CRE 3 - Geometric Centers | Geometry - Triangles
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1. CRE 3 - Geometric Centers | Geometry - Triangles
An equilateral triangle has a side of 18 cm. find the in-radius and circum-radius of the triangle (in cm).
(a) 3, 6
(b) 2√3, 4√3
(c) 3√3, 6√3
(d) None of these
Answer: Option
C
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Explanation :
In an equilateral triangle of side ‘a’.
Inradius = a/2√3
⇒ For the given equilateral triangle
Inradius = 18/2√3 = 3√3
Circumradius = 18/√3 = 6√3
Hence, option (c).
2. CRE 3 - Geometric Centers | Geometry - Triangles
In a triangle with sides 6, 8 and 10. Find the inradius and circumradius.
(a) 3, 6
(b) 4, 10
(c) 3, 5
(d) 2, 5
Answer: Option
D
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Explanation :
In the given triangle, sides form a pythagorean triple i.e., 102 = 62 + 82
∴ Given triangle is a right triangle.
⇒ Inradius = (a + b - c)/2 = (6 + 8 - 10)/2 = 2
⇒ Circumradius = c/2 = 10/2 = 5.
Hence, option (d).
3. CRE 3 - Geometric Centers | Geometry - Triangles
In a triangle ABC, AD is the bisector of ∠A. if AB = 12 cm, AC =16 cm and BD = 3 cm, what is the length of BC?
Answer: 7
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Explanation :
If AD is the angle bisector of ∠A
⇒ AB/AC = BD/CD (Angle bisector theorem)
⇒ 12/16 = 3/CD
⇒ CD = 4
∴ BC = BD + CD = 3 + 4 = 7
Hence, 7.
4. CRE 3 - Geometric Centers | Geometry - Triangles
In triangle PQR, angle bisectors of angles Q and R meet at point I. if ∠P = 80°, find ∠QIR (in degrees).
Answer: 130
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Explanation :
Given, QI is the angle bisector for ∠Q and RI is angle bisector for ∠R.
In ∆PQR, sum of all internal angles = 180°
⇒ 80° + 2x + 2y = 180°
⇒ x + y = 50° …(1)
In ∆QIR, sum of all internal angles = 180°
⇒ x + ∠QIR + y = 180°
⇒ ∠QIR + 50° = 180°
⇒ ∠QIR = 130°
Hence, 130.
5. CRE 3 - Geometric Centers | Geometry - Triangles
In triangle ABC, AD, BE are medians and G is the centroid. ∠AGE = 30°, AD =12, BE =18. Find the area of the triangle.
(a) 36
(b) 72
(c) 144
(d) Can't be determined
Answer: Option
B
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Explanation :
G is the centroid of ∆ABC and centroid divides the median in the ratio of 2 : 1.
∴ G divided AD in the ratio of 2 : 1 ⇒ AG = 2/3 × 12 = 8.
∴ G divided BE in the ratio of 2 : 1 ⇒ GE = 1/3 × 18 = 6.
⇒ Area of ∆ AGE = ½ × AG × GE × Sin30° = ½ × 8 × 6 × ½ = 12.
We also know that centroid divides the triangle in 6 smaller triangles of equal areas.
∴ Area(∆ABC) = 6 × Area(∆AGE) = 6 × 12 = 72.
Hence, option (b).
6. CRE 3 - Geometric Centers | Geometry - Triangles
O is the circumcenter of the triangle ABC with circum-radius 13 cm. Let BC = 24 cm and OD is perpendicular to BC. Then the length of OD is?
Answer: 5
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Explanation :
We know circumcenter is obtained by intersection of perpendicular bisectors of the three sides.
∴ OD is the perpendicular bisector of BC ⇒ BD = CD
Now, ∆BOD is a right triangle.
∴ BO2 = OD2 + BD2
⇒ 132 = OD2 + 122
⇒ OD = 5
Hence, 5.
7. CRE 3 - Geometric Centers | Geometry - Triangles
In ∆ABC, the medians AD and BE meet at G. The ratio of the areas of ∆BDG and the quadrilateral GDCE is?
(a) 3 : 5
(b) 1 : 4
(c) 2 : 3
(d) 1 : 2
Answer: Option
D
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Explanation :
We also that centroid divides the triangle in 6 smaller triangles of equal areas.
