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In \[\Delta {\text{ABC}},{\text{ P}},{\text{ Q}},{\text{ R }}\] are the mid points of the sides \[AB,BC,CA\] respectively. If ABC is \[120\] sq. cm., find \[\Delta {\text{PQR}},{\text{ PQCR}},{\text{ PBCR}}\].

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Last updated date: 13th Jun 2024
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Answer
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Hint: At first, we will find the similarity of all the triangles.
Then, we will find the area of the small triangles.
Using the relation between the area of parallelogram and triangle, we can find the area of the other parallelogram and trapezium.
We know that, if a triangle and parallelogram are on the same base and have the same altitude, the area of the triangle will be half that of the parallelogram.

Complete step-by-step answer:
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It is given that; in \[\Delta {\text{ABC}},{\text{ P}},{\text{ Q}},{\text{ R }}\] are the mid points of the sides \[AB,BC,CA\] respectively. It is also given that ABC is \[120\]sq. cm.
We have to find the area of \[\Delta {\text{PQR}},{\text{ PQCR}},{\text{ PBCR}}.\]
We know that, the line joining the midpoints of the sides of the triangle from four triangles, each of which is similar to the original triangle.
So, \[\Delta ABC \sim \Delta {\text{PQR}}\].
In \[\Delta {\text{ABC}},{\text{ P}}\] and \[{\text{R }}\] are the mid points of \[AB\] and \[CA\] respectively.
So, by mid-point theorem, we get, \[PR\parallel BC\].
In the triangles \[\Delta ABC\] and \[\Delta APR\]
\[\angle A\] is a common angle and \[\angle APR = \angle ABC\] as they are corresponding angles.
 So, by AA similarity condition we get, \[\Delta ABC \sim \Delta {\text{APR}}\]
Again, by mid-point theorem we get, \[PR = \dfrac{1}{2}BC\]
So, \[\Delta ABC \sim \Delta {\text{PQR}}\]
So, we have
\[ \Rightarrow \dfrac{{\Delta ABC}}{{\Delta {\text{PQR}}}} = \dfrac{{B{C^2}}}{{P{R^2}}}\]
Substitute the values we get,
\[ \Rightarrow \dfrac{{\Delta ABC}}{{\Delta {\text{PQR}}}} = \dfrac{{B{C^2}}}{{P{R^2}}} = {(2)^2} = 4\]
Now, area of ABC is \[120\] sq. cm
So, area of \[\Delta PQR\] is \[\dfrac{{120}}{4} = 30\] sq. cm
So, the area of \[\Delta PQR\] is \[30\] sq. cm.
By the mid-point theorem, we get, \[PR\parallel BC;QR\parallel AB;PQ\parallel AC\]. So, \[PQCR\] is a parallelogram.
Here, \[PQCR\] and \[\Delta PQR\] are lying in the same base \[QR\] and the pair of same parallel lines \[QR\& PR\].
So, the area of the parallelogram \[PQCR\] is twice of \[\Delta PQR\].
So, the area of the parallelogram \[PQCR\] is \[2 \times 30 = 60\] sq. cm.
Similarly, we get, the area of the parallelogram \[PBQR\]is \[2 \times 30 = 60\] sq. cm.
Now, to find the area of \[PBCR\].
So, the area of \[PBCR\] is \[Ar(PBQR) + Ar(PQCR) - Ar(PQR)\]
Substitute the values we get,
The area of \[PBCR\] is \[ = 60 + 60 - 30 = 90\]sq. cm.

Hence,
i) The area of \[\Delta PQR\] is \[30\]sq. cm.
ii) The area of the parallelogram \[PQCR\] is \[60\] sq. cm.
iii) The area of \[PBCR\] is \[90\] sq. cm.


Note: If the line segment adjoins midpoints of any of the sides of a triangle, then the line segment is said to be parallel to all the remaining sides, and it measures about half of the remaining sides.
If a triangle and parallelogram are on the same base and have the same altitude, the area of the triangle will be half that of the parallelogram.