Question

the length of two adjacent sides of parellelogram are 17 cm and 12 cm. One of its diagonals is 25 cm . Find the area of parallelogram. Also find the length of altitude from the vertex on the side of length 12 cm

Answer 1

See the attached file

Answer 2

Answer:

$Area \: of \: the \: parallelogram \: ABCD = 180 \: cm^{2}\\Required \: Altitude = 15 \:cm$

Step-by-step explanation:

Given ,

ABCD is a parallelogram,

AB = c = 12 cm,

BC = a = 17 cm,

AC = b = 25 cm

Now,

i) In ∆ABC,

a = 17 , b = 25 and c = 12

$s =\frac{a+b+c}{2}\\=\frac{17+25+12}{2}\\=\frac{54}{2}\\=27$

$s-a=27-17=10,\\s-b=27-25=2,\\s-c=27-12=15$

By Heron's Formula:

$Area\: of \: \triangle ABC \\= \sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{27\times 10\times 2\times 15}\\=\sqrt{3\times 3\times 3 \times 2\times 5 \times 2 \times 3 \times 5 }\\=\sqrt{(3\times 3 )\times (3\times 3)\times (2\times 2)\times(5\times 5)}\\=3\times 3\times 2\times 5\\=90\: cm^{2}$

$Now , \\Area \: of \: the \: parallelogram \: ABCD = 2\times Area \: of \: \triangle ABC \\=2\times 90\: cm^{2}\\=180\: cm^{2}$

$In \: parallelogram \: ABCD , base = CD = 12 \: cm \\Altitude =BE =h$

$Area \: of \: ABCD = 180\:cm^{2}$

$\implies base \times altitude = 180$

$\implies 12 \times h = 180$

$\implies h = \frac{180}{12}$

$\implies h = 15 \: cm$

Therefore,

$Area \: of \: the \: parallelogram \: ABCD = 180 \: cm^{2}\\Required \: Altitude = 15 \:cm$

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