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Published on: 03/08/2019
The Triangle and Its Properties
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Questions + Answers key
Take MCQ Mathematics Test
1.
Draw rough sketch of altitude from A to \(\bar { BC } \) for the following given triangles.
2.
Draw rought sketches of \(\triangle PQR\),where QE is a median.
3.
Show that the sum of the exterior angles of \(\triangle\)ABC as shown in the figure is 360°.
4.
Look at following figures and classify each of the triangles according to its
5.
Verify by drawing a diagram, if the median and altitude of an isosceles triangle can be same.
6.
Least number of possible acute angles in a triangle is:
0
1
2
3
7.
In a triangle ABC,\(\angle \)A+, \(\angle \)B+, \(\angle \)C=
360°
90°
180°
60°
8.
In a\(\triangle ABC\), AD is the bisector of \(\angle A\) meeting BC at D,CF⊥AB and E is the mid-point of AG. Then, median of the triangle is
AD
BE
FC
DE
9.
If the exterior angle of a triangle is 130° and its interior opposite angles are equal, then measure of each interior opposite angle is
55°
65°
50°
60°
10.
The sides of a triangle have length (in cm) 10, 6.5 and a, where a is a whole number. The minimum value that a can take is
6
5
3
4
11.
Two squares are congruent, if they have same_________.
12.
Two rectangles are congruent, if they have same ________ and________.
13.
Every triangle has atmost____________obtuse angles.
14.
If one angle of a triangle is equal to the sum of other two, then the measure of that angle is_____________.
15.
In the following figure, value of x is___________.
16.
The difference between the length of any two sides of a triangle is smaller than the length of third side.
17.
It is possible to have a triangle in which each angle is less than 60°.
18.
It is possible to have a right angled equilateral triangle.
19.
Sum of the measures of three angles of a triangle is greater than 180°.
20.
Sum of any two angles of triangle is always greater than the third angle.
21.
What is the sum of angles of a triangle?
22.
An exterior angle of a triangle is 140°. What is the measure of its interior adjacent angle?
23.
In a right triangle, what is the measure of its greatest angle?
24.
Find angles x and y in each figure.
1.
In the given figure, altitude can be drawn as below:
AL = Altitude from A to BC
2.
In the adjacent figure, we have \(\triangle PQR\) We know that, a median connects a vertex of a triangle to the mid-point of the opposite side. On joining Q and mid-point of PR, i.e. E.We get the required median QE.
3.
Since\(\angle\) b and \(\angle\) d form a linear pair.
\(\therefore\) \(\angle\) b + \(\angle\) d = 180°
Similarly, \(\angle\)c + \(\angle\)e = 180°
and \(\angle\)a + \(\angle\)f = 180°
Adding the angles on both sides, we get
\(\angle\)b + \(\angle\)c + \(\angle\)a + \(\angle\)d +\(\angle\)e + \(\angle\)f
= 180° + 180° + 180°
\(\Rightarrow\) (\(\angle\)a +\(\angle\)b + \(\angle\)c) + (\(\angle\)d + \(\angle\)e + \(\angle\)f) = 540°
But \(\angle\)d + \(\angle\)e + \(\angle\)f= 180°
[\(\because\) Sum of angles of a triangle is 180°.]
\(\therefore\) (\(\angle\)a + \(\angle\)b + \(\angle\)c) + (180°) = 540°
\(\Rightarrow\) (\(\angle\)a + \(\angle\)b + \(\angle\)c) = 540° - 180°
\(\Rightarrow\)\(\angle\)a + \(\angle\)b +\(\angle\)c = 360°
i.e. Sum of exterior angles of \(\triangle\)ABC is 360°.
4.
We know that, On the basis of sides, a triangle is called
(i) scalene triangle, if all three sides of triangle are unequal.
(ii) isosceles triangle, if any rwo sides are equal.
(iii) equilateral triangle, if all three sides are equal.
On the basis of angles, a triangle is called
(i) acute angled triangle, if each angle is less than 90°.
(ii) right angled angled triangle, if one angle is a right angle.
(iii) obtuse angled triangle, if one angle is greater than 90°. Now,
Now,
(i) (a) In \(\triangle ABC\),AC=BC=8 cm
i.e. rwo sides are equal
Therefore,\(\triangle ABC\) is an isosceles triangle.
(b) Also, all the angles of \(\triangle ABC\) are less than 90°.
Therefore,\(\triangle ABC\) is an acute angled triangle.
(ii) (a) In \(\triangle PQR\),\(PQ\neq QR\neq RP \) [given]
i. e. all three sides are unequal.
Therefore,\(\triangle PQR\) is a scalene triangle.
(b) Also,\(\angle R=90^0\) [given]
Therefore, \(\triangle PQR\) is a right angled triangle.
(iii) (a) In \(\triangle LMN\),LN=MN=7cm
i.e. rwo sides are equal.
Therefore,\(\triangle LMN\) is an isosceles triangle.
(b) Also,\(\angle N>90^0\) [given]
Therefore,\(\triangle LMN\) is an obtuse angled triangle.
(iv) (a) In \(\triangle RST\),RS = ST = TR = 5.2 cm [given]
i.e. all three sides are equal.
Therefore,\(\triangle RST\) is an equilateral triangle.
(b) Also, all the angles of \(\triangle RST\) are acute.
Therefore,\(\triangle RST\) is an acute angled triangle.
(v) (a) In \(\triangle ABC\),AB = BC = 3 cm
i.e. rwo sides are equal.
Therefore,\(\triangle ABC\) is an isosceles angled triangle.
(b) Also,\(\angle B>90^0\)
Therefore, \(\triangle ABC\) is an obtuse angled triangle.
(vi) (a) In \(\triangle PQR\),PQ = QR = 6 cm [given]
i.e. rwo sides are equal.
Therefore, \(\triangle PQR\) is an isosceles triangle.
(b) Also,\(\angle Q=90^0\) [given]
Therefore \(\triangle PQR\) is a right angled angled triangle.
5.
Draw a line segment Be. By paper folding locate the perpendicular bisector of BC.The folded crease meets BC at D, its mid-point.
Take any point A on this perpendicular bisector. Join AB and Ae. Thus, the triangle obtained is an isoscelesMBC in which AB = AC.
Since, D is the mid-point of BC, so, AD is its median. Also, AD is perpendicular bisector of Be. So, AD is the altitude of \(\triangle ABC\).
Thus, it is verified that the median and altitude of an isoscelestriangle are same.
6.
(c)
2
7.
(c)
180°
8.
(c)
FC
9.
(b)
65°
10.
(d)
4
11.
( )
Side
12.
( )
Length, breadth
13.
( )
one
14.
( )
90°
15.
( )
90°
16.
(a)
17.
(b)
18.
(b)
19.
(b)
20.
(b)
21.
( )
180°
22.
( )
40°
23.
( )
90°
24.
(i) x + y = 1200 ...(1)
The exterior angle of a triangle is equal to the sum of its two interior opposite angles
x + y + y = 1800
Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of a triangle is 1800
\(\Rightarrow\)x + 2y = 1800 ... (2)
Subtracting equation (1) from equation (2),
y = 60°
Put y = 60° in equation (1),
x + 60° = 120°
\(\Rightarrow\)x = 120° - 60°
\(\Rightarrow\)x = 60°
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