Quadrilateral Word problems

**Page 1**

1.

In the rectangle, $a$ = 5.76 cm. Find the values of S and $x$.

a. | S = 2.4 cm, $x$ = 5.76 cm | ||

b. | $x$ = 2.4 cm, S = 4.8 cm | ||

c. | S = $x$ = 5.76 cm | ||

d. | $x$ = 2.4 cm, S = 5.76 cm |

[From first diagonal.]

S = 2.4 cm

[Simplify.]

[From second diagonal.]

So, S = 2.4 cm and

Correct answer : (1)

2.

Side of a square is $\frac{17}{15}$ cm as shown. Find the length of the diagonal XY.

a. | $\frac{17\sqrt{2}}{15}$ cm | ||

b. | $\frac{34}{15}$ cm | ||

c. | $\frac{17}{15}$ cm | ||

d. | $\frac{17}{15\sqrt{2}}$ cm |

In the right angled triangle XPY , XP

[Pythagorean theorem.]

XY

[Simplify.]

XY =

[Simplify.]

Correct answer : (1)

3.

O is a point on the longer diagonal of the rhombusPQRS so that RS = RO. If $m$$\angle $SRQ = 60, find the ratio of the area of the quadrilateral OQRS to tha area of the quadrilateral OQPS.

a. | $\frac{\sqrt{3}-1}{4}$ | ||

b. | $\frac{\sqrt{3}+1}{2}$ | ||

c. | $\frac{\sqrt{3}-1}{2}$ | ||

d. | $\frac{\sqrt{2}+1}{3}$ |

[Given.]

[Adjacent angles are supplementary.]

[SQ bisects the angles at the vertices.]

ΔSRQ is an equilateral triangle.

[Measure of each angle is 60.]

Area of the rhombus =

[

Area of OQRS =

[SR = OR =

=

[Simplify.]

Area of OQPS = Area of the rhombus - area of OQRS

[Steps 5 and 6.]

=

[

=

Ratio of area of OQRS to area of OQPS =

=

Correct answer : (2)

4.

M is a point on the rectangle PQRS as shown. The areas $a$, $c$ and $b$ of triangles PQM, QMR and RMS respectively are in geometric progression. What is the value of $\frac{\mathrm{S}\mathrm{M}}{\mathrm{P}\mathrm{M}}$?

a. | $\frac{\sqrt{5}-1}{2}$ | ||

b. | $\frac{\sqrt{3}+2}{2}$ | ||

c. | $\frac{\sqrt{3}-1}{2}$ | ||

d. | $\frac{\sqrt{5}+1}{2}$ |

[Given that

[Formula.]

[Formula.]

=

[Formula.]

=

[Formula.]

[Formula.]

[Formula.]

[Formula.]

[Divide by MS x PM.]

[Solve for MS/PM.]

Correct answer : (4)

5.

ABCD is a square and ABM is an equilateral triangle.Find the length of $\stackrel{\u203e}{\mathrm{DM}}$. [Take $a$ =14.]

a. | 14$\sqrt{2+\sqrt{3}}$ | ||

b. | 7$\sqrt{2-\sqrt{3}}$ | ||

c. | 14$\sqrt{2-\sqrt{3}}$ | ||

d. | 28$\sqrt{2-\sqrt{3}}$ |

AB = 14, DN = 7

[DN =

AE = 7

[ΔABM is equilateral.]

ME =

[Altitude of equilateral triangle =

MN = NE - ME = 14 -

[ABCD is a square.]

=

DM

=

= 14² [2 -

DM = 14

Correct answer : (3)

6.

ABCD is a square. ΔABM is an equilateral triangle. Find the measure of $\angle $CDM. [Take $a$ = 7.]

a. | 45 | ||

b. | 18 | ||

c. | 15 | ||

d. | 75 |

AE =

[ΔABM is equilateral.]

ME =

[Height of equilateral triangle is

MN = NE - ME = 7 - 6.06 = 0.94

[ABCD is a square. So, NE = 7.]

Tan (

Correct answer : (3)

7.

Perimeter of the rectangle = 24 cm. Find the area of the portion of the rectangle after a triangle as shown in the figure is removed from it. [Take a = 8.]

a. | 28.54 sq.cm | ||

b. | 26.54 sq.cm | ||

c. | 32 sq.cm | ||

d. | 30.54 sq.cm |

Width =

ABO is a right triangle,

Hypotenuse = AB = 4 cm

OB = 4 cos30 = 4 ×

OA = 4 sin30 =

Area of ΔAOB =

Area of the rectangle = 8 × 4 = 32

Area of the portion of the rectangle after removing the triangle = 32 - 3.46 = 28.54 sq.cm

Correct answer : (1)

8.

A man paved his square courtyard with square tiles. After this, he wanted to increase the size of his courtyard and bought 324 more tiles of the same size. These were added onto the existing courtyard to make a bigger one in square shape. Find the total number of tiles used to make the bigger courtyard.

a. | 900 | ||

b. | 576 | ||

c. | 950 | ||

d. | 324 |

Let the number of tiles after adding 324 more tiles to the original number of tiles be

On observing the above equation, it is known that the pythagorean triplet 18, 24, 30 satisfy the above equation.

So,

So, the number of tiles he used altogether = 30

Correct answer : (1)

9.

The perimeters of two similar triangles are 2 in and 6 in. What is the corresponding side of the second triangle, if one side of the first triangle is 2 in?

a. | 12 in | ||

b. | 7 in | ||

c. | 16 in | ||

d. | 6 in |

Given, the perimeters of the two similar triangles are 2 in and 6 in and one side of the first triangle is 2 in.

Hence, 2 : 6 : : 2 :

[

∴ The corresponding side of the other triangle =

Correct answer : (4)

10.

ABCD is a rectangle with diagonals $\stackrel{\u203e}{\mathrm{AD}}$ and $\stackrel{\u203e}{\mathrm{BC}}$ intersecting at point O. If the length of $\stackrel{\u203e}{\mathrm{AO}}$ is 6 units, find the length of $\stackrel{\u203e}{\mathrm{CB}}$.

a. | 6 units | ||

b. | 18 units | ||

c. | 10 units | ||

d. | 12 units |

[Given ABCD is a rectangle.]

Length of

[As per the question.]

Length of

[Since O is the midpoint.]

Length of

[From step 1.]

Correct answer : (4)