In a ΔABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE ∥ BCAB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm.

Question

0

deepaksoniGuru

Asked: August 14, 20212021-08-14T15:28:32+05:30 2021-08-14T15:28:32+05:30In: CBSE

In a ΔABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE ∥ BCAB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm.

0

Question drawn from very renowned book RD sharma of class 10th, Chapter no. 4
Chapter name:- Triangles
Exercise :- 4.2
This is very important question of triangle.

We have been given a triangle ΔABC,

In which D and E are points on the sides AB and AC respectively.

Now For following case we have to show that DE ∥ BC

Also it is given that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm.

learning CBSE maths in efficient way

RD sharma, DHANPAT RAI publication
Class 10th, triangle

1 Answer

Leave an answer

Leave an answer
Cancel reply

AAREKH · Answer 1 · 2021-08-19T00:23:45+05:30

Given:

Length of side AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm

To show : Side DE is parallel to BC ( DE || BC )

Now,

Length of side BD = AB – AD = 5.6 – 1.4 = 4.2 cm

And,

Length of side CE = AC – AE = 7.2 – 1.8 = 5.4 cm

Now,

AD/BD = 1.4/4.2 = 1/3 – equation 1

AE/CE = 1.8/5.4 =1/3 – equation 2

Now from equation 1 and 2

AD/BD = AE/CE

So, by the converse of Thale’s Theorem

We get, DE ∥ BC.

Hence, Proved.

In a ΔABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE ∥ BCAB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm.

1 Answer

Leave an answerCancel reply

Related Questions

Leave an answer
Cancel reply