Question

As shown in the figure on the right $\Delta ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are indicated in the figure, then the area of the triangle is

• A. $315$
• B. $240$
• C. $56$
• D. $185$

Answer 1

The correct option is C $56$

R.E.F image

Let $p$ be the interior point of $△ABC$ Similarly,point $D,E,F$ are

the points on line $BC,CA$ and $AB$

Let ${}^{\prime }{x}^{\prime }$ be the Area of $△APE$ and $y$ be the area of $△CPD$

Now, from figure,

Altitude $PF$ is the common of altitude of $△APF$ and $△BPF$

$\therefore \frac{A\left(△APF\right)}{A\left(△BPF\right)}=\frac{\frac{1}{2}\times AF\times PF}{\frac{1}{2}\times BF\times PF}$

$\frac{40}{30}=\frac{AF}{BF}$.....(1)

again $CF$ is common altitude of $△ACF$ and $△BCF$

$\therefore \frac{A\left(△ACF\right)}{A\left(△BCF\right)}=\frac{\frac{1}{2}\times AF\times CF}{\frac{1}{2}\times BF\times CF}$

$\frac{84+x+40}{y+35+30}=\frac{AF}{BF}$.........(2)

From equation (1) and (2),

$\frac{124+x}{65+y}=\frac{40}{30}$........(A)

Similarly, In $△ADC$ and $△ADB$, $AD$ is common altitude and In $△PDC$ and

$△PDB$, $PD$ is common altitude,

$\therefore \frac{A\left(△ADC\right)}{A\left(△ADB\right)}=\frac{x+84+y}{10+30+35}=\frac{\frac{1}{2}\times CD\times AD}{\frac{1}{2}\times BD\times AD}=\frac{CD}{BD}$......(3) and

$\frac{A\left(△PDC\right)}{A\left(△PDB\right)}=\frac{y}{35}=\frac{\frac{1}{2}\times PD\times CD}{\frac{1}{2}\times PD\times BD}=\frac{CD}{BD}$.......(4)

From (3) and (4),

$\frac{y}{35}=\frac{x+84+y}{105}$...........(B)

$3y=x+84+y$

$y=\frac{x+84}{2}$..........(5)

substituting the value of $y$ in equation $\left(A\right)$

$\frac{124+x}{65+\frac{x+84}{2}}=\frac{4}{3}$

$3(124+x)=4(65+\frac{x+84}{2})$

$372+3x=260+2x+168$

$x=56$