The document RD Sharma Solutions -Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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*Q 1. If figure, ABCD is a parallelogram, AE âŠ¥ DC and CF âŠ¥ AD. If AB = 16 cm, AE = 8 cm, and CF = 10 cm, Find AD.*

**Solution:**

Given that,

In parallelogram ABCD, CD = AB = 16 cm [âˆµ Opposite side of a parallelogramare equal]

We know that,

Area of parallelogram = Base Ã— Corresponding altitude

Area of parallelogram ABCD = CD Ã— AE = AD Ã— CF

16 cm Ã— cm = AD Ã— 10 cm

cm = 12.8 cm

Thus, The length of AD is 12.8 cm.

*Q 2. In Q 1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, Find AB.*

**Solution:**

We know that,

Area of a parallelogram ABCD = AD Ã— CF â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

Again area of parallelogram ABCD = CD Ã— AEâ‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

Compare equation(1) and equation(2)

AD Ã— CF = CD Ã— AE

â‡’6Ã—10 = DÃ—8

â‡’D = = 7.5 cm

âˆ´ AB = DC = 7.5cm [âˆµOpposite side of a parallelogramare equal]

*Q 3. Let ABCD be a parallelogram of area 124 cm ^{2 }. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.*

**Solution:**

Given,

Area of a parallelogram ABCD = 124 cm^{2}

Construction: Draw APâŠ¥DC

Proof:-

Area of a parallelogram AFED = DF Ã— AP â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

And area of parallelogram EBCF = FC Ã— APâ‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

And DF = FC â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(3) [FisthemidpointofDC]

Compare equation (1), (2) and (3)

Area of parallelogram AEFD = Area of parallelogram EBCF

âˆ´ Area of parallelogram AEFD = = 62cm^{2}

*Q 4. If ABCD is a parallelogram, then prove that*

*Ar (Î”ABD) = Ar (Î”BCD) = Ar (Î”ABC) = Ar (Î”ACD) = **Ar(// ^{gm}ABCD).*

**Solution:**

Given:-

ABCD is a parallelogram,

**To prove :** â€“ Ar (Î”ABD) = Ar (Î”BCD) = Ar (Î”ABC) = Ar (Î”ACD) = Ar(//^{gm}ABCD).

**Proof:-** We know that diagonal of a parallelogram divides it into two equilaterals .

Since, AC is the diagonal.

Then, Ar (Î”ABC) = Ar (Î”ACD) = Ar(//^{gm}ABCD) â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

Since, BD is the diagonal.

Then , Ar (Î”ABD) = Ar (Î”BCD) = Ar(//^{gm}ABCD) â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

Compare equation (1) and (2)

âˆ´ Ar (Î”ABC) = Ar (Î”ACD) = Ar (Î”ABD) = Ar (Î”BCD) = Ar(//^{gm}ABCD)..