RS AGGARWAL CLASS 9 CHAPTER 14 AREAS OF TRIANGLES ,QUADRILATERAL Exercise 14.1

page: 533

Question 1:

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.

Answer 1:

We have:
Base = 24 cm
Height = 14.5 cm

Now,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times 24\times 14.5 \\ =174 {\mathrm{cm}}^2 \end{gathered}$$

Question 2:

The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.

Answer 2:

Let the height of the triangle be h m.
∴ Base = 3h m
Now,
Area of the triangle = $\frac{\mathrm{Total} \mathrm{Cost}}{\mathrm{Rate}}=\frac{783}{58}=13.5 \mathrm{ha} =135000 {\mathrm{m}}^2$
We have:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 135000 {\mathrm{m}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Heigh}t =135000 \\ \Rightarrow \frac{1}{2}\times 3h\times h\mathit{ }=135000 \\ \Rightarrow h^2=\frac{135000\times 2}{3} \\ \Rightarrow h^2=90000 \\ \Rightarrow h\mathit{ }=300 \mathrm{m} \end{gathered}$$

Thus, we have:
Height = h = 300 m
Base = 3h = 900 m

Question 3:

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.

Answer 3:

$$\begin{gathered} \mathrm{Let}: \\ a=42 \mathrm{cm}, b = 34 \mathrm{cm} \mathrm{and} c=20 \mathrm{cm} \\ \therefore s= \frac{a+b+c}{2}=\frac{42+34+20}{2}=48 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{48(48-42)(48-34)(48-20)} \\ =\sqrt{48\times 6\times 14\times 28} \\ =\sqrt{4\times 2\times 6\times 6\times 7\times 2\times 7\times 4} \\ =4\times 2\times 6\times 7 \\ =336 {\mathrm{cm}}^2 \end{gathered}$$

We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 336 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 336 \\ \Rightarrow \mathrm{Height} = \frac{336\times 2}{42}=16 \mathrm{cm} \end{gathered}$$

Question 4:

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.

Answer 4:

$$\begin{gathered} \mathrm{Let}: \\ a=18 \mathrm{cm}, b = 24 \mathrm{cm} \mathrm{and} c=30 \mathrm{cm} \\ \therefore s= \frac{a+b+c}{2}=\frac{18+24+30}{2}=36 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{36(36-18)(36-24)(36-30)} \\ =\sqrt{36\times 18\times 12\times 6} \\ =\sqrt{12\times 3\times 6\times 3\times 12\times 6} \\ =12\times 3\times 6 \\ =216 {\mathrm{cm}}^2 \end{gathered}$$

We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 216 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 216 \\ \Rightarrow \mathrm{Height} = \frac{216\times 2}{18}=24 \mathrm{cm} \end{gathered}$$

Question 5:

Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.

Answer 5:

$$\begin{gathered} \mathrm{Let}: \\ a=91 \mathrm{m}, b = 98 \mathrm{m} \mathrm{and} c=105 \mathrm{m} \\ \therefore s= \frac{a+b+c}{2}=\frac{91+98+105}{2}=147 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{147(147-91)(147-98)(147-105)} \\ =\sqrt{147\times 56\times 49\times 42} \\ =\sqrt{7\times 3\times 7\times 2\times 2\times 2\times 7\times 7\times 7\times 7\times 3\times 2} \\ =7\times 7\times 7\times 2\times 3\times 2 \\ =4116 {\mathrm{m}}^2 \end{gathered}$$

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 4116 {\mathrm{m}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 4116 \\ \Rightarrow \mathrm{Height} = \frac{4116\times 2}{105}=78.4 \mathrm{m} \end{gathered}$$

Question 6:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle.

Answer 6:

Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5$\times$5 = 25 m
12$\times$5 = 60 m
13$\times$5 = 65 m

Now,
$$\begin{gathered} \mathrm{Let}: \\ a=25 \mathrm{m}, b = 60 \mathrm{m} \mathrm{and} c=65 \mathrm{m} \\ \therefore s= \frac{150}{2}=75 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{75(75-25)(75-60)(75-65)} \\ =\sqrt{75\times 50\times 15\times 10} \\ =\sqrt{15\times 5\times 5\times 10\times 15\times 10} \\ =15\times 5\times 10 \\ =750 {\mathrm{m}}^2 \end{gathered}$$

Question 7:

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at ₹ 5 per m².

