How To Solve Geometry Questions Quickly

How To Solve Practical Geometry Problems

From the information provided on this page we will be learning about How to solve Geometry Questions Quickly of various types and how to apply problem solving techniques. Having knowledge about this will help you in solving more numbers of problems in less time. 

How to Solve Geometry Questions Quickly

Steps to Analyze and Solve Problems Quickly.

  • Determine what the problem is asking about and then try to solve the problem.
  • In geometry problems it is very much necessary to draw sketch of the problem solution as it helps you to reach to your solution quickly.
  • Pay attention to units make sure that the units used are converted accordingly and all Numbers have same units.
  • Solve the math you may use pythagoras theorem or other theorem’s to solve it
  • Analyze your results and make sure they are correct.

Type 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
how to solve geometry

(b) 20°
(c) 25°
(d) 30°

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 =
How to solve geometry Questions

(b) 70
(c) 100
(d) None of these


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
how to slove geometry questions

(b) √50
(c) 7
(d) 8


By the pythagoras theorem we get BC= 25. Let BD = x,
Triangle ABD is similar to triangle CBA ⇒
\frac{AD}{15} \ = \frac{x}{15} \
and also triangle ADC is similar to triangle ACB ⇒
\frac{AD}{20} \ = \frac{25-x}{15} \ .
From the 2 equations, we get x= 9  and DC= 16.
We know that area= (semi perimeter) * inradius
For triangle ABD, Area = \frac{1}{2}\times BD\times AD = \frac{1}{2} \times 12\times 9 =54
semi perimeter =\frac{15+9+12}{2} = 18.
On using the above equations we get inradius, r = 3.
PQ = R+r = 7cm.

Type 2: How to solve Geometry Questions of various shapes.

Ques. 4

The lateral area is twice as big as the base area of a cone. If the height of the cone is 9, what is the entire surface area (base area plus lateral area)?

(b) 90π
(c) 54π


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
how to solve geometry questions

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?

(b) 500πcm³
(c) 500/3 πcm³
(d) 50πcm³


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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