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Triangles

Multiple Choice Questions

511. In ∆ABC and ∆DEF, if AB = DF, BC = DE, AC = EF and ∆D = 55°. Then, ∠B =

55°
35°
90°
90°

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512. In the following figure, ∠B = ∠D = 90° and BC = CD. Then, the relation between AB and DE is

AB = DE
AB > DE
AB < DE
AB < DE
AB < DE

163 Views

513. In ∆ABC, AB = AC, BD = EC. Then, ∆ADE is

right angled
scalene
isosceles
isosceles

120 Views

Short Answer Type

514.

Show that ΔABC, where A(–2, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 2) are similar triangles.

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515.

Given $∆ A B C ~ ∆ P Q R, i f \frac{A B}{P Q} = \frac{1}{3}, t h e n f i n d \frac{a r ∆ A B C}{a r ∆ P Q R}$

516.

Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

517.
If the area of two similar triangles are equal, prove that they are congruent.

Given: Let triangles be Δ ABC and ΔDEF both triangles are similar, i.e., ΔABC ~ ΔDEF and also, areas are equal, i.e., area ΔABC = area ΔDEF

To prove: Both triangles are congruent, i.e., ΔABC ≅ ΔDEF

Proof:

As given, ΔABC ~ ΔDEF

Since two triangles are similar therefore the ratio of the area is equal to the square of the ratio of its corresponding side

$\frac{a r e a ∆ A B C}{a r e a ∆ D E F} = {(\frac{B C}{E F})}^{2} = {(\frac{A B}{D E})}^{2} = {(\frac{A C}{D F})}^{2} {(\frac{B C}{E F})}^{2} = {(\frac{A B}{D E})}^{2} = {(\frac{A C}{D F})}^{2} = 1 N o w, t a k i n g a n y o n e c a s e 1 = {(\frac{B C}{E F})}^{2} 1 = \frac{B C}{E F} E F = B C$