Question : In triangle ABC, DE $∥$ BC where D is a point on AB and E is a point on AC. DE divides the area of $Δ A B C$ into two equal parts. Then BD : AB is equal to:
Option 1: $\sqrt{2} : (\sqrt{2} + 1)$
Option 2: $\sqrt{2} : (\sqrt{2} - 1)$
Option 3: $(\sqrt{2} - 1) : \sqrt{2}$
Option 4: $(\sqrt{2} + 1) : \sqrt{2}$

Question

Question : In $△$ ABC, D and E are points on AB and AC, respectively, such that DE || BC and DE divide the $△$ ABC into two parts of equal areas. The ratio of AD and BD is:
Option 1: $1 : 1$
Option 2: $1 : (\sqrt{2} -$ 1)
Option 3: $1 : \sqrt{2}$
Option 4: $1 : (\sqrt{2} +$ 1)

20 Views

Answer 1

Correct Answer: $(\sqrt{2} - 1) : \sqrt{2}$

Solution :

Given: DE $∥$ BC
According to the question,
$a r e a (△ A D E) = a r e a (trapezium BCED)$
⇒ $a r e a (△ A D E) + a r e a (△ A D E) = a r e a (trapezium BCED) + a r e a (△ A D E)$
⇒ $2 a r e a (△ A D E) = a r e a (△ A B C)$ .............(i)
From the figure,
$B D = A B - A D$
⇒ $A D = A B - B D$ ..................(ii)
In $△ A D E$ and $△ A B C$
$∠ D = ∠ D$ (corresponding angle)
$∠ A = ∠ A$ (common angle)
So, $△ A D E \sim △ A B C$
⇒ $\frac{a r e a (△ A D E)}{a r e a (△ A B C)} = \frac{A D^{2}}{A B^{2}}$
From equation (i)
$\frac{a r e a (△ A D E)}{2 a r e a (△ A D E)} = \frac{A D^{2}}{A B^{2}}$
⇒ $\frac{A D^{2}}{A B^{2}} = \frac{1}{2}$
⇒ $\frac{A D}{A B} = \frac{1}{\sqrt{2}}$
⇒ $A B = \sqrt{2} A D$
From equation (ii)
$A B = \sqrt{2} (A B - B D)$
⇒ $\sqrt{2} B D = \sqrt{2} A B - A B$
⇒ $\sqrt{2} B D = A B (\sqrt{2} - 1)$
⇒ $\frac{B D}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}}$
Hence, the correct answer is $(\sqrt{2} - 1) : \sqrt{2}$ .

Exams

Colleges

Predictors

Resources

Quick links

Exams

Colleges & Courses

Predictors

Resources

Exams

Colleges

Predictors & E-Books

Resources

Engineering Preparation

Medical Preparation

Law Preparation

MBA Preparation

Exams

Colleges

Resources

Exams

Colleges

Predictors & Articles

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Upcoming Events

Resources

Exams

Colleges

Resources

Quick Links

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Resources

Diploma Colleges

Exams

Colleges

Resources

Exams

Resources

Top Courses & Careers

Colleges

Exams

Upcoming Events

Resources

Other Exams

Exams

Colleges

Top Countries

Resources

Exams

Top Schools

Products & Resources

NCERT Study Material

Question : In triangle ABC, DE ∥ BC where D is a point on AB and E is a point on AC. DE divides the area of ΔABC into two equal parts. Then BD : AB is equal to:Option 1: 2:(2+1)Option 2: 2:(2−1)Option 3: (2−1):2Option 4: (2+1):2

Related Questions

Know More about

Upcoming Exams

Trending Articles around SSC CGL

Download the Careers360 App on your Android phone

Sign In/Sign Up