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#12.4

A design is made on a rectangular tile of dimensions 50 cm × 70 cm as shown in Figure. The design shows 8 triangles, each of sides 26 cm, 17 cm and 25 cm. Find the total area of the design and the remaining area of the tile.

$[The\; area\; of\; the\; design\; is\; 1632\; cm^{2}\; and\; the\; remaining \; area\; of\; the \; tile \; is\; 1868\; cm^{2}.]$

We have dimensions of rectangle tile as 50 cm × 70 cm

We know that area of rectangle = length × breadth

$Area \; of\; tile = (70 × 50)\; cm^{2} = 3500\; cm^{2}$

Given sides of triangular design: 26 cm, 17 cm, 25 cm

To find the area using Heron’s formula

Let, a = 26 cm, b = 17 cm, c = 25 cm

$S=\frac{a+b+c}{2}=\frac{26+17+25}{2}=\frac{68}{2}=34\; cm$

$Area\; of\; triangle =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{34\left ( 34-26 \right )\left ( 34-17 \right )\left ( 34-25 \right )}$

$= \sqrt{34\times 8\times 17\times 9}$

$= \sqrt{17\times 2\times 2\times 2\times 2\times 17\times 3\times 3}$

$= 2\times 2\times 3\times 17$

Area of $\Delta$ ABC = 204 $cm^{2}$

But we have 8 triangles of equal area

So area of design = 8 × area of one $\Delta$

= 8 × 204 = 1632 $cm^{2}$

Remaining area of tile = Area of tile - Area of design

= (3500 – 1632) $cm^{2}$ = 1868 $cm^{2}$

Hence the area of the design is 1632 $cm^{2}$ and the remaining area of the tile is 1868 $cm^{2}$ .

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#12.4

The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle as shown in the Figure. If the area of the trapezium PQCD is $\frac{5}{6}th$ part of the area of the rectangle, find the lengths QC and PD.

$[QC = 45\; cm, PD = 40\; cm]$

$Given, BC = 51\; cm \; and \; CD = 25\; cm$

$Area \; of\; rectangle\; ABCD = (51 \times 25) cm^{2}$

$In\; trapezium\; PQCD\; , parallel\; sides\; QC\; and\; PD\; are\; in\; the\; ratio\, 9:8$

Let length of QC = 9x and PD = 8x

$We\; have\: , Area\: of\: trapezium\: PQCD = \frac{5}{6}th\: part\; of\; Area(ABCD)$

$\frac{1}{2} \times (Sum\; of\; ||\; sides) \times height =\frac{5}{6} \times length \times breadth$

$\frac{1}{2}\times (PD + QC) \times CD = \frac{5}{6}\times BC \times CD$

$\frac{1}{2} \times (8x + 9x) \times 25 = \frac{5}{6}\times 51 \times 25$

$\frac{1}{2}\times 17x \times 25 = \frac{5}{6}\times 51 \times 25$

$\frac{25}{2}\times 17x = \frac{5}{6}\times 51 \times 25$

$x = \frac{5}{6}\times 51 \times 25\times \frac{1}{17}\times \frac{2}{25}$

$x = 5$

$PD = 8x$

$= 8\times 5$

$= 40\; cm$

$QC = 9x$

= 9 x 5 = 45

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#12.4

In Figure, $\Delta$ ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram, DBCE of same area as that of $\Delta$ ABC is constructed. Find the height DF of the parallelogram.

$[3 cm]$

AB = 7.5 cm, AC = 6.5 cm, BC = 7 cm

Let a = 7.5 cm, b = 6.5 cm, c = 7 cm

$Now, S=\frac{a+b+c}{2}=\frac{7.5+6.5+7}{2}=\frac{21}{2}=10.5\; cm$

$Area\; of\; \Delta ABC,\; By\; heron's\; formula =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{10.5\left ( 10.5-7.5 \right )\left ( 10.5-6.5 \right )\left ( 10.5-7 \right )}$

$=\sqrt{10.5\times 3\times 4\times 3.5}$

$=\sqrt{\frac{105}{10}\times 3\times 4\times \frac{35}{10}}$

$=\sqrt{21\times 3\times 7}=\sqrt{3\times 7\times 3\times 7}=3\times 7=21\; cm^{2}$

$Now,\; we\; find\; the \; length\; DF \; of\; parallelogram\: DBCE$

$Area\; of\; parallelogram = base \times height = BC \times DF$

$Area\; of\; parallelogram = 7DF$

According to question,

$Area\; of\; \Delta ABC = Area \; of\; parallelogram DBCE$

$21 = 7DF$

$\frac{21}{7}= DF$

$DF = 3\; cm$

$Hence\; height\; of\; parallelogram\; is\; 3\; cm.$

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#12.4

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 4 to plough 1 $m^{2}$ of the field, find the total cost of ploughing the field.

$\left [ Rs. \; 19200 \right ]$

Given, ABCD is trapezium having parallel side AB = 90 m, CD = 30 m

Draw DE parallel to CB

So, BE = 30 m

Now, AE = (AB - EB)

AE = (90 - 30) m

AE = 60 m

So, in right triangle $\Delta$ AED

$\left ( AD \right )^{2}= (AE)^{2} + (DE)^{2} \: \; \; \; \; \; [Using Pythagoras theorem]$

$\left ( 100 \right )^{2}= (60)^{2} + (DE)^{2}$

$10000= 3600+(DE)^{2}$

$10000- 3600= (DE)^{2}$

$6400= (DE)^{2}$

Taking square root on both sides

$\sqrt{6400}= \sqrt{(DE)^{2}}$

DE = 80 m

$We\; know \; that\; the \; area \; of \; trapezium \; ABCD = \frac{1}{2} \times (sum \; of \; parallel \; sides) \times height$

$= \frac{1}{2} \times (AB+CD) \times DE$

$= \frac{1}{2} \times (90+30) \times 80$

$= 120\times 40= 4800\; m^{2}$

$\because \; cost \; of\; ploughing \; 1 m^{2} \; field = Rs \; 4$

$\therefore \; cost \; of \; ploughing \; 4800 \; m^{2}\; field = 4800 \times 4 = Rs. \; 19200$

Hence the total cost of ploughing the field is Rs. 19200.

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#12.4

A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed.

$[374\; m^{2}]$

Let ABCD be the rectangular plot,

AB = 40 cm, AD = 15 cm

Given that minimum of 3 m wide space should be left in the front and back

$length \; of \; PQ = [AB - (3 + 3)] \; m$

$PQ = [40 - 6] \; m$

$PQ = 34 \; m$

Similarly, RS = 34 m

Given that 2 m wide space on each of other sides is to be left

$Length\; of \; PS = [AD - (2 + 2)] \; m$

$PS = [15 - 4] \; m$

$PS = 11 \; m \; and$

$QR = 11 \; m$

So here PQRS is another rectangle formed in the rectangle ABCD

So, Area of rectangle PQRS = length $\times$ breadth

$= PQ\times PS= \left ( 34\times 11 \right )\; m^{2}= 374\; m^{2}$

Hence the area of house can be constructed in 374 $m^{2}$

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#12.4

The area of a trapezium is 475 $cm^{2}$ and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

$\left [ 23 \; cm, 27\; cm \right ]$

Let the smaller parallel side be CD = x cm

Then other parallel side AB = (x + 4) cm

Given, area of trapezium = 475 $cm^{2}$

Height DE = 19 cm

$We \; know \; that, \; Area \; of \: trapezium = \frac{1}{2} \times height \times (sum\; of \; parallel \; sides)$

$475= \frac{1}{2}\times DE\times \left ( DC+AB \right )$

$475\times 2= 19\times \left ( x+x+4 \right )$

$\frac{475\times 2}{19}= 2x+4$

25 $\times$ 2 = 2x + 4

50 = 2x + 4

50 - 4 = 2x

46 = 2x

$x= \frac{46}{2}$

x = 23 cm

So the smaller side CD is 23 cm and other parallel side AB is (23 + 4) cm = 27 cm

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#12.4

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.

$\left [ 20\sqrt{30}\; cm^{2} \right ]$

Let the smaller side of the triangle be x cm

Let BC = x cm

According to question,

One side of a triangle is 4 cm longer than the smaller side

Let this side be AC = x + 4

Also, third side is 6 cm less than twice the smaller side

Let this side be AB = (2x - 6) cm

Given perimeter of $\Delta$ ABC = 50 cm

x + x + 4 + 2x - 6 = 50

$4x - 2 = 50$

$4x = 50+2$

$4x = 52$

$x = \frac{52}{4}$

$x = 13 \; cm$

$So\; the\; side \; AC = (x + 4) = (13 + 4) = 17\; cm$

$Side\; AB = \left ( 2x-6 \right )\; cm= \left ( 26-6 \right )\; cm= 20\; cm$

Now in $\Delta$ ABC, a = 13 cm, b = 17 cm, and c = 20 cm

Using Heron’s formula

$S= \frac{a+b+c}{2}= \frac{13+17+20}{2}= \frac{50}{2}= 25\; cm$

$Area \; of\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{25\left ( 25-13 \right )\left ( 25-17 \right )\left ( 25-20 \right )}$

