Tips And Tricks And Shortcuts For Geometry Ques

Tips and Tricks and Shortcuts for Geometry Ques and Answers

From a given perimeter how many triangles with integral sides are possible?

We can solve this manually. But with the help of a tips and tricks and shortcut that is discussed in this page of PrepInsta

  • We can solve this question within seconds. Generally there are two cases for these type of questions.
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Tips and Tricks 1

When Perimeter is odd

Tips and Tricks 2

When perimeter is even

Tips and Tricks for Geometry 1 – Details

  • How many triangles with integral sides are possible for perimeter P where P is even

Solution – In this case, total number of triangles will be the nearest integer to \frac{P^2}{48} \

Tips and Tricks and Shortcuts for Geometry  2 – Details

  • How many triangles with integral sides are possible for perimeter P where P is odd

Solution – In this case, total number of triangles will be the nearest integer to \frac{(P+3)^2}{48} \

tips and tricks and shortcuts for geometry

Tips and Tricks to Solve Geometry

Question 1

ABCD is a square. AD is tangent to circle with radius r and OE = ED. Then what is the ratio of the area of circle to the area of square? tips and tricks for geometry Options a) \frac{Π}{3} \ b) \frac{Πr^2}{3} \ c) \frac{Πr^2}{4} \ d) \frac{2Πr}{4r} \ Explanations OD2 = OA2+ AD2 (2r)2 = r2 + AD2 Thus PQ, which is also the side of square, is equal to r√3. The area of square becomes:  3r2 Hence the ratio of the area of circle to square is: \frac{area  of circle}{area  of square} \ = \frac{πr^2}{3r^2} \ = \frac{π}{3} \ Correct Option (A)

Shortcuts, Tips and Tricks to Solve Geometry

Question 2

If in a triangle ABC, \frac{cos A}{a} \ = \frac{cos B}{b} \ = \frac{cos c}{c} \ , then what can be said about the triangle ?

Options
A)
Right angled triangle
B)
Isosceles triangle
C) Equilateral triangle
D) Nothing can be inferred

Explanations

From the sine rule of triangle we know, \frac{a}{sinA} \ = \frac{b}{sinB} \ \frac{c}{sinC} \ = k

Therefore, a = k(sin A), b = k(Sin B) and c = k(Sin C)

Hence, we can rewrite, \frac{cos A}{a} \ = \frac{cos B}{b} \ = \frac{cos C}{c} \ as \frac{cos A}{k Sin A} \ = \frac{cos A}{k Sin B} \ = \frac{cos C}{k Sin C} \

or Cot A= Cot B= Cot C
A = B = C, Hence the triangle is equal.

Correct Option (C)

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