Ex 9.3, 5 - D, E and F are the mid-points of sides BC, CA - Paralleograms with same base & same parallel lines

Ex 9.3, 5 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

Ex 9.3, 5 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3

Ex 9.3, 5 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 4

Ex 9.3, 5 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 5


Transcript

Question 5 D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. Show that (i) BDEF is a parallelogram. Given: ABC where D, E, F are mid-points of BC, AC & AB respectively To prove: BDEF is a parallelogram Proof : In ABC, F is mid-point of AB, E is mid-point of AC FE BC FE BD Now, FE BD & DE FB In BDEF, both pairs of opposite sides are parallel, BDEF is a parallelogram Hence proved Question 5 D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. Show that (ii) ar (DEF) = 1/4 ar (ABC) In part(i) we proved that BDEF is a parallelogram DBF DEF ar(DBF) = ar(DEF) Similarly, we can prove FDCE is a parallelogram DEC DEF ar(DEC) = ar(DEF) Similarly, we can prove AFDE is a parallelogram AFE DEF ar(AFE) = ar(DEF) From (1),(2) & (3) ar(FBD) = ar(DEC) = ar(AFE) = ar(DEF) Now ar(FBD) + ar(DEC) + ar(AFE) + ar(DEF) = ar(ABC) ar(DEF) + ar(DEF) + ar(DEF) + ar(DEF) = ar(ABC) 4ar(DEF) = ar(ABC) ar (DEF) = 1/4 ar (ABC) Hence proved Question 5 D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. Show that (iii) ar (BDEF) = 1/2 ar (ABC) We proved in part(ii) that ar(DEF) = 1/4 ar(ABC) Multiplying both sides by 2 2 ar(DEF) = 2 1/4 ar(ABC) 2 ar(DEF) = 1/2 ar(ABC) ar(DEF) + ar(DEF) = 1/2 ar(ABC) ar(DEF) + ar(FBD) = 1/2 ar(ABC) ar(BDEF) = 1/2 ar(ABC) Hence proved

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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.