Question

Please answers me fast ​

Answer 1

A N S W E R :

Given :

(i) Length = 5 m and Breadth = 4 m

(ii) Length = 14 cm and Breadth = 10 cm

(iii) Breadth = 9 m and Perimeter = 50 m

(iv) Breadth = 25 cm and Area = 750 cm²

To Find :

(i) Area and Perimeter

(ii) Area and Perimeter

(iii) Length and Area

(iv) Length and Perimeter

Solution :

(i) Calculating area of rectangle :

→ Area of rectangle = Length × Breadth

→ Area of rectangle = 5 × 4

→ Area of rectangle = 20 m²

Calculating Perimeter of rectangle :

→ Perimeter of rectangle = 2(Length + Breadth)

→ Perimeter of rectangle = 2(5 + 4)

→ Perimeter of rectangle = 2(9)

→ Perimeter of rectangle = 18 m

Hence,

• Area of rectangle = 20 m²
• Perimeter of rectangle = 18 m

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(ii) Calculating area of rectangle :

→ Area of rectangle = Length × Breadth

→ Area of rectangle = 14 × 10

→ Area of rectangle = 140 cm²

Calculating Perimeter of rectangle :

→ Perimeter of rectangle = 2(Length + Breadth)

→ Perimeter of rectangle = 2(14 + 10)

→ Perimeter of rectangle = 2(24)

→ Perimeter of rectangle = 48 cm

Hence,

• Area of rectangle = 140 cm²
• Perimeter of rectangle = 48 cm

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(iii) Let Length be 'l' m.

→ Perimeter of rectangle = 2(Length + Breadth)

→ 50 = 2(l + 9)

→ 50 = 2l + 18

→ 50 - 18 = 2l

→ 32 = 2l

→ l = 32 ÷ 2

→ l = 16 m

Calculating area of rectangle :

→ Area of rectangle = Length × Breadth

→ Area of rectangle = 16 × 9

→ Area of rectangle = 144 m²

Hence,

• Length of rectangle = 16 m
• Area of rectangle = 144 m²

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(iv) Calculating Length of rectangle :

→ Area of rectangle = Length × Breadth

→ 750 = Length × 25

→ Length = 750 ÷ 25

→ Length = 30 cm

Calculating Perimeter of rectangle :

→ Perimeter of rectangle = 2(Length + Breadth)

→ Perimeter of rectangle = 2(30 + 25)

→ Perimeter of rectangle = 2(55)

→ Perimeter of rectangle = 110 cm

Hence,

• Length of rectangle = 30 cm
• Perimeter of rectangle = 110 cm

Answer 2

Answer:-

(i) • Area is 20m²

• perimeter is 18 m

(ii) • Area is 140 cm²

• Perimeter is 48 cm

(iii) • Length is 16 m

• Area is 144 m²

(iv) • Length is 30 cm

• Perimeter is 110 cm

Solution:-

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(i) Given:-

• Length = 5 m
• Breadth = 4 m

To find:-

• Area and perimeter

$\green{ \underline{ \large{ \pink{ \mathfrak{ area = l \times b}}}}}$

$\large{ \tt : \implies \: \: \: \: area = 5 \times 4}$

$\large{ \tt : \implies \: \: \: \: area = 20 {m}^{2} }$

$\green{ \underline{ \large{ \pink{ \mathfrak{ \: perimeter = 2(l + b)}}}}}$

$\large{ \tt : \implies \: \: \: \: perimeter =2(5 + 4) }$

$\large{ \tt : \implies \: \: \: \: perimeter =2 \times 9 }$

$\large{ \tt : \implies \: \: \: \: perimeter =18 \: m }$

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(ii) Given :-

• Length = 14 cm
• Breadth = 10 cm

To find :-

• Area and perimeter

$\blue{ \underline{ \large{ \green{ \mathfrak{ area = l \times b}}}}}$

$\large{ \tt : \implies \: \: \: \: area = 14 \times 10}$

$\large{ \tt : \implies \: \: \: \: area =14</strong><strong>0</strong><strong> \: {cm}^{2} }$

$\blue{ \underline{ \large{ \green{ \mathfrak{ \: perimeter = 2(l + b)}}}}}$

$\large{ \tt : \implies \: \: \: \: perimeter =2(14 + 10) }$

$\large{ \tt : \implies \: \: \: \: perimeter =2 \times 24 }$

$\large{ \tt : \implies \: \: \: \: perimeter =48 \: cm }$

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(iii) Given:-

• Breadth = 9 m
• Perimeter = 50 m

To find :-

• Length and area

$\green{ \underline{ \large{ \red{ \mathfrak{ \: perimeter = 2(l + b)}}}}}$

$\large{ \tt : \implies \: \: \: \: 2(l + 9) = 50 }$

$\large{ \tt : \implies \: \: \: \:l + 9 = 25}$

$\large{ \tt : \implies \: \: \: \:l = 25 - 9}$

$\large{ \tt : \implies \: \: \: \:l = 16 \: m}$

$\green{ \underline{ \large{ \red{ \mathfrak{ area = l \times b}}}}}$

$\large{ \tt : \implies \: \: \: \: area = 16 \times 9}$

$\large{ \tt : \implies \: \: \: \: area = 14 4 \: {m}^{2} }$

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(iv) Given:-

• Breadth = 25 cm
• Area = 750 cm²

To find:-

• Length and perimeter

$\pink{ \underline{ \large{ \blue{ \mathfrak{ area = l \times b}}}}}$

$\large{ \tt : \implies \: \: \: \: l \times 25 = 750}$

$\large{ \tt : \implies \: \: \: \: l = \frac{750}{25} }$

$\large{ \tt : \implies \: \: \: \: l = 30 \: cm}$

$\pink{ \underline{ \large{ \blue{ \mathfrak{ \: perimeter = 2(l + b)}}}}}$

$\large{ \tt : \implies \: \: \: \: perimeter =2(30 + 25) }$

$\large{ \tt : \implies \: \: \: \: perimeter =2 \times 55 }$

$\large{ \tt : \implies \: \: \: \: perimeter =110 \: cm }$

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