Math, asked by saritajoshi7260, 10 months ago

Sell Practice 14C
Find the area of the following.
(a) A rectangle 17 cm long and 10 cm wide.
(b) A wall 5 m high and 8 m long.
(c) A field 200 m long and 120 m wide.
(d) A square board of side 1.5 m.
(e) A square lawn of side 6.3 km.
Find the breadth and perimeter of the rectangles.
(a) Area = 90 sq cm; length = 15 cm
Find the length and perimeter of the rectangle.
(a) Area = 42 sq cm; breadth = 6 cm
(b) Area = 117 sq cm; length = 13 cm
(b) Area = 480 sq cm; breadth = 20 cm​

Answers

Answered by XxMissCutiepiexX
17

{\large{\bold{\rm{\red{Given \; that}}}}}

★ Length of rectangle = 17 cm

★ Breadth of rectangle = 13 cm

{\large{\bold{\rm{\green{To \; find}}}}}

★ Area of rectangle.

{\large{\bold{\rm{\orange{Solution}}}}}

★ Area of rectangle =

{\large{\bold{\rm{\pink{Using \; concept}}}}}

★ Area of rectangle formula.

{\large{\bold{\rm{\purple{Using \; formula}}}}}

★ A = L × B

{\large{\bold{\rm{Where,}}}}

★ A denotes area

★ L denotes length

★ B denotes breadth

{\large{\bold{\rm{\color{lime}{Full \; solution}}}}}

★ A = L × B

★ A = 17 × 13

★ A = 221 cm²

{\frak{Henceforth, \: 221 \: cm^{2}}}

{\Large{\bf{\underbrace{\color{lightgreen}{Additional \; information}}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: cylinder \: = \: \sqrt \dfrac{v}{\pi h}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}.

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