Geometry Problems with Solution & Answer - 10th & 12th Standard

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Q.1 Each side of square pyramid is shown below figures 10 inches. The slant height, H, of this pyramid measures 12 inches. 

  .

a.         What is the total surface area of square inches, of the pyramid?

b.         What is the area, in square inches, of the base of the pyramid?

c.         Using the height you determined in part (c), what is the volume, in cubic inches, of the pyramid?

d.         What is h, the height, in inches, of the pyramid? 

Answer 

a) Total Surface Area is = 100 + 4*(1/2)*12*10 = 340 inches squared 

b)  Area of Square is 10*  10 + 100 inches squared 

c) Volume of Pyramid is = (1/3)*100*sqrt(119)

d) height is = sqrt(122 - 52) = sqrt(119)

= 363.6 inches cubed (approximated to 4 decimal digits)

 Q. 2  The parallelogram shown in the figure following has a perimeter of 44 cm and an area of 64 cm2. Find angle T in degrees.

Answer 

44 = 2(3x + 2) + 2(5x + 4) , solve for x

x = 2

height = area / base

= 64 / 14 = 32/7 cm

sin(T) = (32/7) / 8 = 32/56 = 4/7, T = arcsin(4/7) = 34.8 degree

Q. 3  Find the dimensions of a rectangle that has a length 3 meters more that its width & have  a perimeter equal in value to its area?

Answer

1. Let Length be the L & W be the width of the rectangle. L = W + 3
   
   and Perimeter = 2L + 2W = 2(W + 3) + 2W = 4W + 6
   
   Area = L W = (W + 3) W = W² + 3 W
   
   Area and perimeter are equal in value
     Then W² + 3 W = 4W + 6
   
   Solve the above quadratic equation for W and substitute to find L
   
   W = 3 and L = 6

Q. 4 Find the area of the quadrilateral shown in the figure.(NOTE: figure not drawn to scale)

             

Answer

ABD is a right triangle then BD

2 = 152 + 152 = 450

Also Sqart(BC)  + Sqrt(CD)  = 212 + 32 = 450

The above means that triangle BCD is also a right triangle and the total area of the quadrilateral is the sum of the areas of the two right triangles.

Area of quadrilateral = (1/2)*15*15 + (1/2)*21*3 = 144