Find the area of a trapezium whose parallel sides are $24 c m$ and $20 c m$ and the distance between them is $15 c m$

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Solution

We know,
$Area of trapezium = \frac{1}{2} \times Sum of parallel sides \times Distance between them .$

Area of trapezium = $\frac{h}{2} (a + b)$ , where $a$ and $b$ are base and h is height.

Here, $a = 24$ cm, $b = 20$ cm and $h = 15$ cm

Therefore, Area= $\frac{15}{2} (24 + 20) = 330 c m^{2}$

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A parallelogram has 1 set parallel sides = 20 cm and another set = 24 cm. If the distance between the 24 cm sides is 15 cm, then the distance between the 20 cm sides is ____ cm

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Find the area of a trapezium whose parallel sides are 24cm and 20cm and the distance between them is 15cm

We know,Area of trapezium=12×Sum of parallel sides×Distance between them.Area of trapezium = h2(a+b), where a and b are base and h is height.Here, a=24 cm, b=20 cm and h=15 cmTherefore, Area=152(24+20)=330cm2

Find the area of a trapezium whose parallel sides are $24 c m$ and $20 c m$ and the distance between them is $15 c m$

We know,
$Area of trapezium = \frac{1}{2} \times Sum of parallel sides \times Distance between them .$

Area of trapezium = $\frac{h}{2} (a + b)$ , where $a$ and $b$ are base and h is height.

Here, $a = 24$ cm, $b = 20$ cm and $h = 15$ cm

Therefore, Area= $\frac{15}{2} (24 + 20) = 330 c m^{2}$