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Geometry Questions Solving Techniques
From the information provided on this page we will be learning about How to solve Geometry Questions Quickly of various types. For your knowledge
Geometry is all about the shapes and properties for the Plane Geometry and Solid Geometry.


Types 1: How to Solve Geometry Questions Quickly related to Lines and Angles
Ques. 1
In the figure above, AB = BC = CD = DE = EF = FG = GA. Then ∠DAE is approximately
Options
(a) 15°
(b) 20°
(c) 25°
(d) 30°
Explanations
Let us assume, DAE = x
Triangle ABC is isosceles as AB = BC –> BCA = CAB = x
Hence, CBD = CAB + BCA = x + x = 2x ………….. [External angle of triangle ABC]
Triangle BCD is isosceles as BC = CD –> CBD = CDB = 2x
Hence, DCE = DAE + CDA = x + 2x = 3x ………….. [External angle of triangle ACD]
Triangle CDE is isosceles as CD = DE –> DCE = DEC = AED = 3x
Similarly, ADE = EFD = AEF + DAE = EGF + DAE = (DAE + GFA) + DAE = DAE + DAE + DAE = 3x
Hence, in triangle ADE, ADE + DAE + AED = 3x + x + 3x = 7x
Hence, 7x = 180 —> x = 180/7 = 25.7.. ≈ 25
Correct Option (C)
Ques. 2
In triangle DEF shown below, points A, B and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If ∠D=400, then ∠ACB =
Options:
(a) 140
(b) 70
(c) 100
(d) None of these
Explanations
Let the angle E be x in triangle (AEC),
then angle AEB= 180-2*x. Then in triangle DEF, angle F=180-(40+x).
Now in triangle BCF, angle BCF=2*x-100.
Now,angle ACB= 180-(180-2*x+2*x-100)=100
Ques. 3
In the above figure, ACB is a right-angled triangle. CD is the altitude. Circles are inscribed within the ΔACD and ΔBCD. P and Q are the centres of the circles. The distance PQ is
Options
(a) 5
(b) √50
(c) 7
(d) 8
Explanations
semi perimeter =\frac{15+9+12}{2} = 18.
On using the above equations we get inradius, r = 3.
Type 2: How to solve Geometry Questions of various shapes.
Ques. 4
Options
(a) 81π
(b) 90π
(c) 54π
(d) 9π
Explanation:
Lateral Area = LA = π(r)(l) where r = radius of the base and l = slant heightLA = 2B
π(r)(l) = 2π(r2)
rl = 2r2
l = 2r
From the diagram, we can see that r2 + h2 = l2. Since h = 9 and l = 2r, some substitution yields
r2 + 92 = (2r)2
r2 + 81 = 4r2
81 = 3r2
27 = r2
B = π(r2) = 27π
LA = 2B = 2(27π) = 54π
SA = B + LA = 81π
Correct Options (A)
Ques. 5
You are given a right circular cone with height 5 cm. The radius is twice the length of the height. What is the volume?
Options
(a) 100πcm³
(b) 500πcm³
(c) 500/3 πcm³
(d) 50πcm³
Explanation:
You are given a right circular cone with height 5. The radius is twice the length of the height.
Height =5cm. The radius is twice of the height.
You are given a right circular cone with height 5. The radius is twice the height.
Radius = 2* Height
so the radius is 10 cm
Volume = πr² \frac{h}{3} \ = π(10)² \frac{5}{3} \ = \frac{500}{3} \ πr³
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