Question

Long Answer Type Questions :-

1. A wire when bent in the form of a square encloses an area of 484 cm². If the same wire is bent in the form of a circle, find the area of the circle.

2. Two crossroads, each of width 4 m, runs at right angles through the centre of a rectangular park of length 70 m and breadth 45 m and parallel to the sides of the rectangle. Find the area of the road.

3. The circumference of two circles are in the ratio 8 : 6. Find the ratio of their areas.

4. A rectangular piece of a canvas measures 35 m by 18 m. A triangular piece with base 18 m and height 12 m is cut from the canvas. Find the area of the remaining piece.

5. A mirror of length 30 cm and width 25 cm is to be framed with a 4 cm wide frame. Find the area of the frame required. Also find the cost of framing at the rate of ₹ 25 per cm².

6. In the attachment.​

Answer 1

$\huge \sf{ \underline{ \underline{Question 1.}}}$

A wire when bent in the form of a square encloses an area of 484 cm². If the same wire is bent in the form of a circle, find the area of the circle.

$\huge \sf{ \underline{ \underline{Solution:}}}$

Area of the square=484cm²

$\sf \therefore \: Side \: of \: the \: square = \sqrt{484} = 22cm$

$\sf \implies \: Perimeter \: of \: the \: square = 4 \times 22cm$

$\sf = 88cm$

$\because \sf \: (Circumference \: of \: the \: circle )= (perimeter \: of \: square)$

$\sf \implies 2\pi \: r = 88$

$\sf \implies2 \times \frac{22}{7} \times r = 88$

$\sf \therefore \: r = \frac{88 \times 7}{2 \times 22} = 14cm$

$\sf \therefore \: Area \: of \: the \: circle = \pi {r}^{2} = \frac{22}{7} \times 14 {cm}^{2}$

$\sf = 616 {cm}^{2}$

$\huge \sf{ \underline{ \underline{Question 2.}}}$

Two crossroads, each of width 4 m, runs at right angles through the centre of a rectangular park of length 70 m and breadth 45 m and parallel to the sides of the rectangle. Find the area of the road.

$\huge \sf{ \underline{ \underline{Solution}}}$

• width =5 m
• length of first parallel road 70 m

$\sf \: area \: of \: first \: road 70×5=350 sq.m$

$\sf \: length \: of \: second \: road  \: 45 m$

$\sf \: area \: of \: second \: road =45×5=225 sq.m$

$\sf \: area \: of \: the \: common \: part \: of \: cross \: road$

$\sf \: with \:  5 m \:  width \: that  \: lies \: at \: the  \: center \\ \sf \: of \: the \: park \:  5×5 = 225 sq.m$

$\sf \: area \: of \: roads = area \: of \: first \: road  \\ \sf \: +area \: of \: second \: cross \: road−common \: area$

$\sf = (350 + 225) - 25$

$\sf = 575 - 25$

$\sf = 550 \: sq.m$

$\sf \: cost \: of \: the \: construction \: per \:  sq.m =Rs.105$

$\sf \: total \: cost =550×105=Rs 57750$

$\huge \sf{ \underline{ \underline{Question 3.}}}$

The circumference of two circles are in the ratio 8 : 6. Find the ratio of their areas.

$\huge \sf{ \underline{ \underline{Solution}}}$

• Let the radii of two circles be r1 and r2.

Then, according to question, we have.

$\sf = \frac{2\pi \: r1}{2\pi \: r2} = \frac{2}{3}$

$\sf = \frac{r1}{r2} = \frac{2}{3}$

Squaring both the sides, we get,

$\sf \implies\frac{ {r}^{2} 1}{ {r}^{2} 2} = \frac{4}{9}$

$\sf \: So, the \: ratio \: of \: the \: areas \: of \: two \\ \sf circles \: is 4:9$

$\huge \sf{ \underline{ \underline{Question 4.}}}$

A rectangular piece of a canvas measures 35 m by 18 m. A triangular piece with base 18 m and height 12 m is cut from the canvas. Find the area of the remaining piece.

