Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

State True or False
There is a triangle whose sides have lengths 10.2 cm, 5.8 cm and 4.5 cm 

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Suppose such a triangle is possible Then the sum of the lengths of any two side would be greater than the length of the third side  Let us check this
Is 4.5+5.8>10.2  Yes 
Is 5.8+10.2>4.5  Yes
Is 10.2+4.5>5.8  Yes
Therefore the triangle is possible

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

The sum of all the angles of a pentagon are

  1. $360^\circ$
  2. $540^\circ$
  3. $720^\circ$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pentagon is a five sided polygon.

The sum of the interior angles of the pentagon is the sum of interior angles of the three triangles.The sum of interior angles of the three triangles is 180 degree.so the sum of interior angles of the pentagon is 3 times 180 degree which is 540 degree.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

Inscribe a regular pentagon in a circle of radius $3\ cm$. The interior angles of the pentagon are:

  1. $54^\circ$
  2. $60^\circ$
  3. $162^\circ$
  4. $108^\circ$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that internal angle of regular pentagon is $\cfrac{(n-2)}{n}180^{\circ}$ where n = number of sides.

Here, n = 5.
So, interior angle is $\cfrac{(5-2)}{5}180^{\circ} = 108^{\circ}$

So correct answer is option D

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

State true or false:

$ D, E $ and $ F $ are the mid-points of the sides $ AB, BC $ and  $ CA $ of an isosceles $ \bigtriangleup ABC $  in which $ AB= BC $. then
 $ \bigtriangleup DEF $  is also isosceles.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

AB = AC
Hence, $\angle ABC = \angle ACB$ (Isosceles triangle property)
Now, since, D and F are mid point of AB and AC respectively, thus DF II BC (Mid point theorem)
Hence, 
$\angle ADF = \angle ABC$ and $\angle AFD = \angle ACB$ (Corresponding angles)
Thus, $\angle ADF = \angle ABC = \angle AFD = \angle ACB$ 
Now, In $\triangle$ ADF and FEC
$\angle ADF = \angle FEC$ (Corresponding angles of parallel lines EF and AB)
$\angle AFD = \angle ACB $(Corresponding angles of parallel lines DF and BC)
AF = FC (F is the mid point of AC)
Thus $\triangle ADF \cong \triangle FEC$ (AAS rule)
Hence, AD = FE (corresponding sides of congruent triangles)
Similarly, we can prove, AF = DE
Since, AD = AF (half lengths of equal sides, AB and AC)
Thus, EF = DE or $\triangle$ DEF is an isosceles triangle.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

In $\Delta ABC$, D and E are mid points of AB and BC respectively and $\angle ABC=90^o$, then

  1. $AE^2+CD^2=AC^2$
  2. $AE^2+CD^2=\frac {5}{4}AC^2$
  3. $AE^2+CD^2=\frac {3}{4}AC^2$
  4. $AE^2+CD^2=\frac {4}{5}AC^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a right triangle ABC, using the Pythagorean theorem for triangles ABE and BCD, AE^2 = AB^2 + BE^2 and CD^2 = BC^2 + BD^2. Substituting BE = BC/2 and BD = AB/2, we get AE^2 + CD^2 = 5/4(AB^2 + BC^2), which equals 5/4(AC^2).

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

Fill in the blanks:
(i) The ling segment joining a vertex of a triangle to the midpoint of its opposite side is called a $\underline { P } $ of the triangle.
(ii) The perpendicular line segment from a vertex of a triangle to its opposite is called an $\underline { Q } $ of the triangle
(iii) A triangle has $\underline { R } $ altitudes and $\underline { S } $ medians

  1. $P-$ Altitude; $Q$- Median; $R-1$; $S-1$
  2. $P-$ Altitude; $Q$- Median; $R-3$; $S-3$
  3. $P-$ Median; $Q$- Altitude; $R-3$; $S-3$
  4. $P-$ Median; $Q$- Altitude; $R-2$; $S-3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Medians : The line segment from any vertex of a triangle to the midpoint of its opposite side is called medians of triangle. 

Altitude : The perpendicular drawn from a vertex to opposite side is called as altitude. 
A triangle has $3$ altitudes & $3$ medians.