In the figure, what is the area of triangle ADB
[#permalink]
26 Jul 2021, 09:03
Given: \(ar(ACE) = 4, ar(CED) = 3\)
\(ar(ACD) = ar(CED) + ar(DEB) = 4 + 3 = 7\)
Dropping a perpendicular from point \(D \) on base \(CD\), taking the point of intersection as \(X\),
\(ar(CED) = \frac{1}{2} * EX * CD\)......\((2)\)
Also,
\(ar(ACD) = \frac{1}{2} * AC * CD\)......\((3)\)
Dividing \((2)\) by \((3)\),
\(\frac{EX}{AC} = \frac{3}{7}\).....\frac{(4)[}{fraction]
Now, to find the area of triangle \(AEB\), we drop a perpendicular from point \(E \) to base \(AB\), taking the point of intersection as \(Y\),
\(ar(AEB) = [fraction]1/2}AB * EY \)
\(AC = EX + EY \)
Thus, \(EY = AC - EX = AC - \frac{3}{7}AC\).....(from equation \(4\))
\(EY = \frac{4}{7}AC\)
\(ar(AEB) = \frac{1}{2} * AB * \frac{4}{7}AC\)
\(= \frac{4}{7}ar(ABC)\) ....... (since \(ar(ABC) = \frac{1}{2} AB * AC \))
\(ar(ABC) = ar(AEC) + ar(AEB) = 4 + \frac{4}{7}ar(ABC)\)
\(\frac{3}{7}ar(ABC) = 4\)
\(ar(ABC) = \frac{28}{3}\)
Now, triangles \(ABC \) and \(ADB \) have the same base \(AB \) and same height \(AC\), so they have the same area.
Thus, \(ar(ADB) = \frac{28}{3}\)