C-04 — Geometric Constructions

Which one is the correct definition of geometric construction?

Compared to a figure drawn for proving a theorem, a geometric construction must be —

The two essential instruments used in geometric construction are —

In the constructions involving triangles in this chapter, the figure must be drawn —

The data required to draw a triangle in Construction 1 are —

In Construction 1, the segment BM cut on BP equals —

In Construction 1, line MN is drawn —

The third vertex A of the desired triangle in Construction 1 is —

In Construction 1, the height of the resulting triangle ABC equals BM because —

The data given in Construction 2 are —

In Construction 2, the segment DB cut on the ray DE equals —

In Construction 2, $\angle BDF$ is drawn equal to —

In Construction 2, the arc with centre B and radius $a$ —

In Construction 2, why does $\angle BAC = \angle ADC + \angle ACD$?

In Construction 2, $CA + AB$ equals —

Consider the following statements about Construction 2:

The data given in Construction 3 are —

In Construction 3, the segment BP cut from the ray BD equals —

In Construction 3, $\angle DPM$ is drawn equal to —

In Construction 3, since $AP = AC$, we obtain —

In Construction 3, $\angle APC + \angle ACP$ equals —

The data required for Construction 4 are —

In Construction 4, BM is cut on BQ equal to —

In Construction 4, with centre A and radius equal to the median, the arc cuts BE at D. The segment DC is then taken equal to —

According to the remark on Construction 4, the triangle cannot be constructed if —

In Construction 4, D is the —

In Example 1, the given data are base $BC = 5$ cm, $\angle ABC = 60°$ and $AB + AC =$

In Example 1, on the ray BY the segment BD is cut equal to —

In Example 1, the vertex A is located on —

In Example 1, $AD = AC$ because —

In Example 1, $BD = BA + AD$ equals $BA + AC$, hence equals —

In Example 2, the given data are base $BC = 7.5$ cm, $\angle YBC = 45°$ and $AB - AC =$

In Example 2, from the ray BY the segment BD is cut equal to —

In Example 2, RS is the perpendicular bisector of —

To draw a triangle when the perimeter and the two angles adjoining the base are given, the first segment cut on a ray equals —

In a right-angled triangle whose two perpendicular sides are 3 cm and 4 cm, the hypotenuse is —

In Activity 5, the perimeter of $\triangle ABC$ is $12$ cm with $\angle B = 60°$ and $\angle C = 45°$. The third angle $\angle A$ equals —

In any circle construction problem, the centre of the required circle is most often located at the intersection of —

In Construction 5, the line CD is —

In Construction 5, the centre O of the required circle is —

The radius of the circle in Construction 5 is —

Why does any point on CD lie at equal distance from A and B in Construction 5?

In Construction 6, two arcs are drawn with A and B as centres and with radius equal to —

The number of circles obtained in Construction 6 is —

In Construction 6, the centres P and Q lie —

In Construction 6, $PA = PB =$

In Construction 7, the centre O of the required circle is the intersection of —

In Construction 7, the constructed circle passes through Q because —

In Construction 7, the two circles touch each other at the point —

In Construction 8, at the point A on AB the perpendicular drawn is —

In Construction 8, QR is the perpendicular bisector of —

In Construction 8, the centre O is the point of intersection of —

The perpendicular drawn at A to AB serves as —

In Construction 8, the circle touches the line AB at A because —

Consider the following statements about Construction 8:

In Example 3, the centre O of the given circle and the point P are at a distance —

In Example 3, M is —

In Example 3, Q and R are —

In Example 3, the measured length of each tangent PQ and PR is approximately —

By the Pythagorean theorem, the exact tangent length in Example 3 (radius $2$ cm, distance $5$ cm) is —

Consider the following statements about Example 3:

The radius of the incircle of a triangle whose sides are $5, 12, 13$ cm is —

The triangle with sides $5, 12, 13$ cm is —

The centre of the circumcircle of a triangle is the intersection of —

The centre of the incircle of a triangle is the intersection of —

To construct a triangle when two angles and the difference of the lengths of the opposite sides of those angles are given (Exercise 1), the third angle is —

In Exercise 3 — base, vertical angle and the sum of the other two angles given — once the vertical angle is fixed, the sum of the base angles equals —

In Exercise 5, the hypotenuse of the right-angled triangle is —

In Exercise 11, three circles touch each other externally; the distance between any two of the three centres equals —

In Exercise 12, P is a point on a chord AB and the chord CD through P satisfies $CP^2 = AP \cdot PB$. This relation is the —

In Exercise 14, the radius of the circle is $3$ cm and $OT = 5$ cm. By the Pythagorean theorem, the length of each tangent from T is —

In Exercise 14 (3), the sum of the lengths of the two tangents from T to the circle is —

$3.5$ cm, $4.5$ cm and $5.5$ cm are the radii of the three circles that touch each other externally. What is the perimeter of the triangle formed by the three centres of the circles?

In triangle ABC, if $\angle B = 90°$, $\angle A = 60°$ and $BD \perp AC$, then

The base of an isosceles triangle is $6$ cm and each equal side is $5$ cm. The height of the triangle on the base is —

For the same isosceles triangle (base $6$ cm, equal sides $5$ cm), the area is —

The radius of the incircle of the isosceles triangle with base $6$ cm and equal sides $5$ cm is —

In the creative problem, a circle is to be drawn whose diameter equals the base ($6$ cm). Its radius is —

To draw the circumcircle of a triangle with sides $4, 5, 6$ cm, the centre of the circle is the point of intersection of —

To draw a triangle with base $5$ cm, height $4$ cm and angle adjacent to the base $60°$, the construction follows —

To draw two tangents to a circle of diameter $6$ cm from an external point P, one first joins OP (where O is the centre) and constructs —

The length of a tangent drawn from an external point at distance $d$ from the centre of a circle of radius $r$ is given by —