$\huge \sf{ \underline{ \underline{Solution:- }}}$

Let's take out the area of rectangle and triangle.

Area of rectangle = Length × Breadth

=> area of rectangle = 25 × 16 = 400 cm²

And area of triangle = 1/2 × base × height

=> 1/2 × 14 × 12

= 7 × 12

= 84 cm²

We will get the area of remaining portion!

So, area of remaining portion = area of rectangle - area of triangle

=> 400 - 84

= 316 cm²

$\huge \sf{ \underline{ \underline{Question 5.}}}$

A mirror of length 30 cm and width 25 cm is to be framed with a 4 cm wide frame. Find the area of the frame required. Also find the cost of framing at the rate of ₹ 25 per cm².

$\huge \sf{ \underline{ \underline{Solution}}}$

• Cost of framing is Rs. 6840.

• Since we have given that
• Length of mirror frame would be

20+2+2=24 cm

• Width of mirror frame would be

15+2+2=19

$\sf \: Area \: of \: frame \: would \: be = length × breadth$

$\sf = 24 × 19$

$\sf= 456 cm²$

Cost per sq. Cm = Rs. 15

Total cost would be

$\sf456×15 = Rs. 6840$

Hence, cost of framing is Rs. 6840.

$\huge \sf{ \underline{ \underline{Question 6.}}}$

Find the area of the quadrilateral

ABCD. Here AC = 32 cm, BM = 8 cm,

DL=6 cm and BM _|_ AC and DL _|_ AC.

$\sf \huge{ \underline{ \underline{ Solution}}}$

Area of quadrilateral = ar (∆ ABD ) + ar (∆ ADC )  

$\sf \: = \frac{1}{2} × AC× BM + \frac{1}{2} × AC × DAN$

$\sf= \frac{1}{2} × 22 × 3 + \frac{1}{2} × 22 × 3$

=11 × 3 + 11 × 3

= 33 + 33

= 66

$\sf \therefore \: AREA \: OF \: QUADRILATERAL \: ABCD \: IS 66 cm² .$

Answer 2

Solution :

1. A wire when bent in the form of a square encloses an area of 484 cm² .

Let the side of the wire forming the square be x cm .

Thus , area of the square becomes :

=> x² = 484

=> x = 22 cm .

Perimeter of the square :

=> 4x

=> 88 cm .

Thus the entire length of the wire becomes 88 cm .

It is now bent in the form of a circle .

Let us assume that the circle has a radius r .

Circumference :

2 π r = 88 cm

2× 22/7 r = 88

=> r = 14 cm

Area = π r²

=> 22/7 × 14 × 14

=> 44 × 14

=> 616 cm² .

2. The rectangle has a length of 70 m and a breadth of 45 m .

The width of the roads is 4 m .

Area of the road :

=> 70 × 4 + 45 × 4 - 16

=> 115 × 4 - 16

=> 444 m²

3. The circumference of two circles are in the ratio of 8 : 6.

Let the radius be R1 and R2 respectively.

2 π R1/ 2 π R2 = 8 : 6

=> R1 ; R2 = 8 : 6

Thus , the ratio of the areas becomes 64 : 36

=> 16 : 9

4. The rectangular canvas measures 35 m × 18 m .

Area : 630 m² .

Area of the triangular piece :

=> ½ × 18 × 12

=> 9 × 12

=> 108 m² .

Area of the remaining piece :

=> 630 - 108

=> 522 m² .

5. Area of the frame :

=> [ 38 × 33 ] - [ 30 × 25 ]

=> 504 m².

Cost :

=> 504 × 25

=> Rs. 12600

6. Area of the quadrilateral :

=> ½ × [ 6 + 8 ] × 32

=> 7 × 32

=> 224 cm² .

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