In this post we will discuss questions related to area of parallelogram and triangle theorem.
If you want to revisit the theorem, given below is the relevant link.
Question 01
ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Solution
We know that in parallelogram, opposite sides are equal and parallel.
AB = CD = 16 cm
The formula for area of parallelogram is mentioned below;
Area of parallelogram = Base x Altitude
Using the figure above, the formulas can be written as;
Area = CD x AE and Area = AD x CF
CD x AE = AD x CF
Putting the values;
16 x 8 = AD x 10
Solving the equation, we get;
AD = 12.8 cm
Hence, the value of AD is 12.8 cm
Question 02
If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar (EFGH) = 1/2 ar(ABCD)
Solution
Given:
ABCD is a parallelogram.
Points E, F, G & H are the midpoints of their respective slides.
Construction
Draw line HF, parallel to lines AB & CD.
Solution
We know that in parallelogram, opposite sides are equal and parallel.
Since, ABCD is a parallelogram, we can write;
AB = CD and AD = BC
Consider the above equation AD = BC;
1/2 AD = 1/2 BC
Since H & F are midpoint of side AD & BC respectively, we can write above equation as;
AH = BF or HD = FC
Now consider the shape AHFB;
AH = BF (proved above)
AB = HF ( as per construction)
Also, opposite sides are parallel to each other.
Hence, we can say that AHFB is a parallelogram
Using the similar process we can prove that HFCD is a parallelogram.
Now consider parallelogram AHFB and triangle HEF.
Both triangle and parallelogram lie in same base HF and between same parallel lines HF & AB.
Using the triangle – parallelogram area theorem, we can write;
Area (HEF) = 1/2 Area (AHFB) –> eq (1)
Similarly for triangle HFG and parallelogram HFCD, we can write;
Area (HFG) = 1/2 Area (HFCD) –> eq (2)
Now add both the equation (1) and (2), we get;
Area (HEF) + Area (HEG) = 1/2 Area (AHFB) + 1/2 Area (HFCD)
Area (EFGH) = 1/2 Area (ABCD)
Hence Proved.
Question 03
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
Solution
We know that if triangle and parallelogram lie in same base and between same parallel line then area of triangle is half of area of parallelogram.
Both triangle APB and parallelogram ABCD lie between same base and same parallel lines.
Area ( APB ) = 1/2 Area (ABCD) —> eq (1)
Similarly triangle CQB and parallelogram ABCD lie between same base and same parallel lines.
Area ( CQB ) = 1/2 Area (ABCD) —> eq (2)
Combining eq (1) and (2) we get;
Area (APB) = Area (CQB)
Hence Proved
Question 04
P is a point in the interior of a parallelogram ABCD. Show that ;
ar (APB) + ar (PCD) = 1/2 ar (ABCD)
Solution
Construction
Draw line GH parallel to line CD & AB.
So now we have two new parallelogram ABHG and GHCD.
You can also see that both the parallelogram is half of area of ABCD.
i.e. Area (ABHG) = Area (GHCD) = 1/2 ABCD
We know that if triangle and parallelogram in same base and between same parallel lines then area of triangle is half of area of parallelogram.
Now triangle APB and AGHB lie in same base and same parallel lines;
So area (APB) = 1/2 (AGHB) —eq (1)
Similar is the case of triangle CPD and parallelogram GHCD;
So area (CPD) = 1/2 (GHCD) —eq (2)
Adding equation (1) + (2);
Area (APB) + Area (CPD) = 1/2 (AGHB) + 1/2 (GHCD)
Area (APB) + Area (CPD) = 1/2 (AGHB + GHCD)
Area (APB) + Area (CPD) = 1/2 (ABCD)
Hence Proved