$=\sqrt{25\times 12\times 8\times 5}$

$=\sqrt{5\times 5\times 2\times 2\times 3\times 2\times 2\times 2\times 5}$

$=\sqrt{5\times 5\times 5\times 2\times 2\times 2\times 2\times 2\times 3}$

$= 5\times 2\times 2\sqrt{30}$

$Area \; of \; \Delta ABC = 20\sqrt{30}\; cm^{2}$

$Hence\; the \; area\; of\; triangle\; is \; 20\sqrt{30}\; cm^{2}$

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#12.4

How much paper of each shade is needed to make a kite given in Figure, in which ABCD is a square with diagonal 44 cm.

.

$Red = 242 \; cm^{2}$

$Yellow = 484\; cm^{2}$

$Green = 373.14 \; cm^{2}$

$Given,\; ABCD\; is\; a\; square.$

We know that all sides of a square are equal

AB = BC = CD = DA and

$\angle A = \angle B = \angle C = \angle D = 90^{\circ} [\because \; All\; angles\; of\; a\; square\; are\; 90^{\circ}]$

$In \Delta ABC,\; using \; Pythagoras \; theorem$

$(AC)^{2} = (AB)^{2}+ (BC)^{2}$

$(44)^{2} = (AB)^{2}+ (BC)^{2}\; \; \; \; \; \; \; \; [\Theta \; AB = BC\; equal\; sides]$

$44\times 44=2\left ( AB \right )^{2}$

$\frac{44\times 44}{2}=\left ( AB \right )^{2}$

$(AB)^{2} = 22 \times 44$

Taking square root on both sides

$\sqrt{\left ( AB \right )^{2}}=\sqrt{22\times 44}$

$AB=\sqrt{22\times 2\times 22}$

$AB=22\sqrt{2}$

$AB = BC = CD = DA =22\sqrt{2}\; cm$

$Now,\; Area\; of \; square\; ABCD = (side)^{2}$

$=\left ( 22\sqrt{2}\right )^{2}=22\times 22\times \sqrt{2}\times \sqrt{2}$

$= 484\times \sqrt{2\times 2}=484\times 2$

$Area\; of\; square\; ABCD = 968\; cm^{2}$

But square ABCD is divided into four coloured squares.

$So,\; area\; of\; Yellow \; I =\frac{968}{4}= 242 cm^{2}$

$Area\; of\; Yellow \; II =\frac{968}{4}= 242 \; cm^{2}$

$Area\; of\; Green \; III =\frac{968}{4}= 242 \; cm^{2}$

$Area\; of\; Red \; IV =\frac{968}{4}= 242 \; cm^{2}$

$Total\; yellow\; area\; = 242\; cm^{2} + 242\; cm^{2} = 484\; cm^{2}$

We have to find the lower triangle of green colour as well.

$Let\; a = 20\; cm, b = 20\; cm, c = 14\; cm$

$Semi\; perimeter(s) = \frac{a+b+c}{2}$

$= \frac{20+20+14}{2}$

$= \frac{54}{2}=27$

Area of Triangular field:

$By\; heron's\; formula =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{27\left ( 27-20\right )\left ( 27-20 \right )\left ( 27-14 \right )}$

$= \sqrt{3\times 3\times 3\times 7\times 7\times 13}$

$= 21\sqrt{3\times 13}$

$= 131.14\; cm^{2}$

$So\; total\; green\; area\; = 242 + 131.14 = 373.14\; cm^{2}$

$Hence,\; paper\; required$

$Red = 242\; cm^{2}$

$Yellow = 484\; cm^{2}$

$green = 373.14\; cm^{2}$

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#12.3

Find the area of the trapezium PQRS with height PQ given in Figure.

