Grade 10 Mathematics Exercises
Comprehensive practice problems covering circles, similar triangles, heights & distances, and trigonometry
Circle Geometry
Practice problems involving circle theorems, tangents, chords, angles, and cyclic quadrilaterals.
In the diagram below, \( O \) is the center of the circle. Points \( A \), \( B \), and \( C \) lie on the circumference. Angle \( ABC = 50^\circ \), and angle \( OAB = 30^\circ \). Find:
- Angle \( AOC \).
- Angle \( OCB \).
Remember the circle theorem: the angle at the center is twice the angle at the circumference when both angles subtend the same arc. Also, triangle OAB is isosceles because OA and OB are radii.
A circle has a chord \( AB \) of length 12 cm. The tangent at point \( A \) makes an angle of \( 60^\circ \) with the chord \( AB \). Find:
- The radius of the circle.
- The length of the arc \( AB \).
The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. This creates a right triangle that can be used to find the radius.
In a cyclic quadrilateral \( ABCD \), angle \( A = 70^\circ \), angle \( B = 110^\circ \), and angle \( C = 80^\circ \). Find:
- Angle \( D \).
- The measure of the arc \( ADC \).
In a cyclic quadrilateral, opposite angles sum to \(180^\circ\). Also, the measure of an arc is twice the measure of the inscribed angle that subtends it.
Two chords \( AB \) and \( CD \) intersect at point \( P \) inside the circle. If \( AP = 6 \, \text{cm} \), \( PB = 4 \, \text{cm} \), and \( CP = 3 \, \text{cm} \), find the length of \( PD \).
Use the intersecting chords theorem: \( AP \times PB = CP \times PD \).
A tangent \( PA \) and a secant \( PBC \) are drawn to a circle from an external point \( P \). If \( PA = 8 \, \text{cm} \) and \( PB = 4 \, \text{cm} \), find the length of \( BC \).
Apply the tangent-secant theorem: \( PA^2 = PB \times PC \). Then, \( BC = PC - PB \).
In a circle with diameter \( AB \), point \( C \) lies on the circumference such that angle \( ACB = 90^\circ \). If \( AC = 6 \, \text{cm} \) and \( BC = 8 \, \text{cm} \), find:
- The radius of the circle.
- The area of the circle.
Triangle ABC is a right triangle with hypotenuse AB (the diameter). Use the Pythagorean theorem to find AB, then the radius is half of AB.
Two concentric circles have radii \( 5 \, \text{cm} \) and \( 10 \, \text{cm} \). A chord of the larger circle is tangent to the smaller circle. Find the length of the chord.
The radius to the point of tangency is perpendicular to the chord. Use the Pythagorean theorem with the radius of the larger circle as the hypotenuse and the radius of the smaller circle as one leg.
A circle has a radius of \( 7 \, \text{cm} \). A sector of the circle has an angle of \( 120^\circ \). Find:
- The area of the sector.
- The length of the arc of the sector.
Sector area = \(\frac{\theta}{360} \times \pi r^2\) and arc length = \(\frac{\theta}{360} \times 2\pi r\), where \(\theta\) is in degrees.
In a circle, two chords \( AB \) and \( CD \) intersect at point \( E \). If angle \( AEC = 40^\circ \) and arc \( AC = 100^\circ \), find:
- Angle \( BED \).
- The measure of arc \( BD \).
Angle AEC is formed by intersecting chords. It equals half the sum of the measures of arcs AC and BD. Also, vertical angles are equal.
In the diagram below, \( O \) is the center of the circle. \( AB \) is a chord, and \( OC \) is perpendicular to \( AB \), intersecting it at point \( D \). If \( OD = 3 \, \text{cm} \) and \( CD = 4 \, \text{cm} \), find:
- The radius of the circle.
- The length of chord \( AB \).
Join OA to form a right triangle OAD. Note that OC = OD + DC, and OA = OC (both are radii). The perpendicular from the center to a chord bisects the chord.
Similar Triangles
Practice problems involving similar triangles, proportionality, and applications of similarity theorems.
In triangle \( ABC \), \( DE \) is parallel to \( BC \). If \( AD = 4 \, \text{cm} \), \( DB = 6 \, \text{cm} \), and \( DE = 5 \, \text{cm} \), find:
- The length of \( BC \).
- The ratio of the areas of \( \triangle ADE \) to \( \triangle ABC \).
Use the Basic Proportionality Theorem (Thales' theorem) which states that if a line is parallel to one side of a triangle, it divides the other two sides proportionally. The ratio of areas of similar triangles is the square of the ratio of corresponding sides.
In triangle \( PQR \), \( S \) and \( T \) are points on sides \( PQ \) and \( PR \), respectively, such that \( ST \) is parallel to \( QR \). If \( PS = 3 \, \text{cm} \), \( SQ = 2 \, \text{cm} \), and \( QR = 10 \, \text{cm} \), find:
- The length of \( ST \).
- The ratio of the areas of \( \triangle PST \) to \( \triangle PQR \).
Since ST is parallel to QR, triangles PST and PQR are similar. The ratio of corresponding sides is PS:PQ. The ratio of areas is the square of the ratio of corresponding sides.
Heights and Distances
Practice problems involving trigonometry, angles of elevation and depression, and real-life applications.
A person standing on the ground observes the angle of elevation of the top of a tower to be \( 30^\circ \). After walking \( 20 \, \text{meters} \) closer to the tower, the angle of elevation becomes \( 45^\circ \). Find:
- The height of the tower.
- The original distance of the person from the tower.
Let the height of the tower be \(h\) and the original distance be \(d\). Then \(\tan 30^\circ = \frac{h}{d}\) and \(\tan 45^\circ = \frac{h}{d-20}\). Solve these equations simultaneously.
From the top of a cliff \( 100 \, \text{meters} \) high, the angle of depression of a boat at sea is \( 30^\circ \). Find:
- The distance of the boat from the base of the cliff.
- The angle of elevation of the top of the cliff from the boat.
The angle of depression from the cliff top to the boat equals the angle of elevation from the boat to the cliff top. Use \(\tan 30^\circ = \frac{100}{\text{distance}}\) to find the distance.
Trigonometry
Practice problems involving trigonometric ratios, identities, equations, and applications.
In a right triangle \( ABC \), \( \angle B = 90^\circ \), \( AB = 5 \, \text{cm} \), and \( BC = 12 \, \text{cm} \). Find:
- \( \sin A \), \( \cos A \), and \( \tan A \).
- \( \sin C \), \( \cos C \), and \( \tan C \).
First, find the hypotenuse AC using the Pythagorean theorem: \(AC = \sqrt{AB^2 + BC^2}\). Then use the definitions: \(\sin = \frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos = \frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan = \frac{\text{opposite}}{\text{adjacent}}\).
If \( \sin \theta = \frac{3}{5} \), find:
- \( \cos \theta \).
- \( \tan \theta \).
- \( \sin (90^\circ - \theta) \) and \( \cos (90^\circ - \theta) \).
Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find \(\cos \theta\). Remember that \(\sin(90^\circ - \theta) = \cos \theta\) and \(\cos(90^\circ - \theta) = \sin \theta\).
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