Question 23 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at April 16, 2024 by Teachoo

In the given figure, DEFG is a square and ∠BAC = 90°. Show that FG ² = BG × FC

Maths Standard Q 23.jpg

Question 23 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 2

Question 23 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 3

Question 23 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 4

Transcript

Question 23 (OR 1st question) In the given figure, DEFG is a square and ∠BAC = 90° . Show that FG2 = BG × FC Now, DE ∥ GF, So, we can say DE ∥ BC For parallel lines DE and BC, transversal AB ∠ ADE = ∠ GBD For parallel lines DE and BC, transversal AC ∠ AED = ∠ FCE Now, In Δ ADE and Δ GBD ∠ ADE = ∠ GBD ∠ DAE = ∠ DGB Δ ADE ∼ Δ GBD In Δ ADE and Δ FEC ∠ AED = ∠ FCE ∠ DAE = ∠ EFC Δ ADE ∼ Δ FEC So, we can write Δ GBD ∼ Δ FEC Hence, sides of similar triangle are in proportion 𝐺𝐷/𝐹𝐶=𝐺𝐵/𝐹𝐸 GD × FE = FC × GB Since DEFG is a square GD = FE = FG So, our equation becomes FG × FG = FC × GB FG2 = FC × GB Hence proved

Next: Question 23 (OR 2nd question) → Ask a doubt

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular