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 ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm, and CF = 10 cm, Find AD.*

**Solution:**

Given that,

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

We know that,

Area of parallelogram = Base × Corresponding altitude

Area of parallelogram ABCD = CD × AE = AD × CF

16 cm × cm = AD × 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 × CF ⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

Again area of parallelogram ABCD = CD × AE⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

Compare equation(1) and equation(2)

AD × CF = CD × AE

⇒6×10 = D×8

⇒D = = 7.5 cm

∴ AB = DC = 7.5cm [∵Opposite 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 AP⊥DC

Proof:-

Area of a parallelogram AFED = DF × AP ⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

And area of parallelogram EBCF = FC × AP⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

And DF = FC ⋅⋅⋅⋅⋅⋅⋅⋅⋅(3) [FisthemidpointofDC]

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

Area of parallelogram AEFD = Area of parallelogram EBCF

∴ Area of parallelogram AEFD = = 62cm^{2}

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

*Ar (ΔABD) = Ar (ΔBCD) = Ar (ΔABC) = Ar (ΔACD) = **Ar(// ^{gm}ABCD).*

**Solution:**

Given:-

ABCD is a parallelogram,

**To prove :** – Ar (ΔABD) = Ar (ΔBCD) = Ar (ΔABC) = Ar (ΔACD) = Ar(//^{gm}ABCD).

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

Since, AC is the diagonal.

Then, Ar (ΔABC) = Ar (ΔACD) = Ar(//^{gm}ABCD) ⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

Since, BD is the diagonal.

Then , Ar (ΔABD) = Ar (ΔBCD) = Ar(//^{gm}ABCD) ⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

Compare equation (1) and (2)

∴ Ar (ΔABC) = Ar (ΔACD) = Ar (ΔABD) = Ar (ΔBCD) = Ar(//^{gm}ABCD)..

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