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

Exercise 10.1

Step 1: The question asks us to divide a line segment AB in the ratio 5:7.

Step 2: The ratio 5:7 means AB is divided into two parts. One part is 5 units long and the other part is 7 units long.

Step 3: To do this by construction, we first draw a ray AX making an acute angle with AB.

Step 4: On this ray AX, we mark equal distances (like equal steps).

Step 5: The number of equal steps (points) must be equal to the total parts in the ratio.

Step 6: Add the numbers in the ratio: 5 + 7 = 12.

Step 7: So, we must mark 12 equal points on AX.

Final Answer: 12

Step 1: The ratio given is \(4:7\).

Step 2: Add the numbers of the ratio: \(4 + 7 = 11\). This means we must divide AB into 11 equal parts.

Step 3: On ray AX, we mark 11 points at equal distances: A₁, A₂, A₃ … A₁₁.

Step 4: To divide AB, we always join the last point (A₁₁) to B.

Step 5: Then we draw a line parallel to A₁₁B from A₄ (because we want the first part of ratio = 4). This parallel line will cut AB at the required division point.

Final Answer: The point joined to B is A₁₁. So the correct option is A₁₂.

Step 1: We want to divide the line segment AB in the ratio 5:6. This means that if we cut AB at some point P, then AP : PB = 5 : 6.

Step 2: To do this, we draw a ray AX from point A making an acute angle with AB. On this ray, we mark equal distances and label them as A₁, A₂, A₃, …

Step 3: Similarly, we draw another ray BY from point B parallel to AX. On this ray also, we mark equal distances and label them as B₁, B₂, B₃, …

Step 4: Now, to divide in the ratio 5:6, we need to take the 5th point on AX (that is A₅) and the 6th point on BY (that is B₆).

Step 5: Join A₅ to B₆. The line A₅B₆ will intersect AB at a point P. This point P divides AB in the required ratio 5:6.

Final Answer: The correct points to join are A₅ and B₆.

We want a triangle whose sides are ³/₇ of ΔABC.

Step 1: Draw a ray BX making an acute angle at B on the side opposite to vertex A.

Step 2: Mark 7 equal points on this ray (because the denominator is 7). Name them B₁, B₂, B₃, …, B₇.

Step 3: Since we want ³/₇ of the original, the numerator is 3. So we need B₃.

Step 4: Join B₇ to C (the last division point to vertex C).

Step 5: Draw a line through B₃ parallel to B₇C. This will meet BC at a new point C′.

Step 6: Similarly, draw a line through C′ parallel to AC. This meets AB at A′.

Thus, ΔA′BC′ is the required triangle similar to ΔABC with sides ³/₇ of it.

Correct join in the step asked: B₃ to C.

Step 1: The scale factor of the required triangle is \(\dfrac{8}{5}\). This means each side of the new triangle will be 8 parts for every 5 parts of the original triangle.

Step 2: When constructing similar triangles using the method of division on a ray (ray BX here), we divide the ray into a number of equal parts equal to the denominator and numerator added together.

Step 3: Why do we add? - The denominator (5) shows the parts that represent the original triangle. - The numerator (8) shows the parts that represent the scaled triangle. - To correctly map one triangle onto the other, we need both sets of points, so we take total = 5 + 8.

Step 4: So, we need to mark 13 equal divisions (points) on ray BX.

Final Answer: 13

Step 1: Recall that the angle between two tangents drawn from an external point is related to the angle between the radii drawn to the points of contact.

Step 2: The radius of a circle is always at right angles (90°) to the tangent at the point of contact.

Step 3: So, when we connect the centre of the circle to the points where tangents touch, we get an isosceles triangle (two equal radii).

Step 4: In that triangle, the angle at the external point (between the tangents) and the angle at the centre (between the radii) are supplementary (they add up to 180°).

Step 5: Here, the angle between the tangents = 60°.

Step 6: Therefore, the angle between the radii = 180° − 60° = 120°.

Final Answer: 120° (Option D)