Solution to questions on mensuration for SSC CGL Set 27

Submitted by Atanu Chaudhuri on Tue, 14/06/2016 - 21:09

Mensuration for SSC CGL Set 21 questions with solutions

Learn how to solve 10 questions on mensuration for SSC CGL Set 27 in 15 minutes. Mensuration formulas and concepts are used for quick solutions.

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SSC CGL level Question Set 27, Mensuration 2.

Quick solutions to 10 questions on mensuration for SSC CGL Set 27 - Answering time was 15 mins

Problem 1.

The area of a rectangle is thrice that of a square. The length of the rectangle is 20 cm and the breadth $\frac{3}{2}$ times the length of a side of the square. The side length of the square (in cm) is,

Solution 1 - Problem analysis and solving

By given statement area of the rectangle,

$A_r = 20\times{\text{breadth}} = 20\times{\frac{3}{2}a}$, where $a$ is length of a side of the square,

$=30a$

$=3A_s$, where $A_s$, is the area of the square.

So,

$A_s = 10a = a^2$, by definition.

Or, $a = 10$.

Answer: Option a: 10.

Key concepts used: Area of rectangle and square concepts.

Problem 2.

The perimeter of a rhombus is 40 cm and its height is 5 cm. Its area in (cm$^2$) is,

Solution 2 - Problem analysis and solving

Let us see how a rhombus behaves and what are its basic characteristics. In the figure below ABCD is a rhombus all four sides of which are of equal length $a$,

SSC CGL solution set 27 mensuration q2 Rhombus

As length of all sides of a rhombus is equal in length, its side length = $\displaystyle\frac{40}{4} = 10$ cm.

In a rhombus, its area = base side $\times{}$ height = a $\times{}$ h. If you push the rhombus from right to left holding its corner A and keeping its base CD fixed, it adjusts its shape to the rectangle shape of CDGF without losing or adding any area. The area of the triangles $\triangle AGD$ and $\triangle BFC$ are same.

As area of the rectangle = a $\times{}$ h, we get the area of the rhombus as = base $\times{}$ height.

Thus area of the rhombus in our problem = side length $\times{}$ height = 50 cm.

Answer: Option d : 50.

Key concepts used: Basic concepts on area of rhombus.

Problem 3.

The sum of the length, breadth and height of a rectangular parallelepiped is 24 cm and the length of the diagonal is 15 cm. Then its total surface area is,

351 cm$^2$
256 cm$^2$
265 cm$^2$
315 cm$^2$

Solution 3 - Problem analysis and visualization

The following is the figure corresponding to the problem.

SSC CGL solution set 27 mensuration q3 Rectangular parallelepiped

Let the length, breadth and height be, L, B and H. The base diagonal,

$P^2= L^2 + B^2$.

Again, Main diagonal,

$D^2 = P^2 + H^2$

$\hspace{7mm}= L^2 + B^2 + H^2 = 225$.

Given $L + B + H = 24$ cm.

Knowing that the total surface area is, $2(LB +BH+HL)$ using the square of three term sum expression,

$(L +B + H)^2 $

$= L^2 + B^2 + H^2 + 2(LB + BH + HL)$, we get total surface area as,

$2(LB + BH + HL) $

$= 24^2 - 225 $

$= 576 - 225$

$= 351$ cm$^2$.

Answer: a: 351 cm$^2$.

Key concepts used: Diagonal and area concepts on rectangular parallelepiped.

Problem 4.

In the figure given below two shaded regions are formed in a circle of radius $a$.

SSC-CGL question set 27 mensuration q4

The area of the shaded region is,

$\frac{a^2}{2}(\frac{\pi}{2} - 1)$
$a^2(\pi - 1)$
$\frac{a^2}{2}(\pi - 1)$
$a^2(\frac{\pi}{2} - 1)$

Solution 4.

The triangle formed by the horizontal diameter is a right triangle (as all diameters subtend a $90^0$ angle at the periphery) with two inclined sides equal and the vertical radius as the perpendicular bisector of the base.

