Question

prove that length of perpendicular drawn from vertices of equal angle of an isosceles triangle to the opposite sides are equal . show that area of rhombus is half of the product of length of its diagonals.

Answer 1

Hello Mate!

Q. Prove that length of perpendicular drawn from vertices of equal angle of an isosceles triangle to the opposite sides are equal.

Given : In ∆ABC, BE ⊥ AC and CF ⊥ AB where ∠B = ∠C.

To prove : BE = CF

Proof : In ∆BFC and ∆BEC we have,

∠BFC = ∠BEC ( 90° each )

∠B = ∠C ( Given )

BC = BC ( Common )

Hence ∆BFC ≈ ∆BEC by AAS congruency.

BE = CF ( c.p.c.t )

Q.E.D

Q. Show that area of rhombus is half of the product of length of its diagonals.

Given : ABCD is a rhombus

To prove : area of rhombus = ½ × product of diagonals

Proof : Since area of triangle = ½ × base × height

ar(∆ABC) = ½ × AC × OB

Similarly ar(∆ADC) = ½ × AC × OD

Adding eq (i) and (ii) we get,

ar(∆ABC) + ar(∆ADC) = ½ × AC × OB + ½ × AC × OD

ar(ABCD) = ½ × AC( OB + OD )

ar(ABCD) = ½ × AC × BD

Hence, area of rhombus = ½ × product of diagonals.

Q.E.D

Have great future ahead!

Answer 2

Answer:

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