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Question 1: The sides of a right-angled triangle containing the right angle are 4x \ cm and (2x-1) \ cm . if the area of the triangle is 30cm^2      calculate the lengths of its sides.

Answer:

Area = \frac{1}{2} \times base \times height 

30 = \frac{1}{2} \times 4x \times (2x-1) 

60 = 8x^2-4x 

8x^2-4x-60 = 0 \Rightarrow x = 3 cm \ or \ -2.5 cm \ (not \ possible)

Therefore \ the \ sides \ are \ 12 \ cm, \ 5 \ cm \ and \ 13 \ cm.

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Question 2: The hypotenuse of a right-angled triangle is 26 cm and the sum of other two sides is 34 cm. find the lengths of its sides.

 Answer:

Let one side be x. 

Therefore the other side would be (34-x). 

Using Pythagoras theorem

x^2 + (34-x)^2 = 26^2 

x^2 + 1156+x^2-68x = 676 

2x^2-68x+480 = 0 \Rightarrow x = 24\ cm \ or\  10\ cm 

Therefore the sides are 10 \ cm \ and \ 24 \ cm.  

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Question 3: The sides of a right-angled triangle are (x-1) \ cm, \ 3x \ cm \ and \ (3x+1) \ cm . Find: The value of x , The lengths of its sides, and Its area.

Answer:

Using Pythagoras theorem

(x-1)^2+(3x)^2= (3x+1)^2

x^2+1-2x+ 9x^2 = 9x^2+1+6

x^2-8x= 0 \Rightarrow x = 0 \ or \  8

Therefore the length of the sides are 7 cm, 21 cm and 25 cm.

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Question 4: The hypotenuse of a right-angled triangle exceeds one side by 1 cm and the other side by 18 cm; find the lengths of the sides of the triangle.

Answer:

Let the hypotenuse = x \ cm 

The other two sides would then be (x-1) \ and \ (x-18)

Using Pythagoras theorem

(x-1)^2+(x-18)^2= (x)^2

x^2+1-2x + x^2+324-36x = x^2 

x^2-38x+325 = 0 \Rightarrow x = 25 \ cm \ or \ 13 \ cm (not possible) 

Therefore the sides are 25 cm, 24 cm and 7 cm.

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Question 5: The diagonal of a rectangle is 60 m more than its shorter side and the larger side is 30 m more than the shorter side. Find the sides of rectangle.

Answer:

Let the shorter side = x \ cm 

Diagonal = (x+60) \ cm 

Larger sides = (x+30) \ cm 

Using Pythagoras theorem

(x)^2+(x+30)^2= (x+60)^2

x^2+x^2+900+60x = x^2+3600 + 120x

x^2 - 60x - 2700 = 0 \Rightarrow x = 90 \ cm \ or \ -30 \ cm (not \ possible) 

Therefore the sides are 90 cm, 150 cm and 120 cm.

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Question 6: The perimeter of a rectangle is 104 m and its area is 640 m2. Find its length and breadth.

Answer:

Let the length = x \ m

Let the width = y \ m

Therefore

x \times y = 640 ... ... ... ... i)

2(x+y) = 104  ... ... ... ... ii)

\Rightarrow y = 52-x

Substituting in i)

x (52-x) = 640

x^2-52x+640 = 0 \Rightarrow x = 32 \ cm or 20 \ cm

Therefore y = 20 \ cm or 32 \ cm

Hence the sides are 32 cm and 20 cm.

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Question 7: A footpath of uniform width runs around the inside of a rectangular field 32 m long and 24 m wide. If the path occupies 208 m2, find the width of the footpath.

Answer:

Let the width of the footpath = x \ m

Area of the rectangular field = 32 \times 24

Area of the rectangular field not covered by the footpath = (32-2x) \times (24-2x)

Therefore 32 \times 24 - (32-2x) \times (24-2x) = 208

4x^2-112+208 = 0 \Rightarrow x = 2 \ m \ or \  26 \ m (not \ possible) 

Therefore the width of the footpath = 2 m

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Question 8: Two squares have sides x \ cm \ and \ (x+4) \ cm . the sum of their areas is 656 sq.cm. express this as an algebraic equation in x and solve the equation to find the sides of the squares.

Answer:

x^2+(x+4)^2 = 656 

x^2 + x^2+16+8x = 656 

2x^2+8x - 640 \Rightarrow x = 16 cm or -20 cm (not \ possible) 

The sides of the square are 16 cm and 20 cm.

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Question 9: The dimensions of a rectangular field are 50 m by 40 m. a flower bed is prepared inside this field leaving a gravel path of uniform width all around the flower bed and graveling the path at Rs. 30 and Rs. 20 per square meter, respectively, is Rs. 52,000. Find the width of the gravel path.

Answer:

Let the width of the gravel path = x \ m 

Area of the flower bed = (50-2x)(40-2x) = 2000-180x+4x^2 

Are of the gravel path = 50 \times 40 - (50-2x)(40-2x) = 180x-4x^2 

Total cost of work = 52000 \ Rs. 

(2000-180x+4x^2) \times 30 + (180x-4x^2) \times 20 = 52000  

6000-540x+12x^2+360x-8x^2 = 5200 

4x^2-180x+800 = 0 \Rightarrow x = 5 \ m \ or \ 40 \ m (not \ possible) 

Therefore the width of the gravel path = 5 \ m 

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Question 10: An area is paved with square tiles of a certain size and the number required is 128. If the tiles had been 2 cm smaller each way, 200 tiles would have been needed to pave the same area. Find the size of the larger tiles.

Answer:

Let the width of the larger tile = x \ cm 

Width of the smaller tile = (x-2) \ cm 

Thereofore

128x^2 = (x-2)^2 \times 200 

72x^2-800x+800 = 0 \Rightarrow x = 10 \ cm \ or \ 1.111 \ cm \ (not \ possible) 

Therefore the width of the large tile = 10 cm

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Question 11: A farmer has 70 m of fencing, with which he encloses three sides of a rectangular sheep pen; the fourth side being a wall. If the area of the pen is 600 sq.m. find the length of its shorter side.

Answer:

Let the shorter side = x \ m 

Therefore the larger side = (70-2x) \ m 

Therefore x(70-2x)=600 

2x^2-70x+600 = 0 \Rightarrow x = 20 \ m \ or \  15 \ m 

Therefore the larger side = 30 m (when shorter side is 20 m)  or 40 m (when shorter side is 15 m)

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Question 12: A square lawn is bounded on three sides by a path 4 m wide. If the area of the path is 7/8 that of the lawn, find the dimensions of the lawn.

Answer:

Let the width of the path = x \ m 

Given (x+8)(x+4)-x^2 = \frac{7}{8} x^2 

x^2+12x+32-x^2=\frac{7}{8} x^2 

7x^2-96x-256=0 \Rightarrow x = 16 \ m 

Therefore the width of the square = 4 m 

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Question 13: The area of a big rectangular room is 300 m2. If the length were decreased by 5 m and the breadth increased by 5 m; the area would be unaltered. Find the length of the room.

Answer:

Let the length of the room = x \ m 

Let the breadth of the room = y \ m 

Given x \times y = 300 \Rightarrow y = \frac{300}{x} 

Also (x-5)(y+5)= 300 

Substituting (x-5)(\frac{300}{x}+5)= 300 

5x^2-25x-1500 = 0 \Rightarrow x = 20 \ m 

Therefore y = \frac{300}{25} = 12 \ m 

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