Answer 7:

Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25$\times$10 = 250 m
17$\times$10 = 170 m
12$\times$10 = 120 m

Now,
$$\begin{gathered} \mathrm{Let}: \\ a=250 \mathrm{m}, b =170 \mathrm{m} \mathrm{and} c=120 \mathrm{m} \\ \therefore s= \frac{540}{2}=270 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{270(270-250)(270-170)(270-120)} \\ =\sqrt{270\times 20\times 100\times 150} \\ =\sqrt{30\times 3\times 3\times 20\times 20\times 5\times 30\times 5} \\ =30\times 3\times 20\times 5 \\ =9000 {\mathrm{m}}^2 \end{gathered}$$

Cost of ploughing 1 m² field = Rs 5
Cost of ploughing 9000 m² field = $5\times 9000=\mathrm{Rs} 45000$.

Question 8:

Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.

Answer 8:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Let}: \\ a=85 \mathrm{m} \mathrm{and} b = 154 \mathrm{m} \\ \mathrm{Given}: \\ \mathrm{Perimeter} = 324 \mathrm{m} \\ or, a+b+c =324 \\ \Rightarrow c=324-85-154=85 \mathrm{m} \\ \therefore s= \frac{324}{2}=162 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{162(162-85)(162-154)(162-85)} \\ =\sqrt{162\times 77\times 8\times 77} \\ =\sqrt{1296\times 77\times 77} \\ =\sqrt{36\times 77\times 77\times 36} \\ =36\times 77 \\ =2772 {\mathrm{m}}^2 \end{gathered}$$


(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 2772 {\mathrm{m}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 2772 \\ \Rightarrow \mathrm{Height} = \frac{2772\times 2}{154}=36 \mathrm{m} \end{gathered}$$

Question 9:

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.

Answer 9:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ a=13 \mathrm{cm} \mathrm{and} b=20 \mathrm{cm} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{isosceles} \mathrm{triangle}=\frac{b}{4}\sqrt{4a^2-b^2} \\ =\frac{20}{4}\times \sqrt{4(13{)}^2-{20}^2} \\ =5\times \sqrt{676-400} \\ =5\times \sqrt{276} \\ =5\times 16.6 \\ =83.06 {\mathrm{cm}}^2 \end{gathered}$$

Question 10:

The base of an isosceles triangle measures 80 cm and its area is 360 cm². Find the perimeter of the triangle.

Answer 10:

Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} =360 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times QR\times PX = 360 \\ \Rightarrow h =\frac{720}{80}=9 \mathrm{cm} \\ \mathrm{Now},\mathit{ } \\ QX\mathit{ }= \frac{1}{2}\times 80 = 40 \mathrm{cm} \mathrm{and} PX = 9 \mathrm{cm} \end{gathered}$$
Also,
$$\begin{gathered} PQ=\sqrt{Q{X}^2+P{X}^2} \\ a=\sqrt{{40}^2+9^2}=\sqrt{1600+81}=\sqrt{1681}=41 \mathrm{cm} \end{gathered}$$

∴ Perimeter = 80 + 41 + 41  = 162 cm

Question 11:

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find  the area of the triangle.

Answer 11:

The ratio of the equal side to its base is 3 : 2.
$\Rightarrow$Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.
$$\begin{gathered} \Rightarrow 3x+3x+2x=32 \mathrm{cm} \\ \Rightarrow 8x=32 \\ \Rightarrow x=4 \mathrm{cm} \end{gathered}$$
Therefore, the three side of triangle are 3x, 3x, 2x = 12 cm, 12 cm, 8 cm.
Let S be the semi-perimeter of the triangle. Then, $\mathrm{S}=\frac{1}{2}\left(12+12+8\right)=\frac{32}{2}=16$
Area of the triangle will be
$$\begin{gathered} =\sqrt{S\left(S-a\right)\left(S-b\right)\left(S-c\right)} \\ =\sqrt{16\left(16-12\right)\left(16-12\right)\left(16-8\right)} \\ =\sqrt{16\times 4\times 4\times 8} \\ =4\times 4\sqrt{8}=4\times 4\times 2\sqrt{2}=32\sqrt{2} {\mathrm{cm}}^2 \end{gathered}$$
Disclaimer: The answer does not match with the answer given in the book.

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Question 12:

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

Answer 12:

Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x − 6.

