# RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

## 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

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 cm2
Now area of ||gm ABCD = 128 cm2
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
= 6 x 10 = 60 cm2
Again area of ||gm ABCD = 60 cm2
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 cm2
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 cm2
∵ 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 cm2

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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