Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
Share
Notifications

View all notifications

In ∆Abc, Seg Ad ⊥ Seg Bc Db = 3cd. Prove that : 2ab2 = 2ac2 + Bc2 - Geometry

Login
Create free account


      Forgot password?

Question

In ∆ABC, seg AD ⊥ seg BC DB = 3CD. Prove that :
2AB= 2AC+ BC2

Solution

It is given that,
DB = 3 CD

∴ BC = 4 CD        ....(1)

According to Pythagoras theorem,
In ∆ABD

\[{AB}^2 = {AD}^2 + {DB}^2 \]
\[ \Rightarrow {AD}^2 = {AB}^2 - {DB}^2 . . . \left( 2 \right)\]

In ∆ACD

\[{AC}^2 = {AD}^2 + {CD}^2 \]
\[ \Rightarrow {AC}^2 = \left( {AB}^2 - {DB}^2 \right) + {CD}^2 \left( \text{From} \left( 2 \right) \right) \]
\[ \Rightarrow {AC}^2 = {AB}^2 - \left( 3CD \right)^2 + {CD}^2 \left( \text{Given} \right)\]
\[ \Rightarrow {AC}^2 = {AB}^2 - 9 {CD}^2 + {CD}^2 \]
\[ \Rightarrow {AC}^2 = {AB}^2 - 8 {CD}^2 \]
\[ \Rightarrow {AB}^2 = {AC}^2 + 8 {CD}^2 \]
\[ \Rightarrow {AB}^2 = {AC}^2 + 8 \left( \frac{BC}{4} \right)^2 \left( \text{From} \left( 1 \right) \right)\]
\[ \Rightarrow {AB}^2 = {AC}^2 + \frac{{BC}^2}{2}\]
\[ \Rightarrow 2 {AB}^2 = 2 {AC}^2 + {BC}^2\]

Hence, 2AB= 2AC+ BC2.

  Is there an error in this question or solution?

APPEARS IN

 Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 2: Pythagoras Theorem
Problem Set 2 | Q: 13 | Page no. 45
Solution In ∆Abc, Seg Ad ⊥ Seg Bc Db = 3cd. Prove that : 2ab2 = 2ac2 + Bc2 Concept: Pythagoras Theorem.
S
View in app×