Question

find the area of rhombus whose perimeter is 200m and one of diagonal is 80m. according to ch heron's formula

Answer 1

perimeter of rhombus = 4 × side


200= 4× side


side = 200÷4


side = 50 m




in triangle AOD

AD= 50

AO= 40


OD = b


here using Pythagoras theorem

${h}^{2} = {p}^{2} + {b}^{2} \\ \\ {b}^{2} = {h}^{2} - {p}^{2} \\ \\ {b}^{2} = {50}^{2} - {40}^{2} \\ \\ {b}^{2} = 2500 - 1600 \\ \\ b = \sqrt{900} \\ \\ b = 30 \\ \\ \\ now \: we \: have \: all \: the \: side \\ e.i. \\ a = 30 \\ b = 40 \\ c = 50 \\ \\ using \: herons \: formulae \\ \\ s = \frac{perimeter }{2} \\ \\ s = \frac{40 + 50 + 30}{2} \\ \\ s = 60 \\ \\ s - a = 60 - 30 = 30 \\ s - b = 60 - 40 = 20 \\ s - c = 60 - 50 = 10$






now area of trianlge


$= \sqrt{s(s - a)(s - b)( s- c)} \\ \\ = \sqrt{60 \times 30 \times 20 \times10 } \\ \\ = \sqrt{3 \times 2 \times 10 \times 3 \times 10 \times 2 \times 10 \times 10} \\ \\ = \sqrt{ {3}^{2} \times {2}^{2} \times {10}^{2} \times {10}^{2} } \\ \\ = 3 \times 2 \times 100 \\ \\ = 600$


here as we know that the triangles formed by the diagonal in the rhombus are equal


therefore

there are 4 triangles

area of one triangle = 600m sq


area of 4 triangle = 4× 600=2400 m^2





$2400 \: {m}^{2} \: \: \: \: \: \: \: \: ans$

Answer 2

Answer:

Let side of the rhombus be 's' and diagonals be d and d'

Given , perimeter of rhombus=200m

➡4s= 200

➡s=50 m

Also given , d'=80m

We know , diagonals of a rhombus are perpendicular bisectors.

➡(d/2)²+(d'/2)²=s²

➡(d/2)²=900

➡d=2*30=60 m

Area of the rhombus = d*d'/2 = 60*80/2 = 2400 m²