MENSURATION MCQ LIST

 Question 1:

In the figure below, ABCDEF is a regular hexagon and $\angle$ AOF $={90}^{\circ }$ . FO || $ED$. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF?

[1] $\frac{1}{12}$

[2] $\frac{1}{6}$

[3] $\frac{1}{24}$

[4] $\frac{1}{18}$

Answer & Solution

Option # 1

Since FO || ED

$\angle FED+\angle OFE={180}^{\circ }$

In a regular Hexagon each interior angle is ${120}^{\circ }$

$therefore\angle OFE={60}^{\circ }$ and also $\angle AFO+\angle AFE={120}^{\circ }$

$therefore\angle AFO={60}^{\circ }$

Therefore triangle AOF is a 60° – 30° – 90° triangle

Let a be the side of a regular hexagon

Then sin 60 = AO / a

$thereforeAO=a\sin {60}^{\circ }=\frac{a\sqrt{3}}{2}$

similarly OF $=a\sin {30}^{\circ }$ OF $=a/2$ Area of triangle $AOF=\frac{\sqrt{3}{a}^2}{8}$

Area of Hexagon $=\frac{3\sqrt{3}{a}^2}{2}$ . Therefore the ratio of areas $=\frac{\frac{\sqrt{3}}{3\sqrt{3}}{a}^2}{\frac{3\sqrt{3}}{2}{a}^2}=\frac{1}{12}$

Shortcut : Observe that exactly 12 triangles similar to the triangle given can be formed in the given figure

ratio = 1/12.

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Question 2:
ABCDEF is a regular hexagon of side 6 cm. What is the area of triangle BDF?

[1] 32$\sqrt{3}c{m}^2$

[2] 27$\sqrt{3}c{m}^2$

[3] $24c{m}^2$

[4] None of These

Answer & Solution

Option # 2

You should know that

$\frac{\text{Area of}\Delta BDF}{\text{Area of hexagon}}=\frac{1}{2}$

Actually there is a perfect symmetrically.

Therefore, Area of hexagon= $\frac{3\sqrt{3}}{2}\times (6{)}^2=54\sqrt{3}{cm}^2$

$thereforeAreaof\Delta BDF=27\sqrt{3}{cm}^2.$ Ans. $\left(2\right)$

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Question 3:
A cow is tethered to one corner of a square field with side 20 m with a rope equal to the length of a side of the field. Another cow is tethered at the diagonally opposite corner with the grazing area just touching each other. What is the total grazing area?

[1] 200$\pi [2-\sqrt{2}]$

[2] 200$\pi [4-\sqrt{2}]$

[3] 100$\pi [2-\sqrt{2}]$

[4] 150$\pi [2-\sqrt{2}]$

Answer & Solution

Option # 1

Diagonal $=20\sqrt{2}$

length of rope tying the 2nd cow $=20\sqrt{2}-20=20(\sqrt{2}-1)$

grazing area of 1$st$ cow $=\frac{1}{4}\pi {20}^2=\frac{1}{4}\pi 400=100\pi$

grazing area of 2$nd$ cow $=\frac{1}{4}\pi 400(\sqrt{2}-1{)}^2$

Total $100\pi [1+(\sqrt{2}-1{)}^2]=100\pi [1+2+1-2\sqrt{2}]=200\pi [2-\sqrt{2}]$

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Question 4:
What fraction of the area of a regular hexagon is the area of regular hexagon obtained by joining the midpoints of the sides of the first hexagon in order?

[1] $\frac{1}{4}$

[2] $\frac{1}{2}$

[3] $\frac{3}{4}$

[4] $\frac{2}{3}$

Answer & Solution

Option # 3

Side of original hexagon $=2a$

Then a side of the second hexagon $=2\times \frac{a}{2}\sqrt{3}=a\sqrt{3}$

Areas are $6\times \frac{\sqrt{3}}{4}\times 4a^2$ and $6\times \frac{\sqrt{3}}{4}\times 3a^2$

Smaller hexagon is $\frac{3}{4}$ of the bigger one.

