RD Sharma Class 9 Solutions Chapter 14 Exercise 14.2 (Updated for 2023)

RD Sharma Class 9 Solutions Chapter 14 Exercise 14.2: These solutions are created by our experts. You can easily download RD Sharma Class 9 Solutions Chapter 14 Exercise 14.2 PDF. 

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Download RD Sharma Class 9 Solutions Chapter 14 Exercise 14.2

RD Sharma Class 9 Solutions Chapter 14 Exercise 14.2

Access RD Sharma Class 9 Solutions Chapter 14 Exercise 14.2

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 = AreaAltitude = 12810 = 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 cm²
Again area of ||gm ABCD = 60 cm²
Altitude AE = 8 cm
∴ Base AB =AreaAltitude = 608 = 152 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 midpoints of sides AB and CD respectively. E, and F are joined.

Draw DL ⊥ AB
Now the area of ||gm ABCD = Base x Altitude
= AB x DL = 124 cm²
∵ E and F are midpoints of sides AB and CD
∴ AEFD is a ||gm
Now the area of ||gm AEFD = AE x DL
= 12AB x DL [∵ E is the midpoint of AB]
= 12 x area of ||gm ABCD
= 12 x 124 = 62 cm²

Question 4.
If ABCD is a parallelogram, then prove that ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = 12ar( ||^gm ABCD).
Solution:
Given: In ||gm ABCD, BD, and AC are joined

To prove : ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = 12ar(||gm ABCD)
Proof: ∵ Diagonals of a parallelogram bisect it into two triangles equal in the area When BD is the diagonal, then
∴ ar(∆ABD) = ar(∆BCD) = 12ar(||^gm ABCD) …(i)
Similarly, when AC is the diagonal, then
ar(∆ABC) = ar(∆ADC) = 12ar(||^gm ABCD) …(ii)
From (i) and (ii),
ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = 12 ar(||^gm ABCD)

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