Question

ABCD is a square of side 13 units, AE = CF = 12 and BE =DF-5
If the length of EF is equal to avbsuch that a and b are both
positive prime numbers, then (ab) is equal to​

Answer 1

Answer:

GIVEN : Square ABCD, with each side= 13 cm AE = FC = 12cm & BE = DF = 5cm

TO FIND : EF² = ?

CONSTRUCTION: Just extend EA & FD to meet at G. Then extend EB & FC to meet at H.

PROOF & CALCULATION: In triangles FDC & EBA,

Since, sides, , 5,12 & 13 are Pythagorean triplet ( as 13² = 5² + 12² )

=> angle opposite to the longest side = 90°

=> angle F = angle E = 90° ………..(1)

Now, since,

triangle AEB is congruent to triangle CFD ( By SSS Congruence theorem)

=> < 1 = <2 And < 3 = <4 ( cpct) ………..(2)

Now, < 1 + < 3 = 90° ( as angle E= 90°)

And, <1 + <5 = 90° Therefore <3 = <5 ………(3)

Then, < 2 + <4 = 90°

And, <6 + <4 = 90° Therefore < 2 = <6

But <2 = <1 ( proved)

<1 = <6 ……….. (4)

Now by (3) & (4)

Triangle EAB is congruent to triangle GDA ( by ASA Congruence criterion)

=> angle G = angle E = 90° , GA = EB= 5cm And GD = EA = 12 cm ( all by cpct)

Similarly, we can prove for 4th triangle HBC is congruent to triangle EAB

This way we proved that all 4 triangles are congruent to each other.

So, quadrilateral GEHF becomes a quadrilateral with all the sides equal & each angle = 90°

Hence GEHF is a SQUARE

Therefore EF² = FG² + GE²

=> EF² = 17² + 17² = 2 x 289

=> EF² = 578

=> EF = √578 ……….ANS

hope it helpful for you

please mark me as brainlist

Answer 2

Answer:

(ab) = 34

Step-by-step explanation:

GIVEN : Square ABCD, with every side= 13 cm AE = FC = 12cm & BE = DF = 5cm

TO FIND : EF² = ?

CONSTRUCTION: Just amplify EA & FD to fulfil at G. Then amplify EB & FC to fulfil at H.

PROOF & CALCULATION: In triangles FDC & EBA,

Since, sides, , 5,12 & 13 are Pythagorean triplet ( as 13² = 5² + 12² )

⇒ angle opposite to the longest side = 90°

⇒ angle F = angle E = 90° ………..(1)

Now, since,

triangle AEB is congruent to triangle CFD ( By SSS Congruence theorem)

⇒∠1 =∠2  and ∠3 = ∠4 (cpct).......(2)

Now we have,  ∠1 + ∠3 = 90°  ( as angle E= 90°)

And,                  ∠2 + ∠5 = 90° Hence, ∠3 = ∠5

Then,                ∠2 + ∠4 = 90°

And,                  ∠6 + ∠4 = 90° Therefore ∠2 = ∠6

But ∠2 = ∠1 ( proved)

∠1 = ∠6 ……….. (4)

Now by (3) & (4)

Triangle EAB is congruent to triangle GDA ( by ASA Congruence criterion)

⇒ angle G = angle E = 90° , GA = EB= 5cm And GD = EA = 12 cm ( all by cpct)

Similarly, we will show for 4th triangle HBC is congruent to triangle EAB

This manner we proved that each one 4 triangles are congruent to each other.

So, quadrilateral GEHF will become a quadrilateral with all of the sides equal & every angle = 90°

Hence GEHF is a SQUARE

Therefore EF² = FG² + GE²

⇒ EF² = 17² + 17² = 2 x 289

⇒ EF² = 578

⇒EF = 17√2....(5)

comparing equation (5) with a√b

Then we get a = 17 and b = 2 which are actually prime numbers itself.

a × b = 17 × 2 = 34.

#SPJ2