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Question 5:
Find the area of the shaded region in the given figure of square ABCD

[1] 128 cm2

[2] 192 cm2

[3] 148 cm2

[4] 168 cm2

Answer & Solution

Option # 1

Area of shaded region =Area of square - (Area of $\Delta DPQ$ + Area $\Delta QAR+$ Area $\Delta RBC)$

$=256-(24+24+80)$

$\Rightarrow 128.$

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Question 6:
There is a vast grassy farm in which there is a rectangular building of the farmhouse whose length and breadth is 50 m and 40 m respectively. A horse is tethered at a corner of the house with a tether of 80 m long. What is the maximum area that the horse can graze?

[1] $5425\pi$

[2] $5245\pi$

[3] $254\pi$

[4] None of These

Answer & Solution

Option # 1

The length of tether of the horse is 80 m.

Area grazed by horse = $[\pi \times (80{)}^2\times \frac{270}{360}+\pi \times (30{)}^2\times \frac{90}{360}+\pi \times (40{)}^2\times \frac{90}{360}]$

$=\pi (6400\times \frac{3}{4}+900\times \frac{1}{4}+1600\times \frac{1}{4})=\pi \left[\frac{21700}{4}\right]=5425\pi m^2$

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Question 7:
A circular paper is folded along its diameter, then again it folded to form a quadrant. Then it is cut as shown in the figure, after it the paper was reopened in the original circular shape. Find the ratio of the original paper to that of the remaining paper? (The shaded portion is cut off from the  quadrant. The radius of quadrant OAB is 5 cm and radius of each semicircle is 1 cm)

[1] 25 : 16

[2] 25 : 9

[3] 20: 9

[4] None of these

Answer & Solution

Option # 1

When we open the paper after cutting it, we will find it as shown in the following figure.

Radius of the larger circle = 5 cm

Therefore, area of larger circle$=25\pi$ and the radius of each smaller circle is 1$cm$

Therefore, total area of all the 9 circles $=9\times \pi \times (1{)}^2=9\pi$

Therefore, Remaining area$=(25-9)\pi =16\pi$

Hence, the required ratio $=25:16$

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Question 8:
ABCD is a square. A circle is inscribed in the square. Also taking A, B, C, D (the vertices of square) as the centres of four quadrants, drawn inside the circle, which are touching each other on the midpoints of the sides of square. Area of square is 4 cm2 . What is the area of the shaded region?

[1] $(4-\frac{3\pi }{2}){cm}^2$

[2] $(2\pi -4){cm}^2$

[3] $(4-2\pi ){cm}^2$

[4] None of these

Answer & Solution

Option # 3

Area of region x = Area of square – Area of inscribed circle

Area of region y = Area of square – 4 (area of quadrant)

$=4-4\left(\frac{1}{4}\pi \right(1{)}^2)=(4-\pi )$

Therefore,  Required area (of shaded region)=Area of square-[Area of region x + Area of region y ]

$=4-[4-\pi +4-\pi ]=2\pi -4$

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Question 9:
An equilateral triangle circumscribes all the six circles, each with radius 1 cm. What is the perimeter of the equilateral triangle?

[1] 6$(2+\sqrt{3})cm$

[2] 3$(\sqrt{3}+2)cm$

[3] 12$(\sqrt{3}+4)cm$

[4] None of these

Answer & Solution

Option # 1

In $\Delta {BOM}_1$

$\tan {30}^{\circ }=\frac{OM}{BM}$

$\frac{1}{\sqrt{3}}=\frac{1}{BM}$

$BM=\sqrt{3}$

Side of equilateral triangle $=BM+MN+NC$

$=\sqrt{3+4+\sqrt{3}}$

$=4+2\sqrt{3}$

perimeter of equilateral triangle $=3(4+2\sqrt{3})$

$=6(2+\sqrt{3})cm.$

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Question 10:
If the sides of each square is 10. Find the area of shaded region

[1] 75

[2] 80

[3] 100

[4] 120

Answer & Solution

Option # 2

$\Delta ACB\sim \Delta DCE$

$AB=20$

$D$ mid point of EF

$DE=\frac{10}{2}=5$

$\frac{h_1}{h_2}=\frac{DE}{AB}=\frac{1}{4}$

$h_1+h_2=10$

$h_2=8$

Area of ABC$=\frac{1}{2}AB\times h_2$

$=\frac{1}{2}\times 20\times 8=80$