Question

Find the area of the following: (a) A rectangle17cm long and10cm wide​

Answer 1

Answer:

area of rectangle =l×b

here,given

length =17 cm

and breadth =10 cm

Step-by-step explanation:

area =length ×breadth

therefore,17×10=170square cm

Answer 2

Step-by-step explanation:

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

★ 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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