Question

Find the area of the rectangle of length 17 cm and breadth 13 cm. But i don't ask so that i can see if u are correct. Am just asking to see the answer coz i don't know

Answer 1

Answer:

Area of the rectangle = 221 cm²

Step-by-step explanation:

Given:

• Length of the rectangle = 17 cm
• Breadth of the rectangle = 13 cm

To find:

• Area of the rectangle.

Solution:

• Length (l) = 17 cm
• Breadth (b) = 13 cm

We know that,

${\boxed{\sf{Area\:of\: rectangle=Length\times\: Breadth}}}$

Area of the rectangle,

= length × breadth

=( 17×13) cm²

= 221 cm²

Therefore, the area of the rectangle is 221 cm².

_____________________

More info:

• Perimeter of rectangle = 2(length× breadth)
• Diagonal of rectangle = √(length ²+breadth ²)
• Area of square = side²
• Perimeter of square = 4× side
• Volume of cylinder = πr²h
• T.S.A of cylinder = 2πrh + 2πr²
• Volume of cone = ⅓ πr²h
• C.S.A of cone = πrl
• T.S.A of cone = πrl + πr²
• Volume of cuboid = l × b × h
• C.S.A of cuboid = 2(l + b)h
• T.S.A of cuboid = 2(lb + bh + lh)
• C.S.A of cube = 4a²
• T.S.A of cube = 6a²
• Volume of cube = a³
• Volume of sphere = 4/3πr³
• Surface area of sphere = 4πr²
• Volume of hemisphere = ⅔ πr³
• C.S.A of hemisphere = 2πr²
• T.S.A of hemisphere = 3πr²

Answer 2

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

★ Length of rectangle = 17 cm

★ Breadth of rectangle = 13 cm

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

★ Area of rectangle.

${\large{\bold{\rm{Solution}}}}$

★ Area of rectangle =

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

★ Area of rectangle formula.

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

★ A = L × B

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

★ A denotes area

★ L denotes length

★ B denotes breadth

${\large{\bold{\rm{Full \; solution}}}}$

★ A = L × B

★ A = 17 × 13

★ A = 221 cm²

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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