Now,
x + x + 4 + 2x − 6 = 50        (∵ perimeter is 50 cm)
⇒ 4x − 2 = 50
⇒ 4x = 50 + 2
⇒ 4x = 52
⇒ x = 13

∴ The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{13+17+20}{2}=\frac{50}{2}=25 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{25\left(25-13\right)\left(25-17\right)\left(25-20\right)} \\ =\sqrt{25\left(12\right)\left(8\right)\left(5\right)} \\ =20\sqrt{30} {\mathrm{cm}}^2 \end{gathered}$$

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

Question 13:

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield an earning of Rs 2000 per m² a year. A company hired one of its walls for 6 months. How much rent did it pay?

Answer 13:

The sides of the triangle are of length 13 m, 14 m and 15 m.
∴ Semi-perimeter of the triangle is
$s=\frac{13+14+15}{2}=\frac{42}{2}=21 \mathrm{m}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)} \\ =\sqrt{21\left(8\right)\left(7\right)\left(6\right)} \\ =84 {\mathrm{m}}^2 \end{gathered}$$

Now,
The rent of advertisements per m² per year = Rs 2000
The rent of the wall with area 84 m² per year = Rs 2000 × 84
                                                                         = Rs 168000
The rent of the wall with area 84 m² for 6 months = Rs $\frac{168000}{2}$
                                                                                = Rs 84000


Hence, the rent paid by the company is Rs 84000.

Question 14:

The perimeter of an isosceles triangle is 42 cm and its base is $1\frac{1}{2}$ times each of the equal sides. Find (i) the length of each side of the triangle, (ii) the area of the triangle, and (iii) the height of the triangle.

Answer 14:

Let the equal sides of the isosceles triangle be a cm each.
∴ Base of the triangle, b = $\frac{3}{2}$a cm
(i) Perimeter = 42 cm
or, a + a + $\frac{3}{2}$a = 42
or, 2a +$\frac{3}{2}$a= 42

$$\begin{gathered} \Rightarrow 2a+\frac{3}{2}a = 42 \\ \Rightarrow \frac{7a}{2}=42 \\ \Rightarrow a=12 \end{gathered}$$
 
So, equal sides of the triangle are 12 cm each.
Also,
Base = $\frac{3}{2}$a = $\frac{3}{2}\times 12=18 \mathrm{cm}$
(ii)
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{isosceles} \mathrm{triangle} = \frac{b}{4}\sqrt{4a^2-b^2} \\ =\frac{18}{4}\times \sqrt{4(12{)}^2-{18}^2} (a= 12 \mathrm{cm} \mathrm{and} b=18 \mathrm{cm}) \\ =4.5\times \sqrt{576-324} \\ =4.5\times \sqrt{252} \\ =4.5\times 15.87 \\ =71.42 {\mathrm{cm}}^2 \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} =71.42 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 71.42 \\ \Rightarrow \mathrm{Height} = \frac{71.42\times 2}{18}=7.94 \mathrm{cm} \end{gathered}$$

Question 15:

If the area of an equilateral triangle is $36\sqrt{3} {\mathrm{cm}}^2$, find its perimeter.

Answer 15:

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ \Rightarrow \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 =36\sqrt{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow (\mathrm{Side}{)}^2=144 \\ \Rightarrow \mathrm{Side}=12 \mathrm{cm} \end{gathered}$$

Thus, we have:
Perimeter = 3 × Side = 3 × 12 = 36 cm

Question 16:

If the area of an equilateral triangle is $81\sqrt{3} {\mathrm{cm}}^2$, find its height.

Answer 16:

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ \Rightarrow \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 =81\sqrt{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow (\mathrm{Side}{)}^2=324 \\ \Rightarrow \mathrm{Side}=18 \mathrm{cm} \end{gathered}$$

Now, we have:
$$\begin{gathered} \mathrm{Height} =\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ =\frac{\sqrt{3}}{2}\times 18 \\ =9\sqrt{3} \mathrm{cm} \end{gathered}$$

Question 17:

Each side of an equilateral triangle measures 8 cm. Find (i) the area of the triangle, correct to 2 places of decimal and (ii) the height of the triangle, correct to 2 places of decimal. Take $\sqrt{3}=1.732$.

Answer 17:

Side of the equilateral triangle = 8 cm

(i)
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times (8{)}^2 \\ =\frac{1.732\times 64}{4} \\ =27.71 {\mathrm{cm}}^2 \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Height} =\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ =\frac{\sqrt{3}}{2}\times 8 \\ =\frac{1.732\times 8}{2} \\ =6.93 \mathrm{cm} \end{gathered}$$

Question 18:

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take $\sqrt{3}=1.732$.

