

  2. Perimeter of the Isosceles trapezium=AB+BC+CF+FE+ED+AD
    =4xx10+2xx6.61approx53.2cm

  3. Area of the Isosceles trapezium=
    Area of rectangle AEFB+2xxArea of DeltaADE
    (Deltas ADE and BFC are congruent)
    or Area of the Isosceles trapezium =10xx7.5+2xx1/2xx6.61xx7.5approx124.6cm^2