Question

A student construted a triangle ABC with sides AB=5cm, BC=6.5cm and AC=7cm and then constructed a ∆ADE similar to ∆ABC such that each of its sides are 5/7 of the corresponding sides of ∆ABC.The length of AD and AE obtained by

calculation are respectively equal to​

Answer 1

Answer:

Step-by-step explanation:

Draw a line segment BC = 6 cm.

Taking B as centre and radius 5 cm draw an arc.

Taking C as centre and radius 7 cm, draw an arc which cut the previous arc at A.

Join AB and AC.

Thus, ABC is the required triangle.

Answer 2

The schematic of the triangle is given in the figure.

Given:

AB=5cm,

BC=6.5cm

and,

AC=7cm

$\frac{AD}{DB} =\frac{AE}{EC} = \frac{5}{7}$......................1

∆ADE is prepared similar to ∆ABC such that each of its sides are 5/7 of the corresponding sides of ∆ABC

Find:

AD and AE

Solution:

By BPT(Basic proportionality theorem):

we can say that ,

$\frac{AD}{DB} =\frac{AE}{EC}$........................2

Now, AC = AE + EC

or,

EC = AC-AE................3

Putting the value of the equation 1 and 3 in 2

$\frac{5}{7} =\frac{AE}{AC- AE}$

solving this:

AC = 7

5(7-AE) = 7AE

35 - 5AE = 7AE

12AE = 35

AE = 2.9 cm

Similarly solving for AD

$\frac{AD}{DB} =\frac{AE}{EC}$

and,

AB = AD+DB

DB = AB-AD

the equation becomes-

$\frac{AD}{AB-AD} =\frac{5}{7}$

AB = 5cm

$\frac{AD}{5-AD} =\frac{5}{7}$

7AD = 25- 5AD

7AD+5AD = 25

12AD = 25

AD = 25/12

AD = 2.1 cm

Hence the length of AD and AE are AD = 2.1 cm and AE = 2.9 cm respectively

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