Answer 18:

Height of the equilateral triangle = 9 cm
Thus, we have:
$$\begin{gathered} \mathrm{Height} =\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ \Rightarrow 9=\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ \Rightarrow \mathrm{Side} = \frac{18}{\sqrt{3}}=\frac{18}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=6\sqrt{3 }\mathrm{cm} \end{gathered}$$

Also,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times (6\sqrt{3}{)}^2 \\ =\frac{108}{4}\sqrt{3} \\ =27\sqrt{3} \\ =46.76 {\mathrm{cm}}^2 \end{gathered}$$

Question 19:

The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.

Answer 19:

Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,
$$\begin{gathered} PQ=\sqrt{P{R}^2-Q{R}^2} \\ =\sqrt{{50}^2-{48}^2} \\ =\sqrt{2500-2304} \\ =\sqrt{196} \\ =14 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} =\frac{1}{2}\times QR\times PQ \\ =\frac{1}{2}\times 48\times 14 \\ =336 {\mathrm{cm}}^2 \end{gathered}$$

Question 20:

Find the area of the shaded region in the figure given below.

Answer 20:

In right angled ∆ABD,
AB² = AD² + DB²          (Pythagoras Theorem)
⇒ AB² = 12² + 16²
⇒ AB² = 144 + 256
⇒ AB² = 400
⇒ AB = 20 cm

Area of ∆ADB = $\frac{1}{2}\times DB\times AD$
                        = $\frac{1}{2}\times 16\times 12$
                        = 96 cm²               ....(1)

In ∆ACB,
The sides of the triangle are of length 20 cm, 52 cm and 48 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+52+48}{2}=\frac{120}{2}=60 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ACB=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{60\left(60-20\right)\left(60-52\right)\left(60-48\right)} \\ =\sqrt{60\left(40\right)\left(8\right)\left(12\right)} \\ =480 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$


Now,
Area of the shaded region = Area of ∆ACB − Area of ∆ADB
                                          = 480 − 96
                                          = 384 cm²

Hence, the area of the shaded region in the given figure is 384 cm².

Question 21:

The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle. Find its area. (Given, $\sqrt{6}=2.45$).

Answer 21:

In the given figure, ABCD is a quadrilateral with sides of length 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle.

 

Join AC.

In right angled ∆ABC,
AC² = AB² + BC²          (Pythagoras Theorem)
⇒ AC² = 6² + 8²
⇒ AC² = 36 + 64
⇒ AC² = 100
⇒ AC = 10 cm

Area of ∆ABC = $\frac{1}{2}\times AB\times BC$
                        = $\frac{1}{2}\times 6\times 8$
                        = 24 cm²               ....(1)

In ∆ACD,
The sides of the triangle are of length 10 cm, 12 cm and 14 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+12+14}{2}=\frac{36}{2}=18 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{18\left(18-10\right)\left(18-12\right)\left(18-14\right)} \\ =\sqrt{18\left(8\right)\left(6\right)\left(4\right)} \\ =24\sqrt{6} {\mathrm{cm}}^2 \\ =24\left(2.45\right) {\mathrm{cm}}^2 \\ =58.8 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
                                             = (24 + 58.8) cm²
                                             = 82.8 cm²

Hence, the area of quadrilateral ABCD is 82.8 cm².

Question 22:

Find the perimeter and area of a quadrilateral ABCD in which BC = 12 cm, CD = 9 cm, BD = 15 cm, DA = 17 cm and ∠ABD = 90°.

Answer 22:

We know that $△$ABD is a right-angled triangle.
∴ $A{B}^2=\sqrt{A{D}^2-D{B}^2}=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8 \mathrm{cm}$
 $$\begin{gathered} \mathrm{Now,} \\ Area \mathrm{of} \mathrm{triangle} ABD=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ = \frac{1}{2}\times AB\times BD \\ = \frac{1}{2}\times 8\times 15 \\ =60 {\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Let}: \\ a=9 \mathrm{cm}, b = 15 \mathrm{cm} \mathrm{and} c=12 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{9+15+12}{2}=18 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} DBC = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{18(18-9)(18-15)(18-12)} \\ =\sqrt{18\times 9\times 3\times 6} \\ =\sqrt{6\times 3\times 3\times 3\times 3\times 6} \\ =6\times 3\times 3 \\ =54 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of quadrilateral ABCD  = Area of $△$ABD + Area of $△$BCD
                                           = (60 + 54) cm² =114 cm²
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 17 + 8 + 12 + 9 = 46 cm

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Question 23:

Find the perimeter and area of the quadrilateral ABCD in which AB = 21 cm, ∠BAC = 90°, AC = 20 cm, CD = 42 cm and AD = 34 cm.

