Question

Find the area of the following polygons:

(a) A triangle of base 3.5 cm and height 10 cm

b) A rectangle of length 5.5 cm and
breadth 4 cm​

Answer 1

❒ Given :-

Find the area of following polygons :

(a) A triangle of base 3.5 cm and height 10 cm.

(b) A rectangle of length 5.5 cm and

breadth 4 cm.

❒ To Find :-

• ➵ Area = ?

❒ Solution :-

(a.)

⇝We know that :

${\red{\bigstar{\purple{\boxed{\orange{\boxed{\bold{\bf{Area{\small_{(Triangle)}} = \frac{1}{2} \times Base \times Height}}}}}}}}}$

⇝Now the area :

${\large{:{\twoheadrightarrow{\bf{Area = \frac{1}{2} \times b \times h }}}}}$

${\large{:{\twoheadrightarrow{\bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{2} \times 3.5 \times 10 }}}}}$

${\large{:{\twoheadrightarrow{\bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{2} \times 35}}}}}$

${\large{:{\twoheadrightarrow{\bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{\cancel{2}} \times{\cancel {35}}}}}}}$

${\large{:{\twoheadrightarrow{\bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 \times 17.5}}}}}$

${\large{\purple{:{\longmapsto{\underline{\overline{\red{\boxed{\bf{17.5 \: cm²}}}}}}}}}}$

(b.)

⇝We know that :

${\red{\bigstar{\purple{\boxed{\orange{\boxed{\bold{\bf{Area{\small_{(Rectangle )}} =Length \times Breadth }}}}}}}}}$

⇝Now the area :

${\large{:{\twoheadrightarrow{\bf{Area = length \times breadth}}}}}$

${\large{:{\twoheadrightarrow{\bf{Area = 5.5 \times 4}}}}}$

${\large{\purple{:{\longmapsto{\underline{\overline{\red{\boxed{\bf{22 \: cm²}}}}}}}}}}$

❒ Hence :

${\large{\purple{\underline{\red{\underline{\pmb{\pink{\mathfrak{(a.) Area = 17.5 cm²}}}}}}}}}$

${\large{\purple{\underline{\red{\underline{\pmb{\pink{\mathfrak{(b.) Area = 22 \: cm²}}}}}}}}}$

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Answer 2

★ Question :-

⠀⠀⠀⠀⠀⠀⠀

Find the area of the given Polygons :-

1.] A Triangle of Base = 3.5 cm and

Height = 10 cm.

2.] A Rectangle of Length = 5.5 cm and

Breadth = 4 cm.

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★ Answer :-

⠀⠀⠀⠀⠀⠀⠀

1.] The Area of Polygons of Traingle

$\sf\pink{= 17.5\:cm^{2}}$

⠀⠀⠀⠀⠀⠀⠀

2.] The Area of Polygon of Rectangle

$\sf\pink{= 22 cm^{2}}$

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★ Explanation :-

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀ ⠀$\underbrace{\small \maltese \: \: \sf \pink{For\:1\: Equation :-}}\\ \\ \small\sf{A= \frac{1}{2} \times B \times H } \\ \\ \small \sf{A= \frac{1}{ 2} \times 3.5 \times 10 } \\ \\ \small \sf{A= \frac{1}{ \cancel2} \times \cancel{35 }} \\ \\ \boxed{\small \sf \red{A=1.75 \: cm {}^{2} }} \\ \\ \\ \\ \underbrace{ \small \maltese \: \: \sf \pink{For \: 2 \: Equation :-}} \\ \\ \small \sf{A= L\times B } \\ \\ \small \sf{A=5.5 \times 4 } \\ \\ \boxed{ \small \sf\red{A=22 \: cm {}^{2} }}$

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★ Formula Used :-

⠀⠀⠀⠀⠀⠀⠀

1.] $\boxed{ \small \sf \pink{Area \: of \: Traingle} \sf{ \: = \frac{1}{2} \times \: Base × Height }}$

⠀⠀⠀⠀⠀⠀⠀

2.] $\boxed{\small \sf\pink{Area \: of \: Rectangle} \sf{ \: =Lenght × Breadth }}$

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