## RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

Other Exercises

- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1
- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2
- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS
- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Question 1.

In the figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. [NCERT]

Solution:

In ||gm ABCD,

Base AB = 16 cm

and altitude AE = 8 cm

∴ Area = Base x Altitude

= AB x AE

= 16 x 8 = 128 cm^{2}

Now area of ||gm ABCD = 128 cm^{2}

Altitude CF = 10 cm

∴ Base AD = \(\frac { Area }{ Altitude }\) = \(\frac { 128 }{ 10 }\) = 12.8cm

Question 2.

In Q. No. 1, if AD = 6 cm, CF = 10 cm, AE = 8 cm, find AB.

Solution:

Area of ||gm ABCD,

= Base x Altitude

= AD x CF

= 6 x 10 = 60 cm^{2}

Again area of ||gm ABCD = 60 cm^{2}

Altitude AE = 8 cm

∴ Base AB =\(\frac { Area }{ Altitude }\) = \(\frac { 60 }{ 8 }\) = \(\frac { 15 }{ 2 }\) cm = 7.5 cm

Question 3.

Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.

Solution:

Area of ||gm ABCD = 124 cm^{2}

E and F are the mid points of sides AB and CD respectively. E, F are joined.

Draw DL ⊥ AB

Now area of ||gm ABCD = Base x Altitude

= AB x DL = 124 cm^{2}

∵ E and F are mid points of sides AB and CD

∴ AEFD is a ||gm

Now area of ||gm AEFD = AE x DL

= \(\frac { 1 }{ 2 }\)AB x DL [∵ E is mid point of AB]

= \(\frac { 1 }{ 2 }\) x area of ||gm ABCD

= \(\frac { 1 }{ 2 }\) x 124 = 62 cm^{2}

Question 4.

If ABCD is a parallelogram, then prove that ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar( ||^{gm} ABCD).

Solution:

Given : In ||gm ABCD, BD and AC are joined

To prove : ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar(||gm ABCD)

Proof: ∵ Diagonals of a parallelogram bisect it into two triangles equal in area When BD is the diagonal, then

∴ ar(∆ABD) = ar(∆BCD) = \(\frac { 1 }{ 2 }\)ar(||^{gm} ABCD) …(i)

Similarly, when AC is the diagonal, then

ar(∆ABC) = ar(∆ADC) = \(\frac { 1 }{ 2 }\)ar(||^{gm} ABCD) …(ii)

From (i) and (ii),

ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\) ar(||^{gm} ABCD)

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 are helpful to complete your math homework.

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