Question

find the altitude of the trapezium whose areais 65cm2 and whose bases are 13cm and 26cm


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Answer 1

Answer:

area of trapezium =65cm2

first base of trapezium =13cm

2nd base of trapezium =26cm

26cm2 =13cm×26cm×h

h=13cm ×26cm= 13×26 = 1

26cm2 26×26 2

answer is 1/2

Answer 2

Given,

The area of trapezium = 65 cm^2

The first be of the trapezium (a) = 13cm

The second base of the trapezium (b) = 26 cm

To find out,

The altitude of the trapezium.

Solution:

$\huge \: area \: of \: trapezium \: = \frac{1}{2} (a + b)h$

$65 = \frac{1}{2} (13 + 26)h$

$65 = \frac{1}{2} \times 39 \times h$

$65 = 19.5 \times h$

$h = \frac{65}{19.5}$

$h = 3.333..cm$

Therefore the height/altitude of the trapezium is 3.33 cm.

Proof:

$area \: of \: trapezium \: = \frac{1}{2} (13 + 26) \times 3.33$

$area \: of \: trapezium \: = \frac{1}{2} \times 39 \times 3.33$

$area \: of \: trapezium \: = 19.5 \times 3.33$

$area \: of \: trapezium \: = 64.935 \\ so \: approximately \: the \: area \: is \: 65 {cm}^{2}$