Answer 23:

In right angled ∆ABC,
BC² = AB² + AC²          (Pythagoras Theorem)
⇒ BC² = 21² + 20²
⇒ BC² = 441 + 400
⇒ BC² = 841
⇒ BC = 29 cm

Area of ∆ABC = $\frac{1}{2}\times AB\times AC$
                        = $\frac{1}{2}\times 21\times 20$
                        = 210 cm²               ....(1)

In ∆ACD,
The sides of the triangle are of length 20 cm, 34 cm and 42 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+34+42}{2}=\frac{96}{2}=48 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{48\left(48-20\right)\left(48-34\right)\left(48-42\right)} \\ =\sqrt{48\left(28\right)\left(14\right)\left(6\right)} \\ =336 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
                                             = (210 + 336) cm²
                                             = 546 cm²

Also,
Perimeter of quadrilateral ABCD = (34 + 42 + 29 + 21) cm
                                                     = 126 cm

Hence, the perimeter and area of quadrilateral ABCD is 126 cm and 546 cm², respectively.

Question 24:

Find the area of the quadrilateral ABCD in which BCD is an equilateral triangle, each of whose sides is 26 cm, AD = 24 cm and ∠BAD = 90°. Also, find the perimeter of the quadrilateral. (Given: $\sqrt{3}$ = 1.73.)

Answer 24:

We know that $△$BAD is a right-angled triangle.

∴ $AB=\sqrt{B{D}^2-A{D}^2}=\sqrt{{26}^2-{24}^2}=\sqrt{676-576}=\sqrt{100}=10 \mathrm{cm}$

 $$\begin{gathered} \mathrm{Now,} \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} BAD=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ = \frac{1}{2}\times AB\times AD \\ = \frac{1}{2}\times 10\times 24 \\ =120 {\mathrm{cm}}^2 \end{gathered}$$

Also, we know that $△$BDC is an equilateral triangle.
$$\begin{gathered} \therefore \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times (26{)}^2 \\ =\frac{\sqrt{3}}{4}\times 676 \\ =169\sqrt{3} \\ =292.37 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of quadrilateral ABCD = Area of $△$ABD + Area of $△$BDC
                                         = (120 + 292.37) cm² = 412.37 cm²
Perimeter of ABCD = AB + BC + CD + DA = 10 + 26+ 26 + 24 = 86 cm

Question 25:

Find the area of a parallelogram ABCD in which AB = 28 cm, BC = 26 cm and diagonal AC = 30 cm.

Answer 25:

$$\begin{gathered} \mathrm{Let}: \\ a=26 \mathrm{cm}, b =30 \mathrm{cm} \mathrm{and} c=28 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{26+30+28}{2}=42 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} ABC = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{42(42-26)(42-30)(42-28)} \\ =\sqrt{42\times 16\times 12\times 14} \\ =\sqrt{14\times 3\times 4\times 4\times 2\times 2\times 3\times 14} \\ =14\times 4\times 2\times 3 \\ =336 {\mathrm{cm}}^2 \end{gathered}$$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2\times 336=672 {\mathrm{cm}}^2$

Question 26:

Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16 cm. [Given: $\sqrt{3}=1.73$]

Answer 26:

$$\begin{gathered} \mathrm{Let}: \\ a=10 \mathrm{cm}, b =16 \mathrm{cm} \mathrm{and} c=14 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{10+16+14}{2}=20 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} ABC = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{20(20-10)(20-16)(20-14)} \\ =\sqrt{20\times 10\times 4\times 6} \\ =\sqrt{10\times 2\times 10\times 2\times 2\times 3\times 2} \\ =10\times 2\times 2\sqrt{3} \\ =69.2 {\mathrm{cm}}^2 \end{gathered}$$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2\times 69.2 {\mathrm{cm}}^2=138.4 {\mathrm{cm}}^2$

Question 27:

In the given figure ABCD is a quadrilateral in which diagonal BD = 64 cm, AL ⊥ BD and CM ⊥ BD such that AL = 16.8 cm and CM = 13.2 cm. Calculate the area of quadrilateral ABCD.

