Question

12. In Figure 6, ABCD is a parallelogram,
where CE I AB produced and BF I AD.
1 IFAB = 20 cm, AD = 16 cm and CE = 10 cm, then find BF.
it. IfAB= 30 cm, CE=14 cm and BF= 20 cm, then find AD.​

Answer 1

At first,

Area of parallelogram = 1/2 × base × altitude

Area = 1/2 × AB × CE

Area = 1/2 × 30 × 14

★Area = 210 {cm}^{2}Area=210cm

2

★

Secondly,

Area of parallelogram= 1/2 × base × altitude

Area = 1/2 × AD × BF

Area = 1/2 × AD × 20

★Area = 10AD{cm}^{2}Area=10ADcm

2

★

Equaling both we get :

10AD{cm}^{2}=210{cm}^{2}10ADcm

2

=210cm

2

Or, 10AD = 210

Or, AD = 210/10 cm

★ AD = 21 cm★

\large\boxed{AD=21\:cm}

AD=21cm

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Answer 2

Answer:

1) $\large\boxed{BF=12.5\:cm}$

2) $\large\boxed{AD=21\:cm}$

Step-by-step explanation:

1) AB = 20cm, AD = 16cm, CE = 10cm, BF = ?

⇒   Area of parallelogram = 1/2 × base × altitude

⇒   Area = 1/2 × AB × CE

⇒   Area = 1/2 × 20cm × 10cm

⇒   Area = 100${cm}^{2}$

We can also write it as :-

⇒   Area of parallelogram= 1/2 × base × altitude

⇒   Area = 1/2 × AD × BF

⇒   Area = 1/2 × 16cm × BF

⇒   Area = 8BF cm

Equalize both of them :-

⇒   8BF cm = 100${cm}^{2}$

⇒   8BF = 100cm

⇒   BF = 12.5cm

$\large\boxed{BF=12.5\:cm}$

2) AB = 30cm, CE = 14cm, BF = 20cm, AD = ?

⇒   Area of parallelogram = 1/2 × base × altitude

⇒   Area = 1/2 × AB × CE

⇒   Area = 1/2 × 30cm × 14cm

⇒   Area = 210${cm}^{2}$

We can also write it as :-

⇒   Area of parallelogram= 1/2 × base × altitude

⇒   Area = 1/2 × AD × BF

⇒   Area = 1/2 × AD × 20cm

⇒   Area = 10AD cm

Equalize both of them

⇒ 10AD cm = 210${cm}^{2}$

⇒ 10AD = 210cm

⇒ AD = $\frac{210}{10} cm$

⇒ AD = 21 cm

$\large\boxed{AD=21\:cm}$

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