$[114\; cm^{2}]$

$Firstly\; we\; have\; side \; PS = 12 m, SR = 13, QR = 7 m$

Join RT

So here PT = PS – ST

PT = 12 m – 5 m

PT = 7 m

and ST = PS – PT

$ST = \left ( 12-7 \right )m$

ST = 5 m

Now, In $\Delta$ STR, Using Pythagoras theorem

$We\; get, (SR)^{2} = (ST)^{2} + (TR)^{2}$

$(13)^{2} = (5)^{2} + (TR)^{2}$

$169 = 25 + (TR)^{2}\; \; \; \; \; \; \; \; \; \; \; [(13)^{2} = 169 \; and\; (5)^{2} = 25]$

$169 - 25 = (TR)^{2}$

$144 = (TR)^{2}\; \; \; \; \; \; \; \; \; \; \; \; [ \sqrt{144} = 12]$

$TR = 12cm\; \; \; \; \; \; \; \; \; \; \; \; [ \sqrt{12\times 12} = 12]$

Now, we can find the area of trapezium

$Area \; of\; trapezium = \frac{1}{2} \times [sum\; of \; parallel\; sides] \times height$

$= \frac{1}{2}\times \left [ 12+7 \right ]\times 12$

$= 19\times 6\; cm^{2}$

$= 114\; cm^{2}$

$Hence, the\; area\; of\; trapezium\; is\; 114\; cm^{2}.$

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#12.3

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs 5 per $cm^{2}$ . Find the cost of painting.

[Rs. 960]

Let ABCD be a rhombus thus AB = BC = CD = DA = x (Let)

$We \; have\; perimeter \; of \; rhombus = 40 cm$

$\Rightarrow AB + BC + CD + DA = 40 cm$

$\Rightarrow x + x + x + x = 40 cm$

$\Rightarrow 4x = 40 cm$

$\Rightarrow x = \frac{40}{4} cm$

$\Rightarrow x = 10 cm$

$\therefore sides \; of \; rhombus \; AB = BC = CD = DA = 10 cm$

Area of rhombus = 2 $\times$ Ar( $\Delta$ ABC) [diagonal of rhombus divides it into two triangles of equal area]

Now, we find area of triangle using Heron’s formula

$In\; \Delta ABC, S=\frac{a+b+c}{2}= \frac{10+10+12}{2}= \frac{32}{2}= 16cm$

$Area \; of\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{16\left ( 16-10 \right )\left ( 16-10 \right )\left ( 16-12 \right )}$

$= \sqrt{16\times 6\times 6\times 4}$

$= \sqrt{8\times 8\times 6\times 6\times 4}$

$= \sqrt{4\times 4\times 6\times 6\times 4}$

$= 4\times 6\sqrt{2\times 2}$

$= 4\times 6\times 2= 48cm^{2}$

Now, Area of rhombus = 2 $\times$ Ar( $\Delta$ ABC)

$= 2\times 40cm^{2}= 96cm^{2}$

We find the cost of painting

Thus,

$\because cost\; of \; painting\; the\; sheet \; of\; 1 cm^{2} = Rs.\; 5$

$\therefore cost \; of\; painting\; the \; sheet\; of \; 96\; cm^{2}= Rs.\; 96 \times 5 = Rs.\; 480$

Hence, the cost of the painting both sides of the sheet = 2 $\times$ 480 = Rs. 960.

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Get Answers to all your Questions

All Questions

A design is made on a rectangular tile of dimensions 50 cm × 70 cm as shown in Figure. The design shows 8 triangles, each of sides 26 cm, 17 cm and 25 cm. Find the total area of the design and the remaining area of the tile.

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The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle as shown in the Figure. If the area of the trapezium PQCD is part of the area of the rectangle, find the lengths QC and PD.

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In Figure, ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram, DBCE of same area as that of ABC is constructed. Find the height DF of the parallelogram.

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A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 4 to plough 1 of the field, find the total cost of ploughing the field.

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JEE Main high-scoring chapters and topics

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The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle as shown in the Figure. If the area of the trapezium PQCD is $\frac{5}{6}th$ part of the area of the rectangle, find the lengths QC and PD.

In Figure, $\Delta$ ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram, DBCE of same area as that of $\Delta$ ABC is constructed. Find the height DF of the parallelogram.

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 4 to plough 1 $m^{2}$ of the field, find the total cost of ploughing the field.

The area of a trapezium is 475 $cm^{2}$ and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

How much paper of each shade is needed to make a kite given in Figure, in which ABCD is a square with diagonal 44 cm.

.

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs 5 per $cm^{2}$ . Find the cost of painting.