in triangle abc ad is the median and p is a point on ad such that ap : pd=1:2 find the area of triamgle abp
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this question we solve by draw triangle in copy
let a (0,0), b (0, l) and c (l,0) {this point only asumption }
now point d {(0+l)/2, (l+0)/2=(l/2, l/2)
now point p is find by section formula
so, p (l/6, l/6)
hence we have point a (0,0), b (l,0) and p (l/6, l/6)
ar (abp)=l^2/12
but ar (abc)=l^2/2
it means abc is 6 times largar then abp area
let a (0,0), b (0, l) and c (l,0) {this point only asumption }
now point d {(0+l)/2, (l+0)/2=(l/2, l/2)
now point p is find by section formula
so, p (l/6, l/6)
hence we have point a (0,0), b (l,0) and p (l/6, l/6)
ar (abp)=l^2/12
but ar (abc)=l^2/2
it means abc is 6 times largar then abp area
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