∴ Area(∆BDG) = Area(∆DGC) = Area(∆CGE) = a
⇒ Area(∆BDG) : Area(∆GDCE) = a : 2a = 1 : 2
Hence, option (d).
8. CRE 3 - Geometric Centers | Geometry - Triangles
In ∆ABC, AB = 3 cm, AC = 4 cm and BC = 5 cm. What is the length of AD, if D is the midpoint of side BC?
(a) 6.25 cm
(b) √(18.75) cm
(c) 2.5 cm
(d) √(12.5) cm
Answer: Option
C
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Explanation :
Since D is the midpoint of BC ⇒ AD is the median to BC
Using Apollonius theorem, we get
AB2 + AC2 = 2 × (AD2 + BD2 )
∴ 32 + 42 = 2 × (AD2 + BD2 )
∴ AD = 9 + 16 2 - 6.25 12.5 - 6.25 6.25
Alternately,
The sides of the given triangle form a pythagorean triplet.
∴ The given triangle is a right triangle with BC as the hypotenuse.
Since D is the midpoint of the hypotenuse, D is the circum-center of the right triangle hence D is equidistance from all three vertices.
In a right triangle, circumradius = half of the hypoteneuse.
∴ DA = DB = DC = 5/2 = 2.5 cm.
Hence, option (c).
9. CRE 3 - Geometric Centers | Geometry - Triangles
Sides of a triangle are 3, 6 and 9. Its circumcenter lies
Answer: Option
A
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Explanation :
92 > 62 + 32
Only in an obtuse triangle the square of the higest side is greater that the sum of the squares of the other two sides.
Hence, the given triangle is an obtuse triangle.
Cirucmcenter lies
∴ For the given triangle the circumcenter will lie outside the triangle.
Hence, option (a).
10. CRE 3 - Geometric Centers | Geometry - Triangles
In a triangle with sides 9, 12 and 15 cm, find the distance of orthocenter from all three vertices.
(a) 12
(b) 36
(c) 21
(d) None of these
(e) Cannot be determined
Answer: Option
C
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Explanation :
Since, 9, 12 and 15 form a pythagorean triplet, the given triangle is a right triangle.
Also in a right triangle the orthocenter lies on the right vertex.
∴ The distance of orthoceter from 3 vertices will be 9, 0 and 12.
⇒ The sum of the distances = 9 + 0 + 12 = 21
Hence, option (c).
11. CRE 3 - Geometric Centers | Geometry - Triangles
An equilateral triangle has a side of 10 cm. Find the area (in cm2 ) of the triangle formed by the centroid, circumcentre and incentre.
[Type in your answer as the nearest possible integer.]
Answer: 0
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Explanation :
In an equilateral triangle, the centroid, the circumcenter and the incenter are same.
Hence, the triangle formed by these three points will have no area i.e., 0.
Hence, 0.
12. CRE 3 - Geometric Centers | Geometry - Triangles
∆ABC circumscribes a circle of radius 4 cm. If the sides of ∆ABC are 13, 14 and 15 respectively, find the area of ∆ABC.
(a) 84
(b) 42
(c) 21
(d) 96
(e) None of these
Answer: Option
A
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Explanation :
Since ∆ABC circumscribes the circle, it means the circle is the incircle of ∆ABC.
We know that area of a triangle = r × s
where, r is the inradius and s is the semiperimeter of the triangle.
s = (13 + 14 + 15)/2 = 21
⇒ Area = 4 × 21 = 84
Hence, option (a).
13. CRE 3 - Geometric Centers | Geometry - Triangles
Triangle ABD is right-angled at B. On AD there is a point C for which AC = CD and AB = BC. The magnitude of angle DAB, in degrees, is:
(a) 67.5
(b) 60
(c) 45
(d) 30
Answer: Option
B
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Explanation :
As C is the midpoint of the hypotenuse AD, it is the circumcentre.
∴ AC = CD = BC
But AB = BC
∴ In ∆ BAC, AB = AC = BC
∴ ∆ BAC is an equilateral triangle.
∠DAB = ∠BAC = 60°
Hence, option (b).
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