If in Fig. 9.7, PQRS and EFRS are two parallelograms, then ar (MFR) = 1/2 ar (PQRS). Is the given statement true or false and justify your answer.
Solution:
We know that
PQRS and EFRS are on the same base SR and between the same parallels EF and SR
The areas will be equal
ar (PQRS) = ar (EFRS) …. (1)
ar (∆ MFR) = 1/2 ar (EFRS) …. (2)
From both the equations
ar (MFR) = 1/2 ar (PQRS)
Therefore, the statement is true.
✦ Try This: The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 5 cm and 4 cm is :
It is given that
Length of rectangle = 5 cm
Breadth of rectangle = 4 cm
Consider E, F, G and H as the mid-points of sides AB, BC, CD and AD
EFGH is a rhombus
Diagonals are EG and HF
So EF = BC = 5 cm
HF = AB = 4 cm
We know that
Area of rhombus = Product of diagonals/ 2
By further calculation
= (5 × 4)/2
= 20/2
= 10 cm²
Therefore, the figure obtained is a rhombus of area 10 cm².
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9
NCERT Exemplar Class 9 Maths Exercise 9.2 Sample Problem 2
If in Fig. 9.7, PQRS and EFRS are two parallelograms, then ar (MFR) = 1/2 ar (PQRS). Is the given statement true or false and justify your answer.
Summary:
The statement “If in Fig. 9.7, PQRS and EFRS are two parallelograms, then ar (MFR) = 1/2 ar (PQRS)” is true
☛ Related Questions:
- ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm² , then ar (ABC) = 24 cm² . . . .
- PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS . . . .
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