Area of this triangle = $a^2$.

Area of the semi-circle = $\frac{1}{2}\pi{a^2}$.

So, area of the shaded region = $\frac{1}{2}\pi{a^2} - a^2 = a^2(\frac{\pi}{2} - 1)$.

Answer: Option d: $a^2(\frac{\pi}{2} - 1)$.

Key concepts used: Visualization -- Circle and triangle area concept.

Problem 5.

Three circles of radii $a$, $b$ and $c$ touch each other externally. The area of the triangle formed by the three centres is,

$ab + bc + ca$
$(a + b +c)\sqrt{ab + bc + ca}$
$\sqrt{(a + b + c)abc}$
none of the above

Solution 5.

The following figure represents the solution.

SSC CGL solution set 27 mensuration q5

The three sides of the triangle are,

$k = a + b$

$l = b + c$, and

$m = c + a$.

The half perimeter of the triangle is then,

$s = \frac{1}{2}(k + l +m) = a + b + c$.

Using the relation of area of triangle with its half perimeter, area

$A = \sqrt{s(s - x)(s-y)(s-z)}$ where $x$, $y$ and $z$ are the three side lengths.

In our case the area of the triangle is then,

$A = \sqrt{(a + b + c)abc}$.

Answer: Option c: $\sqrt{(a + b + c)abc}$.

Key concept used: Visualization to form length of sides of triangle in terms of radii of the three circles -- using the standard relation between the length of sides of the triangle and its area.

Problem 6.

The circumference of a circle is 11 cm. The area of a sector of the circle subtending an angle of $60^0$ at the periphery (take $\pi = \frac{22}{7})$ is,

$1\displaystyle\frac{29}{48}$ cm$^2$
$2\displaystyle\frac{27}{48}$ cm$^2$
$2\displaystyle\frac{29}{48}$ cm$^2$
$1\displaystyle\frac{27}{48}$ cm$^2$

Solution 6

The visualized figure for the problem is as below.

SSC CGL solution set 27 mensuration q6

The circumference is,

$2\pi{r} = 11$, where $r$ is radius

Or, $2\times{22} r = 7\times{11}$,

Or, $r = \displaystyle\frac{7}{4}$ cm

Area subtended by $60^0$ sector is one-sixth of the total area ($60^0$ is one-sixth of $360^0$ the whole angle covering the circle).

Thus area of the sector = $\displaystyle\frac{1}{6}\times{\frac{22}{7}}\times{\frac{49}{16}} = \displaystyle\frac{77}{48} = 1\frac{29}{48}$ cm$^2$.

Answer: Option a : $1\displaystyle\frac{29}{48}$ cm$^2$.

Key concepts used: Sector and circle area concepts.

Problem 7.

If the difference between the areas of the circumcircle and incircle of an equilateral triangle is 44 cm$^2$, then the area of the triangle (in cm$^2$, take $\pi = \frac{22}{7}$), is,

$28$ cm$^2$
$21$ cm$^2$
$7\sqrt{3}$ cm$^2$
$14\sqrt{3}$ cm$^2$

Solution 7

The corresponding figure is as below.

SSC CGL solution set 27 mensuration q7

The circumcentre and incentre of the equilateral triangle is the same point P and the perpendicular bisector CD and the other two such bisectors from vertices A and B intersect at P which in this case is also the centroid and CD is a median.

Let's assume side length of the triangle as $a$.

So,

$CD = \sqrt{a^2 - \left(\displaystyle\frac{1}{2}a\right)^2} = \displaystyle\frac{\sqrt{3}}{2}a$.

$CD$ being the median, $PD$ is one-third of $CD$, that is,

$PD=\displaystyle\frac{1}{2\sqrt{3}}a$

Similarly, $CP$ is two-third of $CD$, that is,

$CP=\displaystyle\frac{1}{\sqrt{3}}a$.

