NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 10: Construction

Exercise 10.2

Step 1: We are asked to divide a line segment in the ratio \(\sqrt{3} : \dfrac{1}{\sqrt{3}}\).

Step 2: To make the ratio easier to understand, we remove the square root in the denominator. Multiply both sides of the ratio by \(\sqrt{3}\):

\[ \sqrt{3} : \dfrac{1}{\sqrt{3}} = (\sqrt{3} \times \sqrt{3}) : \left(\dfrac{1}{\sqrt{3}} \times \sqrt{3}\right) = 3 : 1 \]

Step 3: Now the ratio is written as \(3:1\). This is a ratio using whole numbers (integers).

Step 4: In geometry, to divide a line segment in the ratio \(m:n\), where \(m\) and \(n\) are whole numbers, we can use the standard geometrical construction method:

• Draw a ray at one end of the line segment.
• Mark \(m+n\) equal parts on that ray (here, \(3+1 = 4\) equal parts).
• Join the last point of the ray to the other end of the line segment.
• Draw a line parallel through the 3rd mark (since the ratio is 3:1).
• This point of intersection divides the line segment in the ratio \(3:1\).

Final Step: Since \(\sqrt{3} : \dfrac{1}{\sqrt{3}}\) becomes \(3:1\), it is possible to divide the line segment using this construction. ✅

Step 1: We want a triangle similar to \(\triangle ABC\) but larger. The scale factor is \(\dfrac{7}{3}\), which means each side of the new triangle should be \(\tfrac{7}{3}\) times the corresponding side of \(\triangle ABC\).

Step 2: To enlarge a triangle from vertex \(B\), we first draw a ray \(BX\) making an acute angle with \(BC\). On this ray, we mark 7 equal divisions: \(B_1, B_2, B_3, ..., B_7\).

Step 3: Why 7 points? Because the numerator of the scale factor is 7. Why do we also consider 3? Because the denominator is 3. To enlarge, we need to use the last point \(B_7\) and also the 3rd point \(B_3\).

Step 4: The correct method is: – Join \(B_7\) to \(C\). – Through \(B_3\), draw a line parallel to \(B_7C\). This line will cut the extended line of \(BC\) at a new point, call it \(C'\).

Step 5: Now, through \(C'\), draw a line parallel to \(AC\). This line meets the extension of \(AB\) at a point \(A'\). Thus, \(\triangle A'BC'\) is the required triangle similar to \(\triangle ABC\) with sides in the ratio \(\dfrac{7}{3}\).

Step 6: In the question, they mistakenly joined \(B_3\) to \(C\) first, instead of \(B_7\) to \(C\). This mixes up the roles of \(B_3\) and \(B_7\). As a result, the construction given is wrong.

Final Check: Since the steps in the question are not the correct method for enlargement, the statement is False.

Step 1: Recall the condition for drawing a tangent from a point to a circle.

A tangent is possible only if the point lies outside the circle. This means the distance of the point from the centre (\(d\)) should be greater than or equal to the radius (\(r\)).

Step 2: Write the given values in SI units (cm).

• Radius of circle, \(r = 3.5\,\text{cm}\)
• Distance of point \(P\) from centre, \(d = 3.0\,\text{cm}\)

Step 3: Compare distance and radius.

Here, \(d = 3.0\,\text{cm}\) and \(r = 3.5\,\text{cm}\).

Since \(d < r\), the point \(P\) is inside the circle.

Step 4: Conclusion.

If the point is inside the circle, no tangent can be drawn. Therefore, the statement is false.

Step 1: When two tangents are drawn from an external point to a circle, they meet at some angle. Let this angle be \(\theta\).

Step 2: The angle between the two radii drawn to the points of contact is always the supplementary angle of \(\theta\). That means, angle between the radii = \(180^\circ - \theta\).

Step 3: Here, \(\theta = 170^\circ\). So, angle between radii = \(180^\circ - 170^\circ = 10^\circ\).

Step 4: A \(10^\circ\) angle at the centre of a circle is possible in geometry (it is a very small but valid angle).

Final Step: Therefore, it is possible to construct such a pair of tangents to the circle.