Geometry Problems with Answers and Solutions - Grade 10

Grade 10 geometry problems with solutions are presented.

Problems

  1. Each side of the square pyramid shown below measures \( 10 \) inches. The slant height, \( H\) , of this pyramid measures \( 12 \) inches.

    pyramid in problem 1.


    1. What is the area, in square inches, of the base of the pyramid?
    2. What is the total surface area, in square inches, of the pyramid?
    3. What is \( h \), the height, in inches, of the pyramid?
    4. Using the height you determined in part (c), what is the volume, in cubic inches, of the pyramid?
  2. The parallelogram shown in the figure below has a perimeter of \( 44 \) cm and an area of 64 cm2. Find angle \( T \) in degrees.

    parallelogram in problem 2.

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

    quadrilateral problem 3.

  4. In the figure below triangle OAB has an area of \( 72 \) units squared and triangle ODC has an area of \( 288 \) units squared . Find \( x \) as the length of segment BC and \( y\) as the length of segment AD.

    triangles in problem 4.

  5. 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?
  6. Find the circumference of a circular disk whose area is \( 100 \pi \) square centimeters.
  7. The semicircle of area \( 1250 \pi \) centimeters is inscribed inside a rectangle. The diameter of the semicircle coincides with the length of the rectangle. Find the area of the rectangle.

Solutions to the Above Problems

  1. a) Area of a square: \( 10 \times 10 = 100 \) inches squared

    b) \( 100 + 4 \times \frac{1}{2} \times 12 \times 10 = 340 \) inches squared

    c) \( h = \sqrt{12^2 - 5^2} = \sqrt{119} \)

    d) Volume \( = \frac{1}{3} \times 100 \times \sqrt{119} = 363.6 \) inches cubed (approximated to 4 decimal digits)


  2. parallelogram problem 2 solution

    \[ 44 = 2(3x + 2) + 2(5x + 4) \] Solve for \( x \): \[ x = 2 \] \[ \text{height} = \frac{\text{area}}{\text{base}} = \frac{64}{14} = \frac{32}{7} \text{ cm} \] \[ \sin(T) = \frac{\text{Opp}}{\text{Hyp}} = \dfrac{H}{3x+2} = \frac{32/7}{8} = \frac{32}{56} = \frac{4}{7} \] \[ T = \arcsin\left(\frac{4}{7}\right) \approx 34.8^\circ \]


  3. quadrilateral problem 3 solution

    Triangle \( \triangle ABD \) is a right triangle; hence \[ BD^2 = 15^2 + 15^2 = 450 \] Also \[ BC^2 + CD^2 = 21^2 + 3^2 = 450 \] This means that triangle \( \triangle BCD \) is also a right triangle, and the total area of the quadrilateral is the sum of the areas of the two right triangles. \[ \text{Area of quadrilateral} = \frac{1}{2} \times 15 \times 15 + \frac{1}{2} \times 21 \times 3 = 144 \]

  4. \[ \text{Area of } \triangle OAB = 72 = \frac{1}{2} \sin(\angle AOB) \cdot OA \cdot OB \] Solve for \( \sin(\angle AOB) \): \[ \sin(\angle AOB) = \frac{2 \cdot 72}{OA \cdot OB} = \frac{1}{2} \] \[ \text{Area of } \triangle ODC = 288 = \frac{1}{2} \sin(\angle DOC) \cdot OD \cdot OC \] Note that: \[ \sin(\angle DOC) = \sin(\angle AOB) = \frac{1}{2}, \quad OD = 18 + y, \quad OC = 16 + x \] Substitute into the area formula: \[ 288 = \frac{1}{2} \cdot \frac{1}{2} \cdot (18 + y)(16 + x) \] \[ 288 = \frac{1}{4}(18 + y)(16 + x) \quad \Rightarrow \quad 1152 = (18 + y)(16 + x) \] \[ \text{Using the intersecting chords theorem:} \quad 16(16 + x) = 14(14 + y) \] \[ \text{Solve the system:} \quad \begin{cases} (18 + y)(16 + x) = 1152 \\ 16(16 + x) = 14(14 + y) \end{cases} \quad \Rightarrow \quad x = 20,\; y = 14 \]

  5. Let \( L \) be the length and \( W \) be the width of the rectangle. Given: \[ L = W + 3 \] Perimeter: \[ \text{Perimeter} = 2L + 2W = 2(W + 3) + 2W = 4W + 6 \] Area: \[ \text{Area} = LW = (W + 3)W = W^2 + 3W \] Since the area and perimeter are equal in value: \[ W^2 + 3W = 4W + 6 \] Solving the quadratic equation: \[ W^2 + 3W - 4W - 6 = 0 \Rightarrow W^2 - W - 6 = 0 \] Factoring: \[ (W - 3)(W + 2) = 0 \] So the solutions are: \[ W = 3 \quad \text{or} \quad W = -2 \] Since width cannot be negative: \[ W = 3 \] Substitute to find the length: \[ L = W + 3 = 3 + 3 = 6 \]

  6. Let \( r \) be the radius of the disk. The area is known and equal to \( 100\pi \); hence, \[ 100\pi = \pi r^2 \] Solve for \( r \): \[ r = 10 \] The circumference is: \[ C = 2\pi r = 20\pi \]

  7. Let \( r \) be the radius of the semicircle. The area of the semicircle is known, so we have: \[ 1250\pi = \frac{1}{2} \pi r^2 \quad \text{(note the \( \frac{1}{2} \) due to the semicircle)} \] Solving for \( r \): \[ r = 50 \] The length of the rectangle is \( 2r = 100 \) (since the semicircle is inscribed). The width of the rectangle is \( r = 50 \) (since the semicircle is inscribed). Thus, the area of the rectangle is: \[ \text{Area} = 100 \times 50 = 5000 \]

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