Question

can anyone tell me the answers ​

Answer 1

1. Given: Length and breadth of a rectangle are 36 cm and 24 cm.

To find: The perimeter, and the area of the rectangle

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$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$

$\star \; \boxed{\pink{\sf Perimeter = 2(length + breath)}}$

Therefore,

$:\implies \sf Perimeter_{\pink{rectangle}} = 2(36+24)\\\\:\implies \sf Perimeter_{\pink{rectangle}} = 2 \times 60\\\\:\implies\underline{ \boxed{\bf Perimeter_{\pink{rectangle}} = 120}}\; \bigstar$

Now, the area of the rectangle.

$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$

$\star \; \boxed{\pink{\sf Area_{\pink{rectangle}} = Length \times Breadth}}$

Therefore,

$:\implies \sf Area _{\pink{ rectangle}} = 36 \times 24\\\\ :\implies \underline{\boxed{\bf Area_{\pink{rectangle}} = 864^2}}\;\bigstar$

$\therefore \sf{\underline{\pink{The \; perimeter \; of \; the \; rectangle \; is \; 120 cm, and \; the \;area\;of \; the \; rectangle \; is \; 864cm^2}}}$

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2. Given: The length and breadth of a rectangular field are equal to 600 m and 400 m.

To Find: The cost of the grass to be planted in it at the rate of ₹ 2.50 per m².

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$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$

$\star \; \boxed{\pink{\sf Area= Length \times Breadth}}$

Therefore,

Area of the field = 600 m × 400 m = 240000 m²

Cost of planting the grass = ₹ 2.50 × 240000 = ₹ 6,00,000

The required cost = ₹ 6,00,000 ★

$\therefore \sf{\underline{\pink{The ~cost ~of ~the ~grass ~to ~be ~planted ~in ~it ~at ~the ~rate ~of ~Rs.~ 2.50 ~per ~m^2.}}}$

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3. Given: A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park.

To Find: The area of the path.

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$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$

$\star \; \boxed{\pink{\sf Area= Length \times Breadth}}$

Therefore,

$:\implies \sf Area ~of ~park = 45 \times 30 = 1350~ cm^2\\\\:\implies \sf Area ~of ~park~with ~path~ is = 45 + 2.5 + 2.5 \times 30 + 2.5 + 2.5 = 1750 cm^2$

$\therefore \sf{\underline{\pink{The~ area ~of ~the ~path~is \; 1750 cm^2.}}}$

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4. Given: Radius is 28 cm.

To Find: The rotation of the wheel to cover a distance of 352m.

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$:\implies \sf Area ~of ~the ~wheel _{(as~ it ~is ~a ~circle)} = 2\pi r \\\\:\implies \sf 2 \times (22 \div 7) \times 28 = 176cm$

• As it must rotate 352m which is given meter's we need to convert it in cm

35200

No of time it rotates = 35200 ÷ 176 = 200 times.

$\therefore \sf{\underline{\pink {The~wheel~should ~ rotate~200~times~to ~cover~a~distance~of~352m.}}}$

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Steps of construction;  

1. Draw the base OR = 3.5 cm
2. At Q construct 40 degrees with protractor
3. Draw a line through it i.e QY
4. At R construct 60 degrees with a compass
5. Draw a line through it i.e RX
6. Join both the lines and name the point of incidence P

{Refer the attachment number 1}

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Steps of construction;  

1. Draw a line segment AB of length 4.5 cms.  
2. Take 4.5 cms as radius and A as center, draw an arc.
4. Take 4.5 cms as radius and B as center, draw and arc.  
5. Let C be the point where the two arcs intersect, join AC and BC and label the sides.

{Refer the attachment number 2}

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(a) m∠A = 80°, m∠B = 60°

m∠A + m∠B = 80° + 60° = 140° < 180°

So, ΔABC can be possible to construct.

(b) PQ = 5 cm, QR = 3 cm, PR = 8.8 cm

PQ + QR = 5 cm + 3 cm = 8 cm < 8.8 cm

or PQ + QR < PR

So, the ΔPQR can not be constructed.

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In isosceles triangle 2 angles are equal.

which means,

$\implies \sf 90^o + x + x = 180^o \\\\ :\implies \sf 90^o + 2x = 180^o \\\\ :\implies \sf 2x = 180-90 \\\\ :\implies \sf 2x = 90^o \\\\ :\implies \underline{\boxed{ \pink{\sf x = 45^o}}} \; \bigstar$

Other two angles are 45°

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Steps of construction;  

1. Draw a line segment AB = 5.8 cm
2. Using suitable compass width and taking A as center draw arcs on both sides of AB
3. Keeping compass width same and taking B as center draw arcs such that it intersects arc draw in previous step at M & N
4. Draw a line passing through M , N and intersecting AB at O
5. Using compass width = OA or OB and taking O as center , draw an arc intersecting line drawn in previous step at C
6. Join AB
7. Join AC

An isosceles right-angled triangle whose hypotenuse is 5.8 cm. is constructed

{Refer to the attachment number 3}