Geometric Constructions
Maharashtra Board · Class 10 · Mathematics
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Quick Quiz: Geometric Constructions
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If triangle ABC is similar to triangle PQR with a ratio of similarity 3:2, and the perimeter of triangle ABC is 30 cm, what is the perimeter of triangle PQR?
In triangle ABC, if AB = 6 cm, BC = 8 cm, and AC = 10 cm, what is the length of the corresponding side to AB in a similar triangle with scale factor 1.5?
If the sides of a triangle are in the ratio 3:4:5, and the shortest side of a similar triangle is 12 cm, find the length of the longest side.
In the construction of a similar triangle, if the original triangle has sides 4 cm, 5 cm, and 6 cm, what will be the area ratio when the scale factor is 2?
Sample Questions
Which of the following methods can be used to construct a tangent to a circle at a point on the circle?
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Using the center of the circle, Without using the center of the circle, Drawing a perpendicular to the radius at the point
Step 1: Method 1 uses the center to draw a radius and then constructs a perpendicular at the point. Step 2: Method 2 uses the tangent-secant angle theorem without locating the center. Step 3: Both methods rely on the fundamental property that a tangent is perpendicular to the radius at the point of contact.
The number of tangents that can be drawn to a circle from an external point is always 2.
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True
Step 1: From any external point, exactly two tangent lines can be drawn to a circle. Step 2: This is a fundamental theorem in geometry. Step 3: The two tangents from an external point are equal in length and make equal angles with the line joining the external point to the center.
Which tools are essential for constructing a tangent to a circle at a given point on the circle?
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Compass, Ruler (straight edge)
Step 1: A compass is needed to draw arcs for construction. Step 2: A ruler (straight edge) is needed to draw straight lines. Step 3: Tangent construction uses basic geometric construction techniques that only require compass and straight edge. Step 4: Protractor and set square are not essential for this construction.
When constructing tangents from an external point P to a circle with center O, a circle is drawn with diameter OP. This construction works because:
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Angle inscribed in semicircle is 90°
Step 1: The construction uses the property that any angle inscribed in a semicircle is a right angle. Step 2: When the auxiliary circle intersects the original circle, it creates right angles at those intersection points. Step 3: Since tangent is perpendicular to radius, these intersection points are the required tangent points. Step 4: Lines from P through these points are the required tangents.
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