Question

13.The framework for a railroad bridge is shown below.You know only that ED∥ BC, AB∥ EF and m∠PER =130°. Please don't spam

Answer 1

Given : framework for a railroad bridge.

ED∥ BC, AB∥ EF , PR ⊥ EB and m∠PER =130°.

To Find : Quadrilateral EPBR

∠P

∠EQP

Quadrilateral formed by join mid points of adjacent sides of a rectangle

Solution:

Opposite sides are parallel hence Parallelogram

also PR ⊥ EB  , hence Diagonal are perpendicular to each other

so Quadrilateral EPBR is a rhombus

Sum of adjacent angle = 180°  ( in a  parallelogram / rhombus)

∠P + ∠PER = 180°

=> ∠P + 150° = 180°

=> ∠P   = 30°

Diagonal of rhombus bisect each other perpendicularly.

∠EQP = 90°

Quadrilateral formed by join mid points of adjacent sides of a rectangle is RHOMBUS

as all sides are Equal to half of diagonal of original rectangle

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Answer 2

Answer:

Given : framework for a railroad bridge.

ED∥ BC, AB∥ EF , PR ⊥ EB and m∠PER =130°.

To Find : Quadrilateral EPBR

∠P

∠EQP

Quadrilateral formed by join mid points of adjacent sides of a rectangle

Solution:

Opposite sides are parallel hence Parallelogram

also PR ⊥ EB , hence Diagonal are perpendicular to each other

so Quadrilateral EPBR is a rhombus

Sum of adjacent angle = 180° ( in a parallelogram / rhombus)

∠P + ∠PER = 180°

=> ∠P + 150° = 180°

=> ∠P = 30°

Diagonal of rhombus bisect each other perpendicularly.

∠EQP = 90°

Quadrilateral formed by join mid points of adjacent sides of a rectangle is RHOMBUS

as all sides are Equal to half of diagonal of original rectangle

Step-by-step explanation: