Geometry Problems with Answers and Solutions - Grade 10
geometry problems with answers are presented.
Each side of the square
pyramid shown below measures 10 inches. The slant height, H, of this pyramid measures 12 inches.
- What is the area, in square inches, of the base of the pyramid?
- What is the total surface area, in square inches, of the pyramid?
- What is h, the height, in inches, of the pyramid?
- Using the height you determined in part (c), what is the volume, in cubic inches, of the pyramid?
shown in the figure below has a perimeter of 44 cm and an area of 64 cm2. Find angle T in degrees.
Find the area of the quadrilateral shown in the figure.(NOTE: figure not drawn to scale)
In the figure below triangle OAB has an area of 72 and triangle ODC has an area of 288. Find x and y.
Find the dimensions of the rectangle that has a length 3 meters more that its width and a perimeter equal in value to its area?
Find the circumference of a circular disk whose area is 100 π square centimeters.
The semicircle of area 1250 π centimeters is inscribed inside a rectangle. The diameter of the semicircle coincides with the length of the rectangle. Find the area of the rectangle.
Answers to the Above Problems
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a) 100 inches squared
b) 100 + 4×(1/2)×12×10 = 340 inches squared
c) h = √(122 - 52) = √(119)
d) Volume = (1/3)×100×√(119)
= 363.6 inches cubed (approximated to 4 decimal digits)
44 = 2(3x + 2) + 2(5x + 4) , solve for x
x = 2
height = area / base
= 64 / 14 = 32/7 cm
sin(T) = Opp/ Hyp = (32/7) / 8 = 32/56 = 4/7
T = arcsin(4/7) = 34.8o
ABD is a right triangle; hence BD2 = 152 + 152 = 450
Also BC2 + CD2 = 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
area of OAB = 72 = (1/2) sin (AOB) × OA × OB
solve the above for sin(AOB) to find sin(AOB) = 1/2
area of ODC = 288 = (1/2) sin (DOC) × OD × OD
Note that sin(DOC) = sin(AOB) = 1/2, OD = 18 + y and OC = 16 + x and substitute in the above to obtain the first equation in x and y
1152 = (18 + y)(16 + x)
We now use the theorem of the intersecting lines outside a circle to write a second equation in x and y
16 × (16 + x) = 14 × (14 + y)
Solve the two equations simultaneously to obtain
x = 20 and y = 14
Let L be the length and W be the width of the rectangle. L = W + 3
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; hence
W2 + 3 W = 4W + 6
Solve the above quadratic equation for W and substitute to find L
W = 3 and L + 6
Let r be the radius of the disk. Area is known and equal to 100π; hence
100π = π r2
Solve for r: r = 10
Circumference = 2 π r = 20 π
Let r be the radius of the semicircle. Area of the semicircle is known; hence
1250π = (1/2) π r2 (note the 1/2 because of the semicircle)
Solve for r: r = 50
Length of rectangle = 2r = 100 (semicircle inscribed)
Width of rectangle = r = 50 (semicircle inscribed)
Area = 100 × 50 = 5000
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