Proof for Validation for Construction of a Triangle with Given Base, Base Angle and Difference between Two Sides

Q. The construction of a triangle ABC, given that BC = 4 cm, $\angle B={30}^{\circ },$ is not possible when difference of AB and AC is equal to

1. 
   3 cm
2. 
   3.5 cm
3. 
   2 cm
4. 
   4 cm

Q. The construction of a triangle ABC, given that BC = 5 cm, ∠B = 45° is not possible when difference of AB and AC is equal to

1. 
   4.9 cm
2. 
   4 cm
3. 
   6 cm
4. 
   3.5 cm

Q.

Given are the steps of construction for constructing a triangle ABC whose base length is given say BC, one of the base angles is given, say $\angle$ B and the difference between the other two sides is also given (AB – AC) considering AB $>$ AC.

Pinku was asked to give the steps of construction for the same. He had mugged up all the steps of construction but forgot one point in between. Following are those steps of construction, let’s see if you can help Pinku with the missing step.

Steps of construction:

Draw the base BC and at point B make an angle say XBC equal to the given angle. Cut the line segment BD equal to (AB – AC) from ray BX. Let it intersect BX at a point A. Join AC

Which of the following do you think is the missing step of this construction?

1. Join D to C and draw a perpendicular bisector of the line AC

2. Join D to C and draw a perpendicular bisector of the line BC

3. Join D to C and draw a perpendicular bisector of the line DC

4. Join D to C and draw an angle bisector of the angle B

Q.

Use ruler and compass only for this question
(i) Construct $\Delta ABC$, where AB = 3.5 cm, BC = 6 cm and $\angle ABC={60}^{\circ }$
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C.

Then the length of PB is

1. 2 cm

2. 2.5 cm

3. 3 cm

4. 3.5 cm

Q.

How will you prove that the construction for a triangle with the given conditions is right?

Given conditions: Base length BC is given, base angle B is given, and difference of the other two sides is given (AB-AC) where AB is greater than AC. For going about the construction, I drew the base length BC, drew the ray BX with angle XBC known to me. Taking B as centre and radius equal to (AB-AC) I cut an arc on the ray BX intersecting it at point D. I then joined D to C. Then drew the perpendicular bisector of the line segment DC and named the point of intersection of this perpendicular bisector and the ray BX as A. Joined A to C and the triangle ABC was ready

Which of the following statements gives the best explanation to this construction?

1. Since the triangles AMD and AMC are congruent, AD = AC and hence the location of A has been plotted correctly

2. the triangles DBC and CAD are congruent the location of A is justified

3. AM is the altitude for the triangle ADC and hence the location of A is justified

4. None of these

Q. Fill in the blanks:
$\left(1\right)$ The point of concurrence obtained by drawing perpendicular bisector of sides of a triangle is called___________.

Q. Choose those times at which the minute hand of a clock is perpendicular to the hour hand.

1. 4:00
2. 9:00
3. 6:30
4. 3:30

Q.

Is it possible to construct a triangle whose sides are $5.5\mathrm{cm}$, $4.5\mathrm{cm}$ and $10\mathrm{cm}$ ? If not , why ?

Q.

The steps to draw a triangle with the base BC, a base angle $\angle$B and the difference of other two sides is given below.
1. Draw the base BC and at point B make an angle say XBC equal to the given angle.
2. Cut the line segment BD equal to AB - AC on the reflection of ray BX (i.e. BX').
3. Join DC and draw the perpendicular bisector, say PQ of DC.​​
The next step will be:

1. Let PQ intersect BX at a point A. Join AC.

2. Draw angle bisector of $\angle$C

3. Draw perpendicular bisector of AD

4. Draw angle bisector of $\angle$ A

Q.

The steps to construct a triangle with given base angles $\angle$ B and $\angle$ C and BC + CA + AB, are:.
A. Draw a line segment, say XY equal to BC + CA + AB.
B. Make $\angle$ LXY equal to $\angle$ B and MYX equal to $\angle$ C.
C. Bisect $\angle$ LXY and $\angle$ MYX. Let these bisectors intersect at a point A.
D. Draw perpendicular bisectors PQ of AX and RS of AY.
E. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC.

1. True

2. False

Q. Question 14
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Q.

The construction of a $∆ABC$ in which $AB=7cm,\angle A=75°$ is possible when $(BC–AC)$ is equal to

1. $7.5cm$

2. $7cm$

3. $8cm$

4. $6.5cm$

Q.

How will you prove that the construction for a triangle with the given conditions is right?

Given conditions: Base length BC is given, base angle B is given, and difference of the other two sides is given (AB-AC) where AB is greater than AC. For going about the construction, I drew the base length BC, drew the ray BX with angle XBC known to me. Taking B as centre and radius equal to (AB-AC) I cut an arc on the ray BX intersecting it at point D. I then joined D to C. Then drew the perpendicular bisector of the line segment DC and named the point of intersection of this perpendicular bisector and the ray BX as A. Joined A to C and the triangle ABC was ready

Which of the following statements gives the best explanation to this construction?

1. Since the triangles AMD and AMC are congruent, AD = AC and hence the location of A has been plotted correctly

2. the triangles DBC and CAD are congruent the location of A is justified

3. AM is the altitude for the triangle ADC and hence the location of A is justified

4. None of these

Q.

Given are the steps of construction for constructing a triangle ABC whose base length is given say BC, one of the base angles is given, say $\angle$ B and the difference between the other two sides is also given (AB – AC) considering AB $>$ AC.

Pinku was asked to give the steps of construction for the same. He had mugged up all the steps of construction but forgot one point in between. Following are those steps of construction, let’s see if you can help Pinku with the missing step.

Steps of construction:

Draw the base BC and at point B make an angle say XBC equal to the given angle. Cut the line segment BD equal to (AB – AC) from ray BX. Let it intersect BX at a point A. Join AC

Which of the following do you think is the missing step of this construction?

1. Join D to C and draw a perpendicular bisector of the line BC

2. Join D to C and draw an angle bisector of the angle B

3. Join D to C and draw a perpendicular bisector of the line AC

4. Join D to C and draw a perpendicular bisector of the line DC

Q.

Which of the following can be the sides of the triangle?

$6\mathrm{cm},8\mathrm{cm}\text{and}7\mathrm{cm}$

Q.

Which of the following can be the sides of the triangle?

$4\mathrm{cm},3\mathrm{cm}\text{and}5\mathrm{cm}$

Q.

Can a triangle be drawn whose sides are -

$\text{3 cm, 4 cm, 7 cm}$

Q.

Can we construct a $∆ABC$ where $AB=5cm$, $\angle A=45°$ and $(BC+AC)=5cm$ ? Justify.