What is the figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8cm and 6cm?

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Rhombus of area 24cm2

Square of area 24cm2

Trapezium of area 24cm2

Diamond of area 24cm2

answer is A.

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Detailed Solution

We have to identify the figure.
Question Image

Rectangle is four -sided-polygon and the opposite sides of the rectangle are equal in length whereas a rhombus is a four-sided quadrilateral in which opposite sides are parallel.
All the sides of a rhombus are equal in length, and the diagonals bisect each other at right angles.
Area of a rhombus

= d 1 × d 2 2

Here, d1 and d2 are the diagonals of a rhombus.
Pythagoras Theorem states that in a right-angled triangle, the square of the length of the hypotenuse equals the sum of the legs of the right triangle.

c 2 = a 2 + b 2

Considering the rectangle ABCD shown below, Question Image

Here, DR, RD, AP, and PB are equal.
Their length is

= 82 = 4 c m

(As P and R are the midpoints of DC and AB, respectively).
Also, AS, SD, BQ, and QC are equal.
Their length

= 62 = 3 c m

(As S and Q are the midpoints of AD and BC, respectively).
Using Pythagoras Theorem,

S R 2 = S D 2 + D R 2 ⇒ S R 2 = 32 + 42 ⇒ S R 2 = 25 ⇒ S R = 5

Similarly, using Pythagoras Theorem for ∆QRC, ∆PBQ and ∆APS
Therefore, RQ, QP, and PS are 5cm.
So, PQRS is a rhombus with a side length of 5cm.
Calculating the area of rhombus PQRS.
Area of rhombus