Class 9 Maths Question Answers - Areas of Parallelograms

Q1. In a parallelogram ABCD, it is being given that AB = 12 cm and the altitude corresponding to the sides AB and AD are DL = 5 cm and BM = 8 cm respectively. Find AD.

∵  Area of a parallelogram = base x height

∴ Area of parallelogram ABCD = AB x DL = 12 cm x 5 cm = 60 cm² Again, area of parallelogram ABCD = AD x BM
∴  AD x BM = 60 cm²                    [∵ ar (parallelogram ABCD) = 60 cm2]
⇒ AD x 8 cm = 60 cm²                 [∵ BM = 8 cm (Given)]
⇒ AD = (60/8) cm
= (15/2) cm = 7.5 cm
Thus, the required length of AD is 7.5 cm.

Q2. Find the area of a trapezium whose parallel sides are 9 cm and 5 cm respectively and the distance between these sides is 8 cm.

Let ABCD be a trapezium such that AD || BC. Let us join BD.

∵ Area of a triangle = (1/2)x base x height
∴ Area of ΔABD = (1/2)x AD x height
= (1/2)x 5 cm x 8 cm = 20 cm²
ar (ΔBCD) = (1/2)x base x height
= (1/2)x BC x height
= (1/2)x 9 cm x 8 cm
= 36 cm² 
Now, ar (trapezium ABCD) = ar (D ABD) + ar (D BCD)
= 20 cm² + 36 cm²
= 56 cm²

Q3. In the adjoining figure, find the area of the trapezium ABCD.

 ∵ DEC is a right triangle.

∴ ar (ΔDEC) = (1/2)x base x height
= (1/2)x 8 x 15 cm²
= 60 cm² ar (rectangle ABED) = length x breadth
= 10 cm x 15 cm = 150 cm²
Now, ar (trapezium ABCD) = ar (rectangle ABED) + ar (ΔDEC)
= 150 cm2 + 60 cm² = 210 cm²
Thus, the required area of trapezium ABCD = 210 cm².

Q 4. In the figure BD || CA, E is mid-point of CA and BD = (1/2) CA.
Prove that ar (ΔABC) = 2 ar (ΔDBC)

Join D and E 

∵ BD = CE and BD || CE
∴ BCED is a parallelogram. ⇒ ΔDBC and ΔEBC are on the same base BC and between the same parallels,

∴ ar (ΔDBC) = ar (ΔEBC)                 ... (1)
In ΔABC, BE is a median,
∴ ar (ΔFBC) = (1/2)ar (ΔABC)
Now, ar (ΔABC) = ar (ΔEBC) + ar (ΔABE)
Also, ar (ΔABC) = 2 ar (ΔEBC)
∴ ar (ΔABC) = 2 ar (ΔDBC)

Q5. In the adjoining figure, ABCD is a quadrilateral in which diagonal BD = 12 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 6 cm and CM = 4 cm, find the area of quadrilateral ABCD.

The diagonal BD divides the given quadrilateral into two triangles BDC and ABD.

Since, area of a triangle = (1/2)x base x altitude
∴ Area of ΔABD = (1/2)x BD x AL
= (1/2)x 12 cm  x 6 cm
= 36 cm²
Area of ΔBCD = (1/2)x base x altitude
=(1/2)x BD x CM
= (1/2)x 12 cm x 4 cm
= 24 cm2.
Now, ar (quadrilateral ABCD) = ar (ΔABD) + ar (ΔBCD)
⇒ ar (quadrilateral ABCD) = 36 cm² + 24 cm²
= 60 cm²
Thus, the required area of quadrilateral ABCD
= 60 cm².

Q6. Find the area of the following quadrilateral.

 ∵ APB is a right triangle.
∴ AB² – BP² = AP²
⇒ 10² – 8² = AP²
⇒ 100 – 64 = AP² 
⇒ AP = 36 = 6 cm
Since, ar (rt ΔAPB) =(1/2)x base x altitude
∴ ar (ΔAPB) =(1/2)x 6 cm x 8 cm
= 24 cm2
Similarly, ar (rt ΔCQD) = 24 cm² Area of rectangle BCQP
= length x breath = 6 cm x 8 cm = 48 cm²
Now, area of quadrilateral ABCD
= ar (rt ΔAPB) + ar (rectangle BCQP) + ar (rt ΔCQD)
= 24 cm² + 48 cm² + 24 cm²
= 96 cm²
Thus, the required area of the quadrilateral ABCD
= 96 cm².