This blog offers important geometry questions with solutions to help students improve their understanding of key concepts and problem-solving skills. The questions cover a range of topics, including Euclidean geometry, coordinate geometry, and trigonometry. Each question comes with a detailed solution, making it an excellent resource for students who want to excel in geometry.
Assignment:1
Que 1: Sides of ∆ABC AB=13 cm, BC=14 cm, CA=15 cm, then find the value of Sin(A/2)×Sin(B/2)×Sin(C/2)
(a) 8/65
(b) 1/20
(c) 4/35
(d) 5/49
Solution: r=4R.Sin(A/2).Sin(B/2).Sin(C/2)
△⇒Area of the triangle
s⇒Semi-Perimeter of the triangle
R⇒Length of Circumradius
r⇒Length of Inradius
s=(13+14+15)/2=21
△2=s.(s-a).(s-b).(s-c)
△= 84 cm2
r=△/s= 4 cm
R=abc/4△= 65/8 cm
Sin(A/2).Sin(B/2).Sin(C/2)=r/4R= 8/65 cm
Option:(a) is correct
Que 2: The lengths of the sides of a triangle are AB = 13 cm, BC = 14 cm, and CA = 15 cm. Find the value of Cos(A/2)× Cos(B/2)× Cos(C/2)
(a) 1
(b) 0.65
(c) 0.81
(d) 0.24
Solution: s=4R.Cos(A/2).Cos(B/2).Cos(C/2)
Cos(A/2).Cos(B/2).Cos(C/2)=s/4R= 42/65= 0.65
Option:(b) is correct
Que 3: In ∆ABC of side lengths a, b, and c. If abc=4s.(s-a) then find the value of a.
Here s is the semi-Perimeter of the triangle.
(a) 3.2
(b) 4.8
(c) 6
(d) Data, not sufficient
Solution: abc=2s.(2s-2a)
abc=(a+b+c)(b+c-a)
abc=[(b+c)2-a2]
abc=b2+c2-a2+2bc
b2+c2-a2=bc(a-2)
\frac{b^2+c^2-a^2}{2bc}=\frac{a-2}{2}.
Cos(A)=\frac{a-2}{2}.
-1<Cos(A)<1
-2<a-2<2
0<a<4
Option:(a) is correct
Que 4: What is the maximum number of isosceles triangles with integer side lengths that can be formed with a perimeter of 23 cm?
(a) 4
(b) 5
(c) 6
(d) 7
Solution: Let a, b, and c be the side lengths of a triangle.
a<P/2
a<11.5
a≥b≥c
11, 11, 1
11, 10, 2
11, 9, 3
11, 8, 4
11, 7, 5
11, 6, 6
10, 10, 3
10, 9, 4
10, 8, 5
10, 7, 6
9, 9, 5
9, 8, 6
9, 7, 7
8, 8, 7
Number of Isosceles triangles= 6
Option:(c) is correct
Que 5: In ∆ABC, BC=a, CA=b, and AB=c. Given 156×(b+c) =143×(c+a) =132×(a+b) then find Cos A: Cos B: Cos C=.
(a) 11:12:13
(b) 9:14:27
(c) 5:12:26
(d) 7:19:25
Solution: 12×13×(b+c) =11×13×(c+a) =11×12×(a+b)
Divide it by 11×12×13
(b+c)/11=(c+a)/12=(a+b)/13=k
b+c=11k
c+a=12k
a+b=13k
Add these three equations:
2(a+b+c)=36k
a+b+c=18k
a=7k
b=6k
c=5k
a:b:c=7:6:5
Cos(A)=\frac{b^2+c^2-a^2}{2bc}=\frac{1}{5}.
Cos(B)=\frac{c^2+a^2-b^2}{2ca}=\frac{19}{35}.
Cos(C)=\frac{a2+b2-c2}{2ab}=\frac{5}{7}.
Cos A: Cos B: Cos C=1/5 : 19/35: 5/7
Multiply it by 35
Cos A: Cos B: Cos C=7:19: 25
Option:(d) is correct
Que 6: What is the value of ∠BAC in △ABC, if point O is the orthocentre and ∠BOC is 110°?
