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# 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.

## 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

**Options(a) **15°

**(b)**20°

**(c)**25°

**(d)**30°

**Explanation**

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=40^{0}, then ∠ACB =

**Options:(a) **140

**(b)**70

**(c)**100

**(d)**None of these

**Explanation**

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

**Explanation**

**area=**(semi perimeter) * inradius

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(r^{2})

rl = 2r^{2}

l = 2r

From the diagram, we can see that r^{2} + h^{2} = l^{2}. Since h = 9 and l = 2r, some substitution yields

r^{2} + 9^{2} = (2r)^{2}

r^{2} + 81 = 4r^{2}

81 = 3r^{2}

27 = r^{2}

B = π(r^{2}) = 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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- Heights and Distances – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
- Perimeter Area and Volume – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
- Coordinate Geometry – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
- Venn Diagrams – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
- Set Theory – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts

- Heights and Distances – Questions |

Formulas |

How to Solve Quickly |

Tricks & Shortcuts - Perimeter Area and Volume – Questions |

Formulas |

How to Solve Quickly |

Tricks & Shortcuts - Coordinate Geometry – Questions |

Formulas |

How to Solve Quickly |

Tricks & Shortcuts - Venn Diagrams – Questions |

Formulas |

How to Solve Quickly |

Tricks & Shortcuts - Set Theory – Questions |

Formulas |

How to Solve Quickly |

Tricks & Shortcuts

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