Answer 27:

$$\begin{gathered} \mathrm{Area} \mathrm{of} ABCD=\mathrm{Area} \mathrm{of} △ABD+\mathrm{Area} \mathrm{of} △BDC \\ =\frac{1}{2}\times BD\times AL+\frac{1}{2}\times BD\times CM \\ =\frac{1}{2}\times BD(AL+CM) \\ =\frac{1}{2}\times 64(16.8+13.2) \\ =32\times 30 \\ =960 {\mathrm{cm}}^2 \end{gathered}$$

Question 28:

The area of a trapezium is 475 cm² and its height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

Answer 28:

In the given figure, ABCD is a trapezium with parallel sides AB and CD.



Let the length of CD be x.
Then, the length of AB be x + 4.

Area of trapezium = $\frac{1}{2}\times \mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}\times \mathrm{height}$
$$\begin{gathered} \Rightarrow 475=\frac{1}{2}\times \left(x+x+4\right)\times 19 \\ \Rightarrow 475\times 2=19\left(2x+4\right) \\ \Rightarrow 950=38x+76 \\ \Rightarrow 38x=950-76 \\ \Rightarrow 38x=874 \\ \Rightarrow x=\frac{874}{38} \\ \Rightarrow x=23 \end{gathered}$$

∴ The length of CD is 23 cm and the length of AB is 27 cm.

Hence, the lengths of two parallel sides is 23 cm and 27 cm.                                                                                                  

Question 29:

In the given figure, a ∆ABC has been given in which AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram DBCE of the same area as that of ∆ABC is constructed. Find the height DL of the parallelogram.

Answer 29:

In ∆ABC,
The sides of the triangle are of length 7.5 cm, 6.5 cm and 7 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{7.5+6.5+7}{2}=\frac{21}{2}=10.5 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{10.5\left(10.5-7.5\right)\left(10.5-6.5\right)\left(10.5-7\right)} \\ =\sqrt{10.5\left(3\right)\left(4\right)\left(3.5\right)} \\ =21 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$

Now,
Area of parallelogram DBCE = Area of ∆ABC
                                               = 21 cm²

Also,
Area of parallelogram DBCE = base × height
$$\begin{gathered} \Rightarrow 21=BC\times DL \\ \Rightarrow 21=7\times DL \\ \Rightarrow DL=\frac{21}{7}=3 \mathrm{cm} \end{gathered}$$

Hence, the height DL of the parallelogram is 3 cm.

Question 30:

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 5 to plough 1 m² of the field, find the total cost of ploughing the field.

Answer 30:

In the given figure, ABCD is a trapezium having parallel sides 90 m and 30 m.



Draw DE perpendicular to AB, such that DE = BC.

In right angled ∆ADE,
AD² = AE² + ED²          (Pythagoras Theorem)
⇒ 100² = (90 − 30)² + ED²
⇒ 10000 = 3600 + ED²
⇒ ED² = 10000 − 3600
⇒ ED² = 6400
⇒ ED = 80 m

Thus, the height of the trapezium = 80 m        ...(1)

Now,
Area of trapezium = $\frac{1}{2}\times \mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}\times \mathrm{height}$
                              = $\frac{1}{2}\times \left(90+30\right)\times 80$
                              = 4800 m²

The cost to plough per m² = Rs 5
The cost to plough 4800 m² = Rs 5 × 4800
                                             = Rs 24000

Hence, the total cost of ploughing the field is Rs 24000.

Page:536

Question 31:

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 the other sides. Find the largest area where house can be constructed.

Answer 31:

Let ABCD be 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, the length of the inner rectangle = 40 − 3 − 3 = 34 m and the breath of the inner rectangle = 15 − 2 − 2 = 11 m.

∴ Area of the inner rectangle PQRS = Length × Breath
                                                          = 34 × 11
                                                          = 374 m²

Hence, the largest area where house can be constructed is 374 m².

Question 32:

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 cm². Find the cost of painting.

Answer 32:

Let the sides of rhombus be of length x cm.



Perimeter of rhombus = 4x
⇒ 40 = 4x
⇒ x = 10 cm

Now,
In ∆ABC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+10+12}{2}=\frac{32}{2}=16 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{16\left(16-10\right)\left(16-10\right)\left(16-12\right)} \\ =\sqrt{16\left(6\right)\left(6\right)\left(4\right)} \\ =48 {\mathrm{cm}}^2 ...\left(1\right) \end{gathered}$$

In ∆ADC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+10+12}{2}=\frac{32}{2}=16 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ADC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{16\left(16-10\right)\left(16-10\right)\left(16-12\right)} \\ =\sqrt{16\left(6\right)\left(6\right)\left(4\right)} \\ =48 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$


∴ Area of the rhombus = Area of ∆ABC + Area of ∆ADC
                                      = 48 + 48
                                      = 96 cm²

The cost to paint per cm² = Rs 5
The cost to paint 96 cm² = Rs 5 × 96
                                        = Rs 480
The cost to paint both sides of the sheet = Rs 2 × 480
                                                                = Rs 960

Hence, the total cost of painting is Rs 960.

