Question

The construction of a triangle DEF in which DE =7 cm, angle D = 75° is possible when (DE-EF) is equal to how many centimeters

Answer 1

Answer:

$\triangle EFD$ is the required triangle.

Step-by-step explanation:

Given,

Construction of $\triangle EFD$  in which $DE=7\ cm\ and \ \angle D=75$°

We know,

The construction of a triangle is not possible if sum of two sides is greater than third side of triangle.

(DE + EF) >DF

=DE>(DF-EF)

=7>(DF-EF)

We can take the value of  is (DF-EF) is 6.5cm

Following steps:

• Draw line segments DE = 7cm using scale.

• At D , draw an angle 75°  like $\angle EDX$ = 75°

• Cuts a point  P on line DX  like DP+6.5cm

• Join  PE

• Bisect the line  PE and the bisecting line cuts the  DX at F.

• Join EF

Hence, $\triangle EFD$ is the required triangle.

Answer 2

Answer:

6.5cm

Step-by-step explanation:

EFD is the required triangle.

Step-by-step explanation:

Given,

Construction of \triangle EFD△EFD in which DE=7\ cm\ and \ \angle D=75DE=7 cm and ∠D=75 °

We know,

The construction of a triangle is not possible if sum of two sides is greater than third side of triangle.

(DE + EF) >DF

=DE>(DF-EF)

=7>(DF-EF)

We can take the value of is (DF-EF) is 6.5cm

Following steps:

Draw line segments DE = 7cm using scale.

At D , draw an angle 75° like \angle EDX∠EDX = 75°

Cuts a point P on line DX like DP+6.5cm

Join PE

Bisect the line PE and the bisecting line cuts the DX at F.

Join EF

Hence, \triangle EFD△EFD is the required triangle.