Question

2. The perimeter of a triangle is 25 cm. If the longest side is double the shortest side
side is 5 more than the shortest side find the sides of the triangle
13. The perimeter of a rectangular garden is 430 m. If the ratio of length and breadth is 3:5. find
the dimensions of the garden
14. The length of a rectangle is 1 com more than the breadth. If the length is increased by 3 cm and
breadth by 2 cm, the area is increased by 38 cm. Find the original dimensions of the rectangle
15. The length of a rectangle is 5 cm more than the breadth. If the length is decreased by 1 cm and
breadth is increased by 2 com, their ratio becomes 7. 6. Find the dimensions of the rectangle.
16. Aditya is now 10 years older than Vijay Five years ago, Aditya was twice as old as Vijay. Find
their present ages
17. Aman is 7 years younger than Saurabh After 20 years, Aman's age will double the present age
of Saurabh. Find their present ages
18. The ages of Ranu and Anu are is the ratio 3:4. Twelve years hence, the ratio of their ages will
be 5: 6. Find their present ages
19. The ages of Priyanka and Sameer are in the ratio 13 : 12. Eight years later, the ratio of their
ages will become 17:16. Find their present ages
20. John and Abhishek start from the two stations A and B respectively, 100 km away from each
other. The speed of John is 10 km/hr more than the speed of Abhishek. If they meet after
hours, find their speeds
od third part​

Answer 1

Answer:

tkztkluxlyzlyslsylyldpd uldludlysylsylzktrktltzyzuxxx

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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