Question 33:

The difference between the semiperimeter and the sides of a ∆ABC are 8 cm, 7 cm and 5 cm respectively. Find the area of the triangle.

Answer 33:

Let the semi-perimeter of the triangle be s.
Let the sides of the triangle be a, b and c.
Given: s − a = 8, s − b = 7 and s − c = 5      ....(1)

Adding all three equations, we get
3s − (a + b + c) = 8 + 7 + 5
⇒ 3s − (a + b + c) = 20
⇒ 3s − 2s = 20                       $\left(\because s=\frac{a+b+c}{2}\right)$
⇒ s = 20 cm                  ...(2)

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{20\left(8\right)\left(7\right)\left(5\right)} \left(\mathrm{from} \left(1\right) \mathrm{and} \left(2\right)\right) \\ =20\sqrt{14} {\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the triangle is $20\sqrt{14} {\mathrm{cm}}^2$.

Question 34:

A floral design on a floor is made up of 16 tiles, each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per sq cm.

Answer 34:

Area of one triangular-shaped tile can be found in the following manner:

$$\begin{gathered} \mathrm{Let}: \\ a=16 \mathrm{cm}, b = 12 \mathrm{cm} \mathrm{and} c=20 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{16+12+20}{2}=24 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{24(24-16)(24-12)(24-20)} \\ =\sqrt{24\times 8\times 12\times 4} \\ =\sqrt{6\times 4\times 4\times 4\times 4\times 6} \\ =6\times 4\times 4 \\ =96 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of 16 triangular-shaped tiles = $16\times 96=1536 {\mathrm{cm}}^2$
Cost of polishing tiles of area 1 cm² = Rs 1
Cost of polishing tiles of area 1536 cm² = $1\times 1536 =\mathrm{Rs} 1536$

Question 35:

An umbrella is made by stitching 12 triangular pieces of cloth, each measuring (50 cm × 20 cm × 50 cm). Find the area of the cloth used in it.

Answer 35:

We know that the triangle is an isosceles triangle.
Thus, we can find out the area of one triangular piece of cloth.
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{isosceles} \mathrm{triangle} = \frac{b}{4}\sqrt{4a^2-b^2} \\ =\frac{20}{4}\times \sqrt{4(50{)}^2-{20}^2} (a= 50 \mathrm{cm} \mathrm{and} b=20 \mathrm{cm}) \\ =5\times \sqrt{10000-400} \\ =5\times \sqrt{9600} \\ =5\times 40\sqrt{6} \\ =200\sqrt{6} \\ =490 {\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of 1 triangular piece of cloth = 490 cm²
Area of 12 triangular pieces of cloth = $12\times 490 =5880 {\mathrm{cm}}^2$

Question 36:

In the given figure, ABCD is a square with diagonal 44 cm. How much paper of each shade is needed to make a kite given in the figure?

Answer 36:

In the given figure, ABCD is a square with diagonal 44 cm.
∴ AB = BC = CD = DA.        ....(1)

In right angled ∆ABC,
AC² = AB² + BC²          (Pythagoras Theorem)
⇒ 44² = 2AB²
⇒ 1936 = 2AB²
⇒ AB² = $\frac{1936}{2}$
⇒ AB² = 968
⇒ AB = $22\sqrt{2}$ cm      ...(2)

∴ Sides of square = AB = BC = CD = DA = $22\sqrt{2}$ cm

Area of square ABCD = (side)²
                                    = ($22\sqrt{2}$)²
                                    = 968 cm²          ...(3)

Area of red portion = $\frac{968}{4}=242 {\mathrm{cm}}^2$
Area of yellow portion = $\frac{968}{2}=484 {\mathrm{cm}}^2$
Area of green portion = $\frac{968}{4}=242 {\mathrm{cm}}^2$

Now, in ∆AEF,
The sides of the triangle are of length 20 cm, 20 cm and 14 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+14}{2}=\frac{54}{2}=27 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆AEF=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{27\left(27-20\right)\left(27-20\right)\left(27-14\right)} \\ =\sqrt{27\left(7\right)\left(7\right)\left(13\right)} \\ =21\sqrt{39} \\ =131.04 {\mathrm{cm}}^2 ...\left(4\right) \end{gathered}$$

Total area of the green portion = 242 + 131.04 = 373.04 cm²

Hence, the paper required of each shade to make a kite is red paper 242 cm², yellow paper 484 cm² and green paper 373.04 cm².

