Question

In DEF, DEF = 900 , DE = EF. DF = 20√ cm. Find perimeter of DEF. Sol. In DEF, DEF = 900 , DE = EF.​

Answer 1

40 + 20√2 cm

Step-by-step explanation:

CORRECT QUESTION :-

In ∆DEF, ∠DEF = 90°, DE = EF, DF = 20√2 cm. Find perimeter of ∆DEF.

__________________________

SOLUTION :-

A.T.Q.,

∠DEF = 90°, so ∆DEF is a right angled triangle whose Hypotenuse (DF) = 20√3 cm.

To get the perimeter of triangle, we have to get the lengths of DE & EF.

We know in a right-angled triangle, square of hypotenuse is equal to sum of squares of other two sides. [Pythagoras theorem]

Here,

both others sides are equal, so let them as x cm.

${h}^{2} = {b}^{2} + {p}^{2} \\ \\ = > {(20 \sqrt{2} )}^{2} = {x}^{2} + {x}^{2} \\ \\ = > 400 \times 2 = 2 {x}^{2} \\ \\ = > \frac{800}{2} = {x}^{2} \\ \\ = > {x}^{2} = 400 \\ \\ = > x = \sqrt{400} = \red{ \underline{20cm}}$

So,

Other sides are of 20 cm.

i.e. DE = EF = 20 cm

Hence Perimeter of ∆DEF

= DE + EF + DF

= 20 + 20 + 20√2 cm

= 40 + 20√2 cm

.

Answer = 40 + 20√2 cm.

_____________________

Hope it helps.

#BeBrainly

Answer 2

Answer:

Answer = 40 + 20√2 cm.

Step-by-step explanation:

CORRECT QUESTION :-

In ∆DEF, ∠DEF = 90°, DE = EF, DF = 20√2 cm. Find perimeter of ∆DEF.

__________________________

SOLUTION :-

A.T.Q.,

∠DEF = 90°, so ∆DEF is a right angled triangle whose Hypotenuse (DF) = 20√3 cm.

To get the perimeter of triangle, we have to get the lengths of DE & EF.

We know in a right-angled triangle, square of hypotenuse is equal to sum of squares of other two sides. [Pythagoras theorem]

Here,

both others sides are equal, so let them as x cm.

\begin{gathered} {h}^{2} = {b}^{2} + {p}^{2} \\ \\ = > {(20 \sqrt{2} )}^{2} = {x}^{2} + {x}^{2} \\ \\ = > 400 \times 2 = 2 {x}^{2} \\ \\ = > \frac{800}{2} = {x}^{2} \\ \\ = > {x}^{2} = 400 \\ \\ = > x = \sqrt{400} = \red{ \underline{20cm}}\end{gathered}

h

2

=b

2

+p

2

=>(20

2

)

2

=x

2

+x

2

=>400×2=2x

2

=>

2

800

=x

2

=>x

2

=400

=>x=

400

=

20cm

So,

Other sides are of 20 cm.

i.e. DE = EF = 20 cm

Hence Perimeter of ∆DEF

= DE + EF + DF

= 20 + 20 + 20√2 cm

= 40 + 20√2 cm

Answer = 40 + 20√2 cm.

_____________________

Hope it helps.