In a $\triangle ABC$,$D$ and $E$ are points on $AB$ and $AC$ respectively such that $DE \parallel BC$. Suppose $BE$ and $CD$ intersect at $O$. If the areas of $\triangle ADE$ and $\triangle ODE$ are $3$ and $1$ respectively,find the area of $\triangle ABC$.
(D) Let $\text{Area}(\triangle ADE) = 3$ and $\text{Area}(\triangle ODE) = 1$.
Since $DE \parallel BC$,$\triangle ADE \sim \triangle ABC$.
Let $\text{Area}(\triangle BOD) = \text{Area}(\triangle COE) = x$ and $\text{Area}(\triangle BOC) = y$.
Since $\triangle BDE$ and $\triangle CDE$ share the same base $DE$ and lie between the same parallels,$\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE)$.
Thus,$x + 1 = x + 1$,which is consistent.
Using the property of areas of triangles with the same altitude,$\frac{\text{Area}(\triangle ODE)}{\text{Area}(\triangle BOD)} = \frac{OE}{OB} = \frac{\text{Area}(\triangle COE)}{\text{Area}(\triangle BOC)} = \frac{x}{y}$.
So,$\frac{1}{x} = \frac{x}{y} \implies y = x^2$.
Also,$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ODE)} = \frac{AD}{DO} \cdot \frac{AE}{EO}$ is not directly useful,but $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{\text{Area}(\triangle ODE)}{\text{Area}(\triangle BOC)} = \frac{1}{y}$ is incorrect; rather $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{DE^2}{BC^2}$.
From similarity,$\frac{DE}{BC} = \frac{OD}{OC} = \frac{OE}{OB}$.
Since $\frac{\text{Area}(\triangle ODE)}{\text{Area}(\triangle BOC)} = \frac{OD \cdot OE}{OB \cdot OC} = \left(\frac{OD}{OC}\right)^2 = \frac{1}{y}$,we have $\frac{OD}{OC} = \frac{1}{\sqrt{y}}$.
Also $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{3}{3 + 2x + y} = \frac{1}{y} \implies 3y = 3 + 2x + y \implies 2y = 2x + 3$ is wrong. Correct approach: $\frac{OD}{OC} = \frac{1}{x}$. Thus $\frac{1}{x^2} = \frac{1}{y}$.
Given $\text{Area}(\triangle ADE) = 3$,$\text{Area}(\triangle ODE) = 1$,$\text{Area}(\triangle BOD) = x$,$\text{Area}(\triangle COE) = x$,$\text{Area}(\triangle BOC) = y$.
We have $x^2 = 3 \cdot y$ is wrong. The correct relation is $x^2 = 3 \cdot 1 = 3$,so $x = \sqrt{3}$.
Then $\frac{1}{x} = \frac{x}{y} \implies y = x^2 = 3$.
Total Area $= 3 + 1 + 2x + y = 4 + 2\sqrt{3} + 3 = 7 + 2\sqrt{3}$.
Since $DE \parallel BC$,$\triangle ADE \sim \triangle ABC$.
Let $\text{Area}(\triangle BOD) = \text{Area}(\triangle COE) = x$ and $\text{Area}(\triangle BOC) = y$.
Since $\triangle BDE$ and $\triangle CDE$ share the same base $DE$ and lie between the same parallels,$\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE)$.
Thus,$x + 1 = x + 1$,which is consistent.
Using the property of areas of triangles with the same altitude,$\frac{\text{Area}(\triangle ODE)}{\text{Area}(\triangle BOD)} = \frac{OE}{OB} = \frac{\text{Area}(\triangle COE)}{\text{Area}(\triangle BOC)} = \frac{x}{y}$.
So,$\frac{1}{x} = \frac{x}{y} \implies y = x^2$.
Also,$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ODE)} = \frac{AD}{DO} \cdot \frac{AE}{EO}$ is not directly useful,but $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{\text{Area}(\triangle ODE)}{\text{Area}(\triangle BOC)} = \frac{1}{y}$ is incorrect; rather $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{DE^2}{BC^2}$.
From similarity,$\frac{DE}{BC} = \frac{OD}{OC} = \frac{OE}{OB}$.
Since $\frac{\text{Area}(\triangle ODE)}{\text{Area}(\triangle BOC)} = \frac{OD \cdot OE}{OB \cdot OC} = \left(\frac{OD}{OC}\right)^2 = \frac{1}{y}$,we have $\frac{OD}{OC} = \frac{1}{\sqrt{y}}$.
Also $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{3}{3 + 2x + y} = \frac{1}{y} \implies 3y = 3 + 2x + y \implies 2y = 2x + 3$ is wrong. Correct approach: $\frac{OD}{OC} = \frac{1}{x}$. Thus $\frac{1}{x^2} = \frac{1}{y}$.
Given $\text{Area}(\triangle ADE) = 3$,$\text{Area}(\triangle ODE) = 1$,$\text{Area}(\triangle BOD) = x$,$\text{Area}(\triangle COE) = x$,$\text{Area}(\triangle BOC) = y$.
We have $x^2 = 3 \cdot y$ is wrong. The correct relation is $x^2 = 3 \cdot 1 = 3$,so $x = \sqrt{3}$.
Then $\frac{1}{x} = \frac{x}{y} \implies y = x^2 = 3$.
Total Area $= 3 + 1 + 2x + y = 4 + 2\sqrt{3} + 3 = 7 + 2\sqrt{3}$.
Explore More
Similar Questions
The vertices of a triangle are $(at_1t_2, a(t_1 + t_2))$,$(at_2t_3, a(t_2 + t_3))$,and $(at_3t_1, a(t_3 + t_1))$. Find the coordinates of its orthocentre.
MediumIIT 1983
View SolutionLet $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15x - y = 82$,$6x - 5y = -4$,and $9x + 4y = 17$. Then $\alpha + 2\beta$ and $2\alpha - \beta$ are the roots of the equation $...........$.
If $P(\sin \alpha, \cos \alpha)$ lies inside the triangle formed by the vertices $(0,0), \left(\frac{\sqrt{3}}{2}, 0\right)$ and $\left(0, \frac{\sqrt{3}}{2}\right)$,then $\alpha$ lies in the interval
Let the circumcentre of a triangle with vertices $A(a, 3)$,$B(b, 5)$,and $C(a, b)$,where $ab > 0$,be $P(1, 1)$. If the line $AP$ intersects the line $BC$ at the point $Q(k_{1}, k_{2})$,then $k_{1} + k_{2}$ is equal to.
DifficultJEE MAIN 2022
View Solution$A$ triangle $ABC$ lying in the first quadrant has two vertices as $A(1, 2)$ and $B(3, 1)$. If $\angle BAC = 90^{\circ}$ and $\text{ar}(\Delta ABC) = 5\sqrt{5}$ sq. units,then the abscissa of the vertex $C$ is
DifficultJEE MAIN 2020
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo