Olympiad Mathematics - Triangles
Exam Duration: 45 Mins Total Questions : 30
Which triangle is formed by AB = 3 cm, BC = 4 cm and AC = 8 cm?
- (a)
A scalene triangle
- (b)
An isosceles triangle.
- (c)
An equilateral triangle.
- (d)
No triangle is formed.
The sum of any two sides is greater than the third side.
Since AB + BC < AC, no triangle is formed.
If one angle is the average of the other two angles and the difference between the greatest and least angles is 60°, which triangle is formed?
- (a)
An isosceles triangle
- (b)
An equilateral triangle.
- (c)
A right angled triangle
- (d)
A right angled isosceles triangle.
Let the least angle be x°,
The greatest angle = x°+ 60°
Third angle =\(={x+x+60^o\over 2}=x+30^o\)
We have,
x + x + 30° + x + 60° = 180°
\(\Rightarrow\)3x + 90° = 180° \(\Rightarrow\) x = 30°
\(\therefore\) The angles are 30°, 60° and 90°.
Since one of the angles is 90°, the triangle formed is a right angled triangle.
If the two legs of a right angled triangle are equal and the square of the hypotenuse is 100 sq units, what is the length of each leg?
- (a)
10 units
- (b)
5\(\sqrt{2}\) units
- (c)
10\(\sqrt{2}\) units
- (d)
15 units
We have, by Pythagoras' theorem,
X2 + X2 = 100
\(\Rightarrow\)x =\( =\sqrt{50}=\sqrt{25\times 2}=5\sqrt2\)
In a \(\triangle\)ABC,if AB + BC= 10 cm,BC+ CA= 12 cm, CA + AB = 16 cm, what is the sum of the lengths of its sides?
- (a)
19 cm
- (b)
17 cm
- (c)
38 cm
- (d)
30 cm
Which of the following are the angles in a right angled triangle other than the right angle?
- (a)
Acute angles
- (b)
Obtuse angles
- (c)
Right angles
- (d)
Reflex angles
In a \(\triangle\) ABC, if \(\angle\)A = \(\angle\)B +\(\angle\)C , what is the measure of \(\angle\)A
- (a)
60°
- (b)
45°
- (c)
90°
- (d)
135°
If the angles of a triangle are in the ratio 1 : 2 : 7, what type of a triangle is it?
- (a)
An acute angled triangle.
- (b)
An obtuse angled triangle.
- (c)
A right angled triangle
- (d)
A right angled isosceles triangle
Let the measures of the angles be 1x, 2x and 7x.
We have, 1x + 2x + 7x = 180°
\(\Rightarrow\) x = 18°
The angles are 18°, 36° and 126°.
\(\therefore\) The triangle is obtuse angled.
In a \(\angle\)ABC, if \(\angle\)B is an obtuse angle, which is the longest side?
- (a)
AB
- (b)
BC
- (c)
AC
- (d)
Either (A) or (B)
The top of a broken tree touches the ground at a distance of 15 m from its base. If the tree is broken at a height of 8 m from the ground, what is the actual height of the tree?
- (a)
20 m
- (b)
25 m
- (c)
30 m
- (d)
17 m
Since\(\triangle\) ABC is right angled,
AC2= AB2+ BC2
\(\Rightarrow\)AC2= 82+ 152= 289
\(\Rightarrow\) AC = \(\sqrt{289}\) = 17 m
\(\therefore\)Actual length of tree
=AB +AC =25 m
What is the ratio in which the centroid of a triangle divides the medians?
- (a)
1 : 2
- (b)
1 : 3
- (c)
2: 1
- (d)
3: 1
The centroid of a triangle is the point of concurrence of which of these?
- (a)
Angle bisectors
- (b)
Perpendicular bisectors
- (c)
Altitudes
- (d)
Medians
In\(\triangle\) ABC, D is the midpoint of BC and G is the centroid of the triangle. If GD = 2 cm, what is the length of AD?
- (a)
4 cm
- (b)
6 cm
- (c)
2 cm
- (d)
8 cm
The centroid divides the median in the ratio 2 : 1.
\(\therefore\) AG:GD =2: 1
So, AG = 2 X GD = 2 x 2 = 4cm
AD = AG + GD = 4 + 2 = 6 cm
In \(\triangle\)XYZ, XP is the median.Which of the following is correct?
- (a)
XP = XY
- (b)
YP = PZ
- (c)
XP = XZ
- (d)
XY = XZ
Since XP is the median, P is the midpoint of YZ. So,YP= PZ.
The exterior angle of a triangle is 110o. If one of the interior opposite angles is 55o, what is the measure of the other?
- (a)
45o
- (b)
65o
- (c)
55o
- (d)
35o
In \(\triangle\) PQR.\(\angle\)Q = 90o. Which of the following is the longest side?
- (a)
RQ
- (b)
PQ
- (c)
PR
- (d)
Either (A) or (B)
From the following figure, what are the respective values of x and y?
- (a)
80°,60°
- (b)
60°,40°
- (c)
60°,80°
- (d)
40°,60°
600 & yare vertically opposite angles which are equal
\(\Rightarrow\)y = 600
In the triangle,
x +60°+40° = 180°
(Angle sum property)
\(\Rightarrow\)x = 180° -100° = 80°
Angles of a triangle are (x + 10o), (x + 40o) and (2x - 30°) . What is the value of x?
- (a)
30°
- (b)
40°
- (c)
20°
- (d)
10°
Given that the angles of the triangle are (x + 10°), (x + 40°) and (2x - 30°).
Sum of the angles of a triangle = 1800
\(\Rightarrow\)x + 10°+ x + 40° + 2x - 30° = 180°
\(\Rightarrow\)4x = 160°
\(\Rightarrow\) x = 40°
In the figure given, what are the values of \(\angle\) b, \(\angle\)c and \(\angle\)a respectively?
- (a)
18°,70° and 92°
- (b)
92°,70° and 18°
- (c)
70°,92° and 18°
- (d)
70°,18° and 92°
Given AB = BD
\(\Rightarrow\) \(\angle\)BAD = \(\angle\)BDA = 35°
\(\angle\)b = \(\angle\)BDA +\(\angle\)BAD
\(\Rightarrow\) \(\angle\)b = 35° + 35° = 70°
Also given AC = CE
\(\Rightarrow\)\(\angle\)CAE = \(\angle\)CEA = 46°
Using exterior angle property,
\(\Rightarrow\)\(\angle\)c = \(\angle\)CAE + \(\angle\)CEA
= 46° + 46° = 92°
In \(\triangle\)ABC, \(\angle\)a+\(\angle\)b+\(\angle\)c= 180°
\(\Rightarrow\) \(\angle\)a = 1800-\(\angle\)b-\(\angle\)c
\(\Rightarrow\) \(\angle\)a = 180°-70°-92° = 18°
\(\therefore\) \(\angle\)a = 18°, \(\angle\)b = 70° and \(\angle\)c = 920
Find the angles x and y respectively.
- (a)
x = 47°, Y = 25°
- (b)
x = 27°, Y = 45°
- (c)
x = 45°,Y = 27°
- (d)
x = 25°, Y = 47°
Given BE IICD, X° = 45°.
(Corresponding angles)
In \(\triangle\)ACD, X° + y° + 108° = 180°
(Angle sum property.)
\(\Rightarrow\)45°+y+108° = 180°
\(\Rightarrow\)y = 180° -153° = 27°
\(\therefore\) x = 45°, y = 27°
Find the value of x.
- (a)
106°
- (b)
53°
- (c)
26°
- (d)
52°
In \(\triangle\)ABC, \(\angle\)ACB =\(\angle\)1800-
(35°+ 39°) = 180° -74° = 106°
AE and BD intersect at C
\(\Rightarrow\)\(\angle\)DCE = \(\angle\)ACB = 106°
(Vertically opposite angles)
\(\Rightarrow\)x = 180°-(106°+48°) = 26°
In the figure (not drawn to scale), ADF and BEF are triangles and EC = ED, find y.
- (a)
90°
- (b)
91°
- (c)
92°
- (d)
93°
In\(\triangle\)CED,
CE= ED
\(\therefore\) \(\angle\)EDC = \(\angle\)ECD
[Angles opposite to equal sides are equal]
\(\Rightarrow\) \(\angle\)ECD = 28°
Also, \(\angle\)ECD = \(\angle\)BCA (Vertically opposite angles)
\(\Rightarrow\) \(\angle\)BCA = 28°
In \(\triangle\)BCA,
y = 62° + 28° [Exterior angle property]
\(\Rightarrow\)y = 90°
In a \(\triangle\) ABC, which of the given conditions holds?
- (a)
AB - BC > CA
- (b)
AB + BC < CA
- (c)
AB - BC < CA
- (d)
AB + CA < BC
In the figure (not drawn to scale), EFA is a right-angled triangle with \(\angle\)EFA = 90° and FGB is an equilateral triangle, find y - 2x.
- (a)
2°
- (b)
8°
- (c)
17°
- (d)
20°
In \(\triangle\) FGC, \(\angle\)GCF = 92° (given)
As we know, \(\angle\)CGF = 60°
(Angle of equilateral triangle)
\(\therefore\) x + 60° + 92° = 180°
\(\Rightarrow\) x = 180° - 152° = 28°
Now, in\(\triangle\)BCF, \(\angle\)CBF = 60°
\(\angle\)FCB = 180° - 92° (Linear pair)
\(\Rightarrow\) \(\angle\)FCB = 88°
\(\therefore\) \(\angle\)BFC + 88° + 60° = 180°
(Angle sum property)
\(\Rightarrow\)\(\angle\)BFC = 180° - 148° = 32°
And, \(\angle\)AFE = 90°
\(\Rightarrow\)Y + 32° = 90°\(\Rightarrow\) y = 90° - 32° = 58°
\(\therefore\) y - 2x = 58° - 2 x 28° = 58° - 56° = 2°
In the figure (not drawn to scale), ABC and DEF are two triangles, CA is parallel to FD and CFBE is a straight line. Find the value of x + y
- (a)
185°
- (b)
134°
- (c)
148°
- (d)
176°
\(\angle\)FCA = \(\angle\)BFD (Corresponding angles)
\(\Rightarrow\)x = 51°
Now, in \(\triangle\)ABC
y = 51 ° + 83°(Exterior angle property)
\(\Rightarrow\) y = 134°
SO,x + y = 51° + 134° = 185°
In a \(\triangle\)ABC, if AB + BC = 10 cm, BC + CA = 12 cm, CA + AB = 16 cm, then the perimeter of the triangle is_________________
- (a)
19 cm
- (b)
17 cm
- (c)
28 cm
- (d)
22 cm
It is given that,
AB + BC = 10 cm..................(i)
BC + CA = 12 cm..................(ii)
CA + AB = 16 cm...................(iii)
Adding (i), (ii) and (iii); we get,
2(AB + BC + CA) = 10 + 12 + 16
\(\Rightarrow\)AB + BC + CA = 19 cm.
A 26 m long ladder reached a window 24 m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall
- (a)
10m
- (b)
20m
- (c)
5 m
- (d)
25 m
In \(\triangle\)PRQ,
PR2 = PQ2 + QR2
(By Pythagoras theorem)
(26)2 = (24)2 + QR2
or QR2 = 676 - 576 = 100
\(\Rightarrow\)QR = \(\sqrt{100}\Rightarrow\) QR = 10
\(\therefore\)The distance of the foot of the ladder from the wall is 10 m.
A 34 m long ladder reached a window 16 m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.
- (a)
40 m
- (b)
30 m
- (c)
50 m
- (d)
10m
Let AB = length of the ladder, AC = height of the window
In \(\triangle\)ABC,
(AB)2 = (AC)2 + (BC)2
\(\Rightarrow\) (34)2= (16)2+ BC2
or BC2 = (34)2 - (16)2
\(\Rightarrow\) BC2 = 1156 - 256 = 900
\(\therefore\) BC = \(\sqrt{900}\) = 30 m
Mrs Kaushik gives a problem to her students. Find the perimeter of a rectangle whose length is 28 cm and diagonal is 35 cm
What will be the correct answer?
- (a)
90 cm
- (b)
45 cm
- (c)
89 cm
- (d)
98 cm
ABCD is a rectangle
In \(\triangle\)ACB,
AC2 - AB2 + BC2
(By Pythagoras theorem)
(35)2 = (28)2 + BC2 or BC2
= (35)2 - (28)2
\(\Rightarrow\) BC2 = 1225 - 784 \(\Rightarrow\) BC2 = 441
\(\therefore\) BC =\(\sqrt{144}\) = 21 cm
\(\therefore\) Perimeter of rectangle = 2 x (28 + 21) cm
= 2 x (49) cm = 98 cm
If Y is five times of x, find the values of x,y and z.
- (a)
x y z 20° 80° 140° - (b)
x y z 30° 80° 140° - (c)
x y z 20° 100° 160° - (d)
x y z 30° 100° 160°
As, Y = 5x
\(\therefore\)In\(\triangle\)RQS,
x + Y + 60° = 180° (Angle sum property)
\(\Rightarrow\) x + 5x + 60° = 180°
\(\Rightarrow\) 6x = 180° - 60° = 120° \(\Rightarrow\)x = \({120^o\over 6}\) = 200
\(\therefore\)Y = 5 x 20° = 100°
Also \(\angle\)QRS + \(\angle\)QSR = z
(Exterior angle property)
\(\Rightarrow\)z = 60° + 100° = 160°
State 'T' for true and 'F' for false.
(i) In the given right-angled triangle ABC, \(\angle\)B = 65°, \(\angle\)C = 25°, then AB2= BC2 + CA2.
(ii) The length of the third side of a triangle cannot be smaller than the difference of the lengths of the other two sides.
(iii) A triangle can have only one median.
- (a)
(i) (ii) (iii) F F T - (b)
(i) (ii) (iii) F T F - (c)
(i) (ii) (iii) F T T - (d)
(i) (ii) (iii) F F F
(i) In the given right angled triangle,
BC2 = AB2 + AC2
(iii) A triangle can have three medians.