Question

I need these answer's please ​

Answer 1

$\huge{\color{blue}{\underline{\underline{Solution:-}}}}$

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$\large 1.$

Circumference of a circle = 2πr= 88

$\bf =2 × (\dfrac{22}{7}) × \dfrac{r (88 × 7)}{(2 × 22) }$

$\bf {\boxed{ r = 14 cm}}$

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$\large 2.$

$\bigstar$Given :-

Perimeter of the rectangle = 120 cm

Length of the rectangle =25 cm

$\bigstar$ To Find :-

Breadth and area of the rectangle

$\bigstar$ Formula required :-

Breadth of the rectangle :-

$\tt \dashrightarrow \large{ \orange{ \boxed{ \tt\gray{ \dfrac{Perimeter}{2} - Length }}}}$

Area of the rectangle :-

$\tt \dashrightarrow \large{ \orange{ \boxed{ \tt \gray{Length \times Breadth}}}}$

Breadth :-

$\tt \longrightarrow \dfrac{120}{2} - 25$

$\tt \longrightarrow \cancel \dfrac{120}{2} - 25 = \boxed{ \tt60 - 25}$

$\tt \longrightarrow35 \sf \: cm$

Area of the rectangle :-

$\tt \longrightarrow 35 \times 25$

$\tt \longrightarrow875 \: { \sf{cm}}^{2}$

$\therefore$The Breadth of the rectanglular sheet is 35 cm and area of the rectanglular sheet is 875 cm².

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$\large 3.$

⇒ AB=b=4.2 and height=h

⇒ Area of parallelogram =b×h.

⇒ 36=9×h.

⇒ h=$\bf \dfrac{36}{9}$

⇒ h=4cm.

∴ Height of parallelogram is 4cm.

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$\large 4.$

$\blue\bigstar$Base of traingle = 14 cm

$\blue\bigstar$ Height of triangle = 20 cm

$\blue\bigstar Area \:of \:triangle = \dfrac{1}{2}×b×h$

= $\bf \dfrac{1}{2} ×14×20$

= 14 ×10

= $\bf 140cm^2$

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$\blue\bigstar$ Hope it helps u mate

Answer 2

\huge{\color{blue}{\underline{\underline{Solution:-}}}}

Solution:−

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\large 1.1.

Circumference of a circle = 2πr= 88

\bf =2 × (\dfrac{22}{7}) × \dfrac{r (88 × 7)}{(2 × 22) }=2×(

7

22

)×

(2×22)

r(88×7)

\bf {\boxed{ r = 14 cm}}

r=14cm

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\large 2.2.

\bigstar★ Given :-

Perimeter of the rectangle = 120 cm

Length of the rectangle =25 cm

\bigstar★ To Find :-

Breadth and area of the rectangle

\bigstar★ Formula required :-

Breadth of the rectangle :-

\tt \dashrightarrow \large{ \orange{ \boxed{ \tt\gray{ \dfrac{Perimeter}{2} - Length }}}}⇢

2

Perimeter

−Length

Area of the rectangle :-

\tt \dashrightarrow \large{ \orange{ \boxed{ \tt \gray{Length \times Breadth}}}}⇢

Length×Breadth

Breadth :-

\tt \longrightarrow \dfrac{120}{2} - 25⟶

2

120

−25

\tt \longrightarrow \cancel \dfrac{120}{2} - 25 = \boxed{ \tt60 - 25}⟶

2

120

−25=

60−25

\tt \longrightarrow35 \sf \: cm⟶35cm

Area of the rectangle :-

\tt \longrightarrow 35 \times 25⟶35×25

\tt \longrightarrow875 \: { \sf{cm}}^{2}⟶875cm

2

\therefore∴ The Breadth of the rectanglular sheet is 35 cm and area of the rectanglular sheet is 875 cm².

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\large 3.3.

⇒ AB=b=4.2 and height=h

⇒ Area of parallelogram =b×h.

⇒ 36=9×h.

⇒ h=\bf \dfrac{36}{9}

9

36

⇒ h=4cm.

∴ Height of parallelogram is 4cm.

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\large 4.4.

\blue\bigstar★ Base of traingle = 14 cm

\blue\bigstar★ Height of triangle = 20 cm

\blue\bigstar Area \:of \:triangle = \dfrac{1}{2}×b×h★Areaoftriangle=

2

1

×b×h

= \bf \dfrac{1}{2} ×14×20

2

1

×14×20

= 14 ×10

= \bf 140cm^2140cm

2

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\blue\bigstar★ Hope it helps u mate