Question

Question #8c23f

Answer 1

As below

Explanation:

Given
ABCD is an Isosceles trapezium
Length of its minor base `AB=10 cm`
`sinD=3/4`
The diagonal DB bisects `/_ADC`

ABCD being an Isosceles trapezium
`/_ADC=/_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"`
and
`AB||ED`
being two parallel sides of a trapezium.
So ABDE is a parallelogram.
Now by given condition diagonal
DB bisects`/_ADE`,
Hence `/_ADB=/_BDE`
Again `/_BDE="alternate"/_ABD`
`:.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^@`

Answer 2

`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:

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