Thus difference in the areas of the circles,

$A_c - A_i = \pi(CP^2 - PD^2) $

$\hspace{18mm}= \pi{a^2}\left(\displaystyle\frac{1}{3} - \displaystyle\frac{1}{12}\right)$

$\hspace{18mm}=\displaystyle\frac{1}{4}\pi{a^2}$

$\hspace{18mm}=44$

Or, $a^2 = 56$.

The area of the equilateral triangle with side length $a$ is,

$A_t = \displaystyle\frac{\sqrt{3}}{4}a^2 $

$\hspace{6mm}= \displaystyle\frac{\sqrt{3}}{4}56$

$\hspace{6mm}=14 \displaystyle\sqrt{3}$ cm$^2$.

Answer: d: $14\sqrt{3}$ cm$^2$.

Key concepts used: Concepts of circumcentre, incentre, median and areas of circles and equilateral triangles.

Problem 8.

If area of an equilateral triangle is $A$ and its height is $b$, the value of $\displaystyle\frac{b^2}{A}$ is,

$\displaystyle\frac{1}{3}$
$3$
$\sqrt{3}$
$\displaystyle\frac{1}{\sqrt{3}}$

Solution 8

The following figure depicts the problem situation.

SSC CGL solution set 27 mensuration q8

If the side length of an equilateral triangle is $a$, its height is,

$b^2 = a^2 - \displaystyle\frac{1}{4}a^2 = \displaystyle\frac{3}{4}a^2$,

And its area is,

$A = \displaystyle\frac{\sqrt{3}}{4}a^2$,

So desired expression,

$\displaystyle\frac{b^2}{A} = \displaystyle\frac{\displaystyle\frac{3}{4}a^2}{\displaystyle\frac{\sqrt{3}}{4}a^2} = \sqrt{3}$.

Answer: Option c: $\sqrt{3}$.

Key concepts used: Expressing both height $b$ and area $A$ in terms of link variable or intermediate variable side length $a$ simplifies the procedure.

Problem 9.

ABCD is a parallelogram. P and Q are the mid-points of sides BC and CD respectively. If the area of the $\triangle ABC$ is 12 cm$^2$, the area of $\triangle APQ$ is,

9 cm$^2$
12 cm$^2$
10 cm$^2$
8 cm$^2$

Solution 9 - Problem analysis

The following figure corresponds to visualization of the problem.

SSC CGL solution set 27 mensuration q9

P being midpoint of BC, it divides the $\triangle ABC$ into two equal parts (height being same, base being half of larger base), that is,

Area of $\triangle APC = \displaystyle\frac{1}{2}$ of area of $\triangle ABC = 6$ cm$^2$.

Similarly, area of $\triangle AQC$ is half of area of $\triangle ACD = 6$ cm$^2$.

So area of quadrilateral $APCQ = 12$ cm$^2$.

Now we only have to find the area of $\triangle PCQ$ and subtract it from this area of the quadrilateral to get the desired area of $\triangle APQ$.

Solution 9 - Problem solving

The other diagonal BD also divides the area of the parallelogram into two equal parts and so, area of the $\triangle BCD = 12$ cm$^2$.

Again $BD||PQ$ and P and Q are the midpoints of the other two sides BC and CD of the $\triangle BCD$. Thus these two triangles $\triangle PCQ$ and $\triangle BCD$ are similar and each side including the height of the smaller triangle is half its corresponding side and the height of the larger triangle.

This makes the area of the $\triangle PCQ = \displaystyle\frac{1}{4}$th of the area of $\triangle BCD = 3$ cm$^2$.

Finally then, the area of $\triangle APQ = 12 - 3 = 9$ cm$^2$.

Answer: Option a: 9 cm$^2$.

Key concepts used: Visualization of relationships between the important areas of triangles -- parallelogram concepts.

Problem 10.

A wire when bent in the form of a square encloses an area of 484 sq cm. What will be the enclosed area when the same wire is bent into the form of a circle?