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Let P, Q, R be the mid-point of the sides AB, BC & CA respectively of a triangle ABC. If the area of the triangle ABC is 5 square units, then the area of the triangle PQR
- a)5/3 square units
- b)5/2√2 square units
- c)5/4 square units
- d)1 square units
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let P, Q, R be the mid-point of the sides AB, BC & CA respectively...
Since P, Q & R are the mid points of AB, BC & CA respectively, they divide the triangle ABC into 4 equal triangles: APR, BPQ, CQR & PQR;
∴ Area of the triangle PQR = ¼ × (Area of the triangle ABC) = 5/4 square units
Most Upvoted Answer
Let P, Q, R be the mid-point of the sides AB, BC & CA respectively...
And CA of triangle ABC, respectively.
First, we need to find the coordinates of points P, Q, and R.
Let's assume that the coordinates of A, B, and C are (x1, y1), (x2, y2), and (x3, y3) respectively.
The coordinates of P can be found by finding the average of the x-coordinates and the average of the y-coordinates of A and B.
So, the x-coordinate of P is (x1 + x2)/2, and the y-coordinate of P is (y1 + y2)/2.
Similarly, the coordinates of Q can be found by finding the average of the x-coordinates and the average of the y-coordinates of B and C.
So, the x-coordinate of Q is (x2 + x3)/2, and the y-coordinate of Q is (y2 + y3)/2.
Finally, the coordinates of R can be found by finding the average of the x-coordinates and the average of the y-coordinates of C and A.
So, the x-coordinate of R is (x3 + x1)/2, and the y-coordinate of R is (y3 + y1)/2.
Therefore, the coordinates of P are ((x1 + x2)/2, (y1 + y2)/2),
the coordinates of Q are ((x2 + x3)/2, (y2 + y3)/2),
and the coordinates of R are ((x3 + x1)/2, (y3 + y1)/2).
First, we need to find the coordinates of points P, Q, and R.
Let's assume that the coordinates of A, B, and C are (x1, y1), (x2, y2), and (x3, y3) respectively.
The coordinates of P can be found by finding the average of the x-coordinates and the average of the y-coordinates of A and B.
So, the x-coordinate of P is (x1 + x2)/2, and the y-coordinate of P is (y1 + y2)/2.
Similarly, the coordinates of Q can be found by finding the average of the x-coordinates and the average of the y-coordinates of B and C.
So, the x-coordinate of Q is (x2 + x3)/2, and the y-coordinate of Q is (y2 + y3)/2.
Finally, the coordinates of R can be found by finding the average of the x-coordinates and the average of the y-coordinates of C and A.
So, the x-coordinate of R is (x3 + x1)/2, and the y-coordinate of R is (y3 + y1)/2.
Therefore, the coordinates of P are ((x1 + x2)/2, (y1 + y2)/2),
the coordinates of Q are ((x2 + x3)/2, (y2 + y3)/2),
and the coordinates of R are ((x3 + x1)/2, (y3 + y1)/2).
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Let P, Q, R be the mid-point of the sides AB, BC & CA respectively of a triangle ABC. If the area of the triangle ABC is 5 square units, then the area of the triangle PQRa)5/3 square unitsb)5/2√2 square unitsc)5/4 square unitsd)1 square unitsCorrect answer is option 'C'. Can you explain this answer?
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