Rd sharma solution 2019 class 9 chapter 14 Areas of paralllelograms and triangles

Exercise 14.1    Exercise 14.2

Exercise 14.2


QUESTION 1 

lf the given f‌igure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm. AE = 8 cm and CF =10 cm, f‌ind AD.

Sol :

Given: Here in the question it is given
(1) ABCD is a parallelogram,

(2) AE ⊥ DC and

(3) CF ⊥ AD , AB = 16 cm

(4) AE = 8 cm

To find : AD = ?

Calculation : We know that formula for calculating the

Area of parallelogram = base × height

Therefore,

Area of paralleogram ABCD = DC × AB (Taking base as DC and Height as AE)

Area of parallelogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)

Therefore ,

Area of paralleogram ABCD =16 × 8 …(l)

Taking the base of Parallelogram ABCD as AD we get

Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)

Area of paralleogram ABCD = AD × 10 …(2)

Since equation (1) and (2) both represent the Area of the same Parallelogram ABCD , both should be equal.

Hence from equation (1) and (2)

This means that ,

16 × 8 = AD × 10

\text{AD}=\dfrac{16\times 8}{10}

AD = 12.8 cm

Hence , we get the result as AD = 12.8 cm


QUESTION 2

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

Sol :

Given : Here in the question it is given that

(1) ABCD is a parallelogram,

(2) AE ⊥ DC and

(3) CF ⊥ AD

(4) AD = 6 cm

(5) AB = 8 cm

(6) CF = 10 cm

To find : AB = ?

Calculation :  We know that formula for calculating the

Area of paralleogram = base × height

Therefore,

Area of paralleogram ABCD = DC × AB (taking base as DC and Height as AE)

Area of parallelogram ABCD = AB × AE (AB = DC as opposite side o f the parallelogram are equal)

Therefore, Area of paralleograrm ABCD = 16 × 8

Area of Parallelogram ABCD = AB × 8 …(l)

Taking the base of Parallelogram ABCD as AD we get

Area of paralleogram  ABCD = AD × CF (taking base as AD and height as CF)
Area of paralleogram ABCD = 6 × 10 …(2)

Since equation 1 and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.

Hence equation (1) is equal to equation (2)

Which means that ,

AB × 8 = 6 × 10

AB = \dfrac{6 \times 10}{8}

AD = 7.5 cm

Hence we got the measure of AB equal to 7.5 cm

 


QUESTION 3

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

Sol :

Given : Here in the equation it is given that

(1) Area of paralleogram ABCD = 124 cm2

(2) E is the mid-point of AB, which means AE =\dfrac{1}{2}(AB)

(3) F is the midpoint of CD, which means DF =\dfrac{1}{2}(CD)

To find : Area of parallleogram AEFD

Calculation : We know that formula for calculating the

Area of parallelogram = base × height

Therefore ,

Area of paralleogram ABCD = AB × AD (Taking base as AB and Height as AD) …(1)

Area of paralleogram ABFD = AE × AD (Taking base as AB and Height as AD) …(2)

=\dfrac{1}{2}AB \times AD \left(AE=\dfrac{1}{2}\right)

=\dfrac{1}{2}\times 124

= 62 cm2

here we got the result Area of parallelogram AEFD = 62 cm2


QUESTION 4

If ABCD is a parallelogram, then prove that

area(ΔABD) = area(ΔBCD) = area(ΔABC) = area(ΔACD) =\dfrac{1}{2}  \text{parallelogram ABCD}

Sol :

Given : Here in the question it is given that

(1) ABCD is parallelogram

To prove :

(1) Area of ΔADC = \dfrac{1}{2}\left(\text{area of parallelogram ABCD}\right)

(2) Area of ΔBCD = \dfrac{1}{2}\left(\text{area of parallelogram ABCD}\right)

(3) Area of ΔABC = \dfrac{1}{2}\left(\text{area of parallelogram ABCD}\right)

(4) Area of ΔABD = \dfrac{1}{2}\left(\text{area of parallelogram ABCD}\right)

Construction : Draw AE ⊥ CD

Calculation : We know that formula for calculating the

Area of parallelogram =  base × height

Area of paralleogram ABCD = BC × AE (Taking base as BC and Height as AE ) …(l)

We know that formula for calculating the

Area of triangle =\dfrac{1}{2} \left(\text{Base}\times \text{Height}\right)

Area of ΔADC =\dfrac{1}{2} \left(\text{Base}\times \text{Height}\right)

=\dfrac{1}{2} \left(\text{AD}\times \text{AE}\right) (AD is the base of ΔADC and AE is the height of ΔADC)

=\dfrac{1}{2}\text{Area of parallelogram ABCD } [From equation-1]

Area of ΔADC =\dfrac{1}{2}\left(\text{area of parallelogram ABCD}\right)

Similarly we can show that

(2) Area of ΔBCD=\dfrac{1}{2}\left(\text{area of parallelogram ABCD}\right)

(3) Area of ΔABC =\dfrac{1}{2}\left(\text{area of parallelogram ABCD}\right)

(4) Area of ΔABD =\dfrac{1}{2}\left(\text{area of parallelogram ABCD}\right)


 

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