## GMAT - Data Sufficiency - Geometry

LEAP Administrator
11 months ago
**Data Sufficiency** |
**Geometry**

### The figure above represents an L-shaped garden. What is the value of k? (more)

The figure above represents an L-shaped garden. What is the value of k?

(1) The area of the garden is 189 square feet.

(2) The perimeter of the garden is 60 feet.

Area of the shaded portion = area of larger square - area of smaller square = 15^{2} - (15 - k)^{2} = 30k - k^{2} .... (a)

Perimeter of shaded portion = 2*15 + 2*k + 15 - k + 15 - k = 60

Perimeter is independent of k.. (b)

(1) The area of the garden is 189 square feet.

from equation (a)

189 = 30k - k^{2}

k^{2} - 30k + 189 = 0

k^{2} - 21k - 9k + 189 = 0

k (k - 21) - 9 (k - 21) = 0

k = 21 or k = 9

but k has to be less than 15

thus k = 9

sufficient

(2) The perimeter of the garden is 60 feet.

from equation (b)

perimeter is independent of k. Insufficient.

LEAP Administrator
11 months ago
**Data Sufficiency** |
**Geometry**

### The length of the edging that surrounds circular garden K is 1/2 the length of the edging that surrounds circular garden G. What is the area of garden K (Assume that the edging has negligible width.) (more)

The length of the edging that surrounds circular garden K is 1/2 the length of the edging that surrounds circular garden G. What is the area of garden K (Assume that the edging has negligible width.)

(1) The area of G is 25π square meters.

(2) The edging around G is 10π meters long.

Given

Circumference of Garden K = 1/2 Circumeference of Garden G

2πr_{k} = 1/2 2π r_{g}

r_{k} = 1/2 r_{g} .. (i)

Statement 1:

The area of G is 25π square meters.

πr_{g}^{2} = 25π

r_{g} = 5

r_{k} = 5/2 = 2.5

area of k = πr_{k}^{2} = π2.5^{2} = 6.25π

Sufficient

Statement 2:

The edging around G is 10π meters long.

2πr_{g} = 10π

r_{g} = 5

r_{k} = 2.5

Same answer as statement 1

sufficient

LEAP Administrator
11 months ago
**Data Sufficiency** |
**Geometry**

### If arc PQR above is a semicircle, what is the length of diameter PR (more)

If arc PQR above is a semicircle, what is the length of diameter PR?

(1) a=4

(2) b=1

In triangle PQR, angle Q = 90^{0} (Angle subtended by the diameter at the circumference)

Using pythagorus theorem in all three triangles

PQ^{2} + QR^{2} = PR^{2} .. (i)

PQ^{2} = a^{2} + 2^{2} = a^{2} + 4 ... (ii)

QR^{2} = b^{2} + 2^{2} = b^{2} + 4 ... (iii)

replacing PQ and QR in equation 1, and replacing PR = a + b

a^{2} + 4 + b^{2} + 4 = (a + b)^{2}

a^{2} + b^{2} + 8 = a^{2} + b^{2} + 2ab

ab = 4 ... (iv)

Statement 1:

a = 4

from (iv)

b = 4/a = 4/4 = 1

diameter = a+b = 1 + 4 = 5

sufficient

Statement 2:

b = 1

from (iv)

a = 4/b = 4/1 = 4

diameter = a+b = 4 + 1 = 5

sufficient

LEAP Administrator
11 months ago
**Data Sufficiency** |
**Geometry**

### The hypotenuse of a right triangle is 10 cm (more)

The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.

(2) The 2 legs of the triangle are of equal length.

Given hypotenuse = 10cm

let other sides of the triangle be x and y,

By Pythagorus Theorem -

Statement 1:

The area of the triangle is 25 square centimeters

Since we know that in a right angle triangle area can be written as = 1/2 * product of two sides that are not hypotenuse

we can calculate the values of x and y from the equation using a calculator which would be

Perimeter would be

sufficient

Statement 2:

The 2 legs of the triangle are of equal length.

We can find out the lengths of 2 sides by using pythagorus theorem and get the same result in statement 1

Sufficient

LEAP Administrator
11 months ago
**Data Sufficiency** |
**Geometry**

### What is the area of triangular region ABC above? (more)

What is the area of triangular region ABC above?

(1) The product of BD and AC is 20.

(2) x = 45

Statement 1:

The product of BD and AC is 20.

i.e. BD*AC = 20

Area of a triangle = 1/2 * base * height = 1/2 * BD*AC = 1/2*20 = 10

Sufficient

Statement 2:

x = 45

No information about the size of the triangle is given, hence insufficient.