Filter By

## 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.

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

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 $cm^{2}$.

#### 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

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### 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,

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

$21 = 7DF$

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

$DF = 3\; cm$

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

#### 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.

## JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

#### 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 $m^{2}$

#### 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

## NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

#### 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

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

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

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

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

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

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

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

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

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

#### 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:

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

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

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

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

$= 131.14\; cm^{2}$

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

$Hence,\; paper\; required$

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

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

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

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### 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

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

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

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

$= 114\; cm^{2}$

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

#### 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.