# How To Solve Geometry Questions Quickly

## 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

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.
Volume = πr²$\frac{h}{3} \$ = π(10)²$\frac{5}{3} \$ = $\frac{500}{3} \$πr³