Question

1. The parallel sides AB and DC of a trapezium ABCD are 51 cm and 30 cm respectively. If
the other sides AD and BC are 20 cm and 13 cm respectively. Find the distance between
the parallel sides and the area of the trapezium.​

Answer 1

Answer:

Draw CE∥DA and CF⊥AB.

Therefore, ADCE is a parallelogram having AE∥CD and CE∥DA.

⇒AD=CE=39cm,DC=AE=30cm and BE=75−30=45cm

In △BCE, by heron's formula we have

a=39cm,b=45cm and c=42cm

⇒s=

2

a+b+c

=

2

39+42+45

=63cm

∴ Area of ΔBEC=

s(s−a)(s−b)(s−c)

=

63(63−39)(63−42)(63−45)

cm

2

=

63×24×21×18

cm

2

=756cm

2

Also, Area of ΔBEC=

2

1

×base×height

⇒

2

1

×45×h=756cm

2

⇒h=33.6cm

So, Area of trapezium:-

=

2

1

×(AB+CD)×height

=

2

1

×(75+30)×33.6=1764cm

2

Answer 2

Answer:

A big question, but see the explanation for the solution

Step-by-step explanation:

Construction: Draw a line through B parallel to AC which intersects CD at E. Also draw BP perpendicular to CD . Now BP is also perpendicular to ED.

Now AC || EB,

AB || EC.

→ ABEC is a parallelogram.

Now Carefully look at the base of the ∆BED,

ED = CD-CE

ED=51-30

ED=21.

Now by using the heron's formula, we can find out the area of ∆BED.

s=(20+13+21)/2

s=27

$area(bed) = \sqrt{s(s - a)(s - b)(s - c)}$

So by the formula,

$area = \sqrt{(27)(7)(6)(14) }$

$area \: of \: triangle \: bed = 126$

Now also area of triangle BED = (bh)/2

Here,

base = 21

height = h

Also the area is 126

So,

126 = 21×h/2

(126×2)/21=h

h=12 cm

So height of trapezium is 12 cm .

Area of trapezium = area of parallelogram + area of triangle

→ b × h + 126

→ 30 ×12+126

360 + 126

→ 486

So the area of trapezium is 486cm²