Question

find the area of a trapezium whose parallel sides are 24cm and 29 cm and the distance between them is 15cm​

Answer 1

Answer:

397.5cm^2

Step-by-step explanation:

Area of trapezium = 1/2 h (a+b)

Two parralel sides = 24cm  and 29 cm

a= 24  b= 29

Distance = height

h=15cm

Area of trapezium = 1/2 x 15 (24+29)

                             = 15/2 (53)

                            = 397.5cm^2

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Answer 2

Answer:

Area of trapezium is $\rm\blue { {397.5cm}^{2}}$

Step-by-step explanation-

ABCD is a trapezium in which AB and CD are two parallel sides of trapezium.

Given-

• AB = 24 cm

• CD = 29 cm

• Height (Distance between them) = 15 cm

To find-

• Area of the trapezium

Solution-

We know ,

Area of trapezium= $\frac{1}{2}$×( sum of parallel sides) × (height)

Area of trapezium=

$\huge\rm\red { { \frac{1}{2} \times( AB + CD) \times h}}$

$\rm\blue {[ \frac{1}{2} \times (24 + 29) \times 15] {cm}^{2} }$

$\rm\blue { (\frac{1}{2} \times 53 \times 15) {cm}^{2}}$

$\rm\green { (\frac{53 \times 15}{2} ) {cm}^{2} = \frac{795}{2} {cm}^{2} = 397.5 {cm}^{2}}$

Area of trapezium = 397.5 centimetre square.