# How To Solve Geometry Questions Quickly

## How To Solve Practical Geometry Problems ## 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=400, 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

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)?

Options
(a)
81π
(b) 90π
(c) 54π
(d)

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.
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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