# Tips And Tricks And Shortcuts For Geometry

## Geometry Tips and Tricks and Shortcuts.

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.
•

#### Tips and Tricks 1

When Perimeter is odd

#### Tips and Tricks 2

When perimeter is even

## Tricks and Tips 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 on 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? Options
a)
$\frac{Π}{3} \$

b) $\frac{Πr^2}{3} \$

c) $\frac{Πr^2}{4} \$

d) $\frac{2Πr}{4r} \$

Explanations

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)

## Geometry Shortcuts Tips and Tricks and Shortcuts

### 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)