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3 years ago

In the diagram below, MNPQ is a parallelogram whose diagonals are perpendicular. Prove: MNPQ is a rhombus.

Mathematics

1 answer:

katen-ka-za [31]3 years ago

4 0

Answer:

Given : MNPQ is a parallelogram whose diagonals are perpendicular.

To prove : MNPQ is a rhombus.

Proof:

In parallelogram MNPQ,

R is the intersection point of the diagonals MP and NQ( shown in below diagram)

$\implies MR\cong RP$ (Because, the diagonals of parallelogram bisects each other)

$\angle MRQ\cong \angle QRP$ (Right angles )

$QR\cong QR$ (Reflexive)

Thus, By SAS postulate of congruence,

$\triangle MRQ\cong \triangle PRQ$

By CPCTC,

$MQ\cong QP$

Similarly,

We can prove, $\triangle MRN\cong \triangle PRN$

By CPCTC,

$MN\cong NP$

But, By the definition of parallelogram,

$MN\cong QP$ and $MQ\cong NP$

⇒ $MN\cong NP\cong PQ\cong MQ$

All four side of parallelogram MNQP are congruent.

⇒ Parallelogram MNPQ is a rhombus.

Hence, proved.

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3 years ago

At a zoo, the lion pen has a ring-shaped sidewalk around it. The outer edge of the sidewalk is a circle with a radius of 11 m. T

elixir [45]

Answer:

$\text{Exact area of the sidewalk}=40 \pi\text{ m}^2$

$\text{Approximate area of the sidewalk}=125.6\text{ m}^2$

Step-by-step explanation:

We have been given that at a zoo, the lion pen has a ring-shaped sidewalk around it. The outer edge of the sidewalk is a circle with a radius of 11 m. The inner edge of the sidewalk is a circle with a radius of 9 m.

To find the area of the side walk we will subtract the area of inner edge of the side walk of lion pen from the area of the outer edge of the lion pen.

$\text{Area of circle}=\pi r^2$ , where r represents radius of the circle.