How To Solve Geometry Ques Quickly

August 19, 2019

How to Solve Geometry Ques. Quickly

Geometry is all about the shapes and properties for the Plane Geometry and Solid Geometry.

How to solve page of PrepInsta is the best pages for preparation for Geometry.
For formula of Geometry please visit formula page.

Types 1: How to Solve Geometry Ques. Quickly

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} \ * BD* AD = \frac{1}{2} \ * 12* 9 =54<br />semi perimeter = \frac{15+9+12}{2} \

= 18.
On using the above equations we get inradius, r = 3.

PQ = R+r = 7cm.

Type 2: Solve Geometry Prob's Quickly

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) 9πExplantions
ateral 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

Explanations

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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