Question

In Fig. 20.38, a parallelogram is drawn in a trapezium, the area of the parallelogram is 80 cm² ,find the area of the trapezium. pls reply fast.​

Answer 1

Answer:

• Area of trapezium is 128 cm².

Step-by-step explanation:

Given that,

Area of parallelogram is 80 cm².

Height of parallelogram and height of trapezium will be same.

So,

Let, The height of parallelogram be h.

Base of parallelogram = 10 cm.

Because opposite sides of parallelogram are equal and parallel.

Area of parallelogram = Base × height

Put the values :

⇒80 = 10 × h

⇒80/10 × h

⇒h = 8

Height of parallelogram is 8 cm.

Then,

Height of trapezium will be also 8 cm.

Now,

We know that,

Area of trapezium = (a + b)/2 × h

Where,

• a and b are parallel sides of trapezium.
• h is height of trapezium.

Put, a, b and h in formula :

⇒Area = (10 + 22)/2 × 8

⇒Area = 32/2 × 8

⇒Area = 16 × 8

⇒Area = 128

Therefore,

Area of trapezium is 128 cm².

Answer 2

$\huge\tt { \underline \red{given}} \hookrightarrow$

• The area of the parallelogram ➲ 80 cm².

• The length of the parallel sides of the trapezium ➲ 22 cm and 10 cm.

• The base of the parallelogram ➲ 10 cm.

$\huge\tt { \underline \red{to \: find}} \hookrightarrow$

(The height of the parallelogram and the height of the trapezium is same.)

• Height of the trapezium .

• Area of the trapezium.

$\huge\tt { \underline \red{formula \: used \: }} \hookrightarrow$

$\\ \large \bigstar\tt area \: of \: the \: parallelogram \: = \boxed{\red \tt{ \tt \: base} \: \times \tt\blue { height} }\\ \\ \large \bigstar \tt \: area \: of \: the \: trapezium \: = \boxed{ \pink \tt{\frac{(a + b)}{2} } \times \tt \: \blue {height }\: }$

( here a and b are the length of the parallel sides of the trapezium )

$\huge\tt { \underline \red{solution}} \hookrightarrow$

( now we will find the height of the trapezium / parallelogram by applying the 1st formula )

$\\ \large \tt \: area \: of \: the \: parallelogram \: = \: base \: \times \: height \: \\ \\ \large \tt \: 80cm \: = \: 10 \: \times \: height \: \\ \\ \large \tt \: \: 10 \: \times \: height \: = 80cm \\ \\ \large \tt \: height \: = \frac{8 \cancel0}{1 \cancel0} \\ \\ \large \tt \: height \: = \: 8cm$

$\large \tt \red{\: \therefore \: the \: height \: of \: the \: trapezium \: is \: 8cm \: }$

( now we will find the area of the trapezium by applying the 2nd formula .)

$\large \tt \: let \: \: \: \red a \: = 22 \: and \: \red b\: = 10 \\ \\ \large \tt \: area \: of \: the \: trapezium \: = \: \frac{1}{2} (a + b) \times h \\ \\ \large \tt \: area \: of \: the \: trapezium \: = \: \frac{1} {\cancel{2}} (22 + 10) \times \cancel 8 {}^{4} \\ \\ \large \tt \: area \: of \: the \: trapezium \: = \:32 \times 4 \\ \\ \large \tt \: area \: of \: the \: trapezium \: = 128 \: cm {}^{2}$