﻿ Quadrilateral Word problems | Problems & Solutions

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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

#### Solution:

In a rectangle, diagonals bisect each other. From the given rectangle S2 = 5.76 cm
[From first diagonal.]

S = 2.4 cm
[Simplify.]

x = 5.76 cm
[From second diagonal.]

So, S = 2.4 cm and x = 5.76 cm.

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

#### Solution:

Diagonal XY divides square PXRY into two right angled triangles XPY and XRY.

In the right angled triangle XPY , XP2 + PY2Ã‚Â = XY2
[Pythagorean theorem.]

XY2 = 2 × 289225Þ = XY = 578225
[Simplify.]

XY = 172 15 cm
[Simplify.]

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. b. c. d.

#### Solution:

mSRQ = 60
[Given.]

mPQR = 120

mSQR = 60
[SQ bisects the angles at the vertices.]

ΔSRQ is an equilateral triangle.
[Measure of each angle is 60.]

Area of the rhombus = 23a24
[a = side of the rhombus; use formula.]

Area of OQRS = 2a2Sin302
[SR = OR = a, mSRO = 30, use formula.]

= a22
[Simplify.]

Area of OQPS = Area of the rhombus - area of OQRS
[Steps 5 and 6.]

= 23a24-a22
[a = side of the rhombus; use formula.]

= (3-1)a22

Ratio of area of OQRS to area of OQPS = 13-1

= 3 + 12

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 $\frac{SM}{PM}$?

 a. $\frac{\sqrt{5}-1}{2}$ b. c. $\frac{\sqrt{3}-1}{2}$ d. $\frac{\sqrt{5}+1}{2}$

#### Solution:

c2 = ab
[Given that a, c and b are in G.P.]

a = PQ×PM2
[Formula.]

b = PQ×QR2
[Formula.]

= PQ×PS2
[Formula.]

= PQ×(PM + MS)2
[Formula.]

c = PQ×MS2
[Formula.]

PQ2×MS24 = PQ×(PM + MS)2×PQ×PM2
[Formula.]

MS2 =PM2 + MS×PM
[Formula.]

MS / PM =PM / MS + 1
[Divide by MS x PM.]

MS / PM =5+12
[Solve for MS/PM.]

5.
ABCD is a square and ABM is an equilateral triangle.Find the length of $\stackrel{‾}{\mathrm{DM}}$. [Take $a$ =14.]

 a. 14 b. 7 c. 14 d. 28

#### Solution:

AB = 14, DN = 7
[DN = 142.]

AE = 7

mMAE = 60
[ΔABM is equilateral.]

ME = 32 × 14
[Altitude of equilateral triangle = 32 × side.]

MN = NE - ME = 14 - 32 × 14
[ABCD is a square.]

= 2 -32 × 14

DM2 = DN2 + MN 2 = 14²4 + (2 -3)24 × 14²

= 1424[1 + 4 + 3 - 43]

= 14² [2 - 3]

DM = 142-3

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

#### Solution:

AB = 7, DN = 72 = 3.5

AE = 72 = 3.5

mMAE = 60
[ΔABM is equilateral.]

ME = 32 × 7 = 6.06
[Height of equilateral triangle is 32 × side.]

MN = NE - ME = 7 - 6.06 = 0.94
[ABCD is a square. So, NE = 7.]

Tan (CDM) = MNDN = 0.943.5 = 0.26

mCDM = Tan-1(0.26) = 15

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

#### Solution:

Perimeter = 24 cm.

Width = 24 - 2 × 82 = 24 - 162 = 4 cm

ABO is a right triangle, AOB = 90°

Hypotenuse = AB = 4 cm

OB = 4 cos30 = 4 × 32 = 3.46

OA = 4 sin30 = 42 = 2

Area of ΔAOB = 1 / 2 × 3.46 × 2 = 3.46

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

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

#### Solution:

At first, the number of tiles used by man = x2.

Let the number of tiles after adding 324 more tiles to the original number of tiles be y2.

x2 + 324 = y2

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

So, x = 24 and y = 30.

So, the number of tiles he used altogether = 302 = 900.

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

#### Solution:

The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.

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 : x
[x is the corresponding side of the second triangle.]

∴ The corresponding side of the other triangle = 6×2 / 2 = 6 in.

10.
ABCD is a rectangle with diagonals $\stackrel{‾}{\mathrm{AD}}$ and $\stackrel{‾}{\mathrm{BC}}$ intersecting at point O. If the length of $\stackrel{‾}{\mathrm{AO}}$ is 6 units, find the length of $\stackrel{‾}{\mathrm{CB}}$.

 a. 6 units b. 18 units c. 10 units d. 12 units

#### Solution:

Diagonals of a rectangle are of equal length.
[Given ABCD is a rectangle.]

Length of AO is 6 units.
[As per the question.]

Length of AD is 12 units.
[Since O is the midpoint.]

Length of CB = Length of AD = 12 units.
[From step 1.]