# RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

In this chapter, we provide RD Sharma Solutions for Chapter 14 Quadrilaterals Ex 14.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 14 Quadrilaterals Ex 14.2 Maths pdf, free RD Sharma Solutions for Chapter 14 Quadrilaterals Ex 14.2 Maths book pdf download. Now you will get step by step solution to each question.

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
= 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 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
= 12AB x DL [∵ E is mid point 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 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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