Find the area of the parallelogram ABCD in which BC = 12 cm, CD = 17 cm and BD = 25 cm. Also, find the length of the altitude AE from vertex A on the side BC.

Explanation

In `Delta`BCD, we have
BC = 12 cm, CD = 17 cm and BD = 25 cm.
Let these sides be denoted by a, b, c respectively.
Then, a = 12 cm, b = 17 cm and c = 25 cm.
`therefore" "s=(1)/(2)(12+17+25)cm =((1)/(2)xx54)cm = 27 cm`.
`therefore" "(s-a)=(27-12)cm = 15 cm, (s-b)=(27-17)cm = 10 cm and (s-c)=(17-25)cm = 2 cm`.
Now, area `(Delta BCD)=sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(27xx15xx10xx2)cm^(2) = sqrt(81xx25xx4)cm^(2)`
`=sqrt(9xx5xx2)cm^(2)=90 cm^(2)`.
Area of || gm ABCD = 2 `xx`(area `Delta`BCD)
`=(2xx90)cm^(2) = 180 cm^(2)`
Draw AE `bot` BC. Let AE = h cm. Then,
Area of ||gm ABCD = (base `xx` altitude) = `(12 xx h) cm^(2)`.
`therefore" "12 xx h = 180 rArr h = (180)/(12) = 15 rArr AE = "h cm" = 15 cm`.
Hence, the altitude of the given parallelogram is 15 cm.