The document Long Answer Type Questions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.

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**Question 1. 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. Solution: **∵ Area of a parallelogram = base x height

∴ Area of parallelogram ABCD = AB x DL = 12 cm x 5 cm = 60 cm^{2} Again, area of parallelogram ABCD = AD x BM

∴ AD x BM = 60 cm^{2} [∵ ar (parallelogram ABCD) = 60 cm2]

⇒ AD x 8 cm = 60 cm^{2} [∵ BM = 8 cm (Given)]

⇒ AD = (60/8) cm

= (15/2) cm = 7.5 cm

Thus, the required length of AD is 7.5 cm.

**Question 2. 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. Solution:** 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 cm^{2}

ar (ΔBCD) = (1/2)x base x height

= (1/2)x BC x height

= (1/2)x 9 cm x 8 cm

= 36 cm^{2 }

Now, ar (trapezium ABCD) = ar (D ABD) + ar (D BCD)

= 20 cm^{2} + 36 cm^{2}

= 56 cm^{2}

**Question 3. In the adjoining figure, find the area of the trapezium ABCD. Solution: **∵ DEC is a right triangle.

∴ ar (ΔDEC) = (1/2)x base x height

= (1/2)x 8 x 15 cm^{2}

= 60 cm^{2} ar (rectangle ABED) = length x breadth

= 10 cm x 15 cm = 150 cm^{2}

Now, ar (trapezium ABCD) = ar (rectangle ABED) + ar (ΔDEC)

= 150 cm2 + 60 cm^{2} = 210 cm^{2}

Thus, the required area of trapezium ABCD = 210 cm^{2}.

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

Solution.

∴ 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)

**Question 5. 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. Solution:** 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 cm^{2}

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 cm^{2} + 24 cm^{2}

= 60 cm^{2}

Thus, the required area of quadrilateral ABCD

= 60 cm^{2}.

**Question 6. Find the area of the following quadrilateral.**

**Solution**: ∵ APB is a right triangle.

∴ AB^{2} – BP^{2 }= AP^{2}

⇒ 10^{2} – 8^{2 }= AP^{2}

⇒ 100 – 64 = AP^{2 }

⇒ 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 cm^{2} Area of rectangle BCQP

= length x breath = 6 cm x 8 cm = 48 cm^{2}

Now, area of quadrilateral ABCD

= ar (rt ΔAPB) + ar (rectangle BCQP) + ar (rt ΔCQD)

= 24 cm^{2} + 48 cm^{2} + 24 cm^{2}

= 96 cm^{2}

Thus, the required area of the quadrilateral ABCD

= 96 cm^{2}.

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