Free Constructions 01 Practice Test - 9th Grade
In which of the following situations an angle bisector cannot be constructed to bisect the angle formed between two given lines?
Solution : B
Since two parallel lines do not intersect each other, an angle cannot be formed between them. Thus in this case, angle bisection is not possible.
In which of the following quadrilaterals, a diagonal is not an angle bisector?
Solution : B
In the case of a rectangle, the diagonal is not an angle bisector.
Consider a rectangle ABCD.
Here, ∠ ADB = ∠ DBC (Alternate angles)
But, ∠ DBC is not equal to ∠ BDC (Angles opposite to unequal sides of a triangle are unequal)
∴ ∠ ADB is not equal to ∠ BDC
So, diagonal DB does not bisect ∠ D.
Each point on a/an _________ is such that it forms an isosceles triangle with the end points of the given line segment.
Solution : A
Consider the above figure.
Here, XY is the perpendicular bisector of a line AB.
Let P be any random point on XY.
In △ PMA ≅ △ PMB
AM = BM (Perpendicular bisector divides a line segment into two equal halves)
∠ PMA = ∠ PMB = 90∘
Also PM is common side
So △ PMA ≅ △ PMB (SAS Rule)
∴ PA = PB (CPCT)
Hence, △ PAB is an isosceles triangle.
So, the given statement is true.
For which of the following can a perpendicular bisector be drawn?
Solution : C
A perpendicular bisector can be drawn only if a figure has end points. Only a line segment has a definite length and hence it can be bisected by a perpendicular bisector.
Which among the following angles cannot be constructed just by using a ruler and a compass?
Solution : C
All the standard angles and the angles which can be obtained by bisecting standard angles can be constructed just by using a ruler and compass. Here except 70∘ all other angles can be constructed using ruler and compass.
For constructing a right angle without a protractor, which of the following method/s is/are used?
Solution : A and B
To construct a right angle triangle, the following steps should be followed:
1. Draw a line OB of given length.
2. With O as centre, make an arc of any radius intersecting OB at X.
3. With X as a centre, draw another arc keeping the radius same intersecting the previous arc at D.
4. ∠DOB will be 60∘.
5. With D as a centre and keeping the radius same, draw another arc intersecting the first arc at C.
6. ∠COB will be 120∘ .
7. Bisect COD drawing two equal arcs from each points intersecting at E.
8. Join EO and extend till A. ∠AOB will be 90∘.
So, it can be seen that both 60∘ and angle bisector construction is used in the process.
Two sides of a triangle have lengths of 5 units and 6 units respectively. Which of the following is a possible length for the third side?
Solution : D
In a triangle, the sum of the lengths of any 2 sides of a triangle must be greater than the third side. The third side can measure anything less than 11 units. Hence, the third side can be 10 units.
For constructing a triangle with a given base, a base angle and difference between the other two sides, a
The procedure for Triangle Construction 2 is:
1. Draw the base BC of ∆ABC as given and construct ∠XBC of the required measure at B as shown.
2. From the ray, BX cut an arc equal to AB – AC at point P and join it to C as shown
3. Draw the perpendicular bisector of PC and let it intersect BX at point A as shown:
4. Join AC, ∆ABC is the required triangle.
Hence, it is clear that a perpendicular bisector is required in the process.
A triangle ABC has base angle 45∘. It's perimeter is [2+√]. What type of triangle is it?
Solution : C
The information given above is insufficient to construct a triangle. So, the triangle cannot be constructed and its characteristics can’t be determined.
The construction of ΔABC, given that BC = 12 cm, is possible when the difference of AB and AC equals
Solution : B and C
The difference between two sides of a triangle should be smaller than the third side. So only 8 cm and 10 cm can be the possible difference.