(a) 110°
(b) 70°
(c) 100°
(d) 90°
Solution: ∠BAC+∠BOC=180°
∠BAC=70°
Option:(b) is correct
Que 7: What type of triangle is defined by the equation a2 + b2 + c2 = ab + bc + ca, where a, b, and c are the lengths of the sides of the triangle?
(a) Isosceles
(b) Equilateral
(c) Scalene
(d) Right-angled
Solution: 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
(a-b)2+(b-c)2+(c-a)2=0
a-b=b-c=c-a=0
a=b=c (Equilateral Triangle)
Option:(b) is correct
Que 8: What is the measure of ∠BAD in an isosceles △ABC, where AB = AC and ∠B= 35°, and AD is the median to the base BC?
(a) 70°
(b) 35°
(c) 110°
(d) 55°
Solution: AD is the angle bisector of ∠A
∠B=∠C=35°
∠A+∠B+∠C=180°
∠A=110°
∠BAD=55°
Option:(d) is correct
Que 9: What is the ratio AP: PB in an isosceles triangle ABC, where AB = AC and a circle passing through B and touching AC at its midpoint intersects AB at point P?
(a) 2:5
(b) 1:3
(c) 4:1
(d) 3:4
Solution: Let AB=AC=a
AM=a/2
AM2=AP.AB
a2/4=a.x
x=a/4
AP: PB=a/4:a-a/4=a/4:3a/4=1:3
Option:(b) is correct
Que 10: What is the value of sin 75° in an isosceles right triangle ABC, where ∠B=90° and D is a point inside the triangle with perpendiculars PD and QD drawn to sides AB and AC, respectively, such that AP = a cm, AQ = b cm, and ∠BAD=15°?
(a) 2b/√3a
(b) √3b/2a
(c) √3a/2b
(d) 2a/√3b
Solution: AB=BC, so ∠A=∠C=45°
In △AQD, Cos 30°=AQ/AD
AD=2b/√3
In △APD, Sin 75°=AP/AD= √3a/2b
Option:(c) is correct
Que 11: What type of triangle is it if the angle bisector of a vertex of the triangle bisects the opposite side?
(a) Right angled
(b) Isosceles
(c) Equilateral
(d) None
Solution: It is a property of an isosceles triangle that if an angle bisector is drawn from a vertex whose adjacent sides are equal, then it will bisect the opposite side.
This condition is also true for Equilateral Triangle. So options b and c both are correct.
Option:(d) is correct
Que 12: What is the value of AD2 + BD2 in an isosceles right-angled △ABC with ∠C=90°, where D is any point on the side AB?
(a) CD2
(B) 2CD2
(c) 3CD2
(d) 4CD2
Solution: Let AC=BC=a
Draw CE⟂AB, so AE=EB
AC2+BC2=AB2
2a2=(AD+BD)2
AD2+BD2+2.AD.BD=2a2 …….(i)
In △CED, CD2=CE2+DE2
In △CEA, AC2=CE2+AE2
AC2-CD2=AE2-DE2
⇒(AE+DE)(AE-DE)
Since AE=EB
⇒(BE+DE).AD
BD.AD=a2-CD2 ……..(ii)
From equations (i) and (ii):
AD2+BD2+2.AD.BD=2a2
AD2+BD2+2.(a2-CD2)=2a2
AD2+BD2=2.CD2
Option:(b) is correct
Que 13: What is the measure of ∠RPS in an acute-angled △PQR, where O and C are the orthocentre and circumcentre, respectively, P and O are joined and produced to meet the side QR at S, ∠PQS=60°, and ∠QCR=130°?
(a) 30°
(b) 35°
(c) 100°
(d) 60°
Solution: O is the intersection point of altitudes and C is the center of the circumscribed circle of the △PQR.
∠P=130°/2=65°
∠QPS=30°
∠RPS=65°-30°=35°
Option:(b) is correct
Que 14: What type of triangle is it if the circumcentre of the triangle lies outside it?
(a) Equilateral
(b) Acute angled
(c) Right angled
(d) Obtuse angled
Solution: Obtuse Angle triangle
Option:(d) is correct
Que 15: How many triangles can be formed by taking any three of the line segments out of segments of length 2 cm, 3 cm, 5 cm, and 6 cm?
(a) 3
(b) 2
(c) 1
(d) 4
Solution: a+b>c<a-b
Two Triangles⇒(3,5,6) and (2,5,6)
Option:(b) is correct