In a - ABC - - A -80- and - B -20- and in - DEF - E -80- and - F -20- - Is - ABC - EF D -A- TrueB- False

The correct option is A True In the Δ ABC and Δ EFD ∠ ABC=∠ EFD=20^∘ ∠ CAB=∠ DEF= 80^∘ So, Δ ABC ∼Δ EFD (By AA similarity criteria) ...

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Byju's Answer

Standard X

Mathematics

Perpendicular from Right Angle to Hypotenuse Divides the Triangle into Two Similar Triangles

In a Δ ABC , ...

Question

In a $Δ A B C, ∠ A = 80^{\circ} a n d ∠ B = 20^{\circ}$ and in $Δ D E F ∠ E = 80^{\circ} a n d ∠ F = 20^{\circ}$ . Is $Δ A B C \sim Δ E F D$ ?

True

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False

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Solution

The correct option is A
True

In the $Δ A B C$ and $Δ E F D$

$∠ A B C = ∠ E F D = 20^{\circ}$

$∠ C A B = ∠ D E F = 80^{\circ}$

So, $Δ A B C \sim Δ E F D$ (By AA similarity criteria)

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Similar questions

$I n Δ A B C a n d Δ D E F, it is given that ∠ B = ∠ E a n d ∠ C = ∠ F . In order that Δ A B C ≅ Δ D E F, we must have (a) A B = D F (b) A C = D E (c) B C = E F (d) ∠ A = ∠ D$

Q. In

Δ A B C, ∠ B = 30^{\circ}, ∠ C = 80^{\circ}

and

∠ A = 70^{\circ}

then,

Δ A B C

and

Δ D E F

are similar and

∠ A = 40^{˚}, ∠ E + ∠ F =

In a $Δ A B C, i f A B = A C and BC is produced$

$to D such that ∠ A C D = 100^{\circ}, t h e n ∠ A =$

Question 40

In $Δ A B C$ , $∠ A = 100^{\circ}$ , AD bisects $∠ A$ and $A D ⊥ B C$ . Then, $∠ B$ is equal to

a) $80^{\circ}$
b) $20^{\circ}$
c) $40^{\circ}$
d) $30^{\circ}$

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In a ΔABC, ∠A=80∘ and ∠B=20∘ and in ΔDEF ∠E=80∘ and ∠F=20∘. Is ΔABC∼ΔEFD?

The correct option is A True In the ΔABC and ΔEFD ∠ABC=∠EFD=20∘ ∠CAB=∠DEF=80∘ So, ΔABC∼ΔEFD (By AA similarity criteria)

In a $Δ A B C, ∠ A = 80^{\circ} a n d ∠ B = 20^{\circ}$ and in $Δ D E F ∠ E = 80^{\circ} a n d ∠ F = 20^{\circ}$ . Is $Δ A B C \sim Δ E F D$ ?

The correct option is A
True

In the $Δ A B C$ and $Δ E F D$

$∠ A B C = ∠ E F D = 20^{\circ}$

$∠ C A B = ∠ D E F = 80^{\circ}$

So, $Δ A B C \sim Δ E F D$ (By AA similarity criteria)