Page:537

Question 37:

A rectangular lawn, 75 m by 60 m, has two roads, each road 4 m wide, running through the middle of the lawn, one parallel to length and the other parallel to breadth, as shown in the figure. Find the cost of gravelling the roads at Rs 50 per m².

Answer 37:

Area of rectangle ABCD = Length × Breath
                                        = 75 × 4
                                        = 300 m²

Area of rectangle PQRS = Length × Breath
                                       = 60 × 4
                                       = 240 m²

Area of square EFGH = (side)²
                                    = (4)²
                                    = 16 m²

∴ Area of the footpath = Area of rectangle ABCD + Area of rectangle PQRS − Area of square EFGH
                                     = 300 + 240 − 16
                                     = 524 m²

The cost of gravelling the road per m² = Rs 50
The cost of gravelling the roads 524 m² = Rs 50 × 524
                                                                = Rs 26200

Hence, the total cost of gravelling the roads at Rs 50 per m² is Rs 26200.

Question 38:

The shape of the cross section of a canal is a trapezium. If the canal is 10 m wide at the top, 6 m wide at the bottom and the area of its cross section is 640 m², find the depth of the canal.

Answer 38:

The top and the bottom of the canal are parallel to each other.
Let the height of the trapezium be h.

Area of trapezium = $\frac{1}{2}\times \mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}\times \mathrm{height}$
⇒ 640 = $\frac{1}{2}\times \left(10+6\right)\times h$
⇒ 640 = $8\times h$
⇒ h = $\frac{640}{8}$
⇒ h = 80 m

Hence, the depth of the canal is 80 m.

Question 39:

Find the area of a trapezium whose parallel sides are 11 m and 25 m long, and the nonparallel sides are 15 m and 13 m long.

Answer 39:

In the given figure, ABCD is the trapezium.

 

Draw a line BE parallel to AD.

In ∆BCE,
The sides of the triangle are of length 15 m, 13 m and 14 m.
∴ Semi-perimeter of the triangle is
$s=\frac{15+13+14}{2}=\frac{42}{2}=21 \mathrm{m}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆BCE=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-15\right)\left(21-13\right)\left(21-14\right)} \\ =\sqrt{21\left(6\right)\left(8\right)\left(7\right)} \\ =84 {\mathrm{m}}^2 ...\left(1\right) \end{gathered}$$

Also,
Area of ∆BCE = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$
$$\begin{gathered} \Rightarrow 84=\frac{1}{2}\times 14\times \mathrm{Height} \\ \Rightarrow 84=7\times \mathrm{Height} \\ \Rightarrow \mathrm{Height}=\frac{84}{7} \\ \Rightarrow \mathrm{Height}=12 \mathrm{m} \end{gathered}$$

∴ Height of ∆BCE = Height of the parallelogram ABED = 12 m

Now,
Area of the parallelogram ABED = Base × Height
                                                     = 11 × 12
                                                     = 132 m²                     ...(2)

∴ Area of the trapezium = Area of the parallelogram ABED + Area of the triangle BCE
                                        = 132 + 84
                                        = 216 m²

Hence, the area of a trapezium is 216 m².

Question 40:

The difference between the lengths of the parallel sides of a trapezium is 8 cm, the perpendicular distance between these sides is 24 cm and the area of the trapezium is 312 cm². Find the length of each of the parallel sides.

Answer 40:

Let the length of the parallel sides be x and x − 8.
The height of the trapezium = 24 cm

Area of trapezium = $\frac{1}{2}\times \mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}\times \mathrm{height}$
⇒ 312 = $\frac{1}{2}\times \left(x+x-8\right)\times 24$
⇒ 312 = 12(2x − 8)
⇒ 2x − 8 = $\frac{312}{12}$
⇒ 2x − 8 = 26
⇒ 2x = 26 + 8
⇒ 2x = 34
⇒ x = 17 cm

Hence, the lengths of the parallel sides are 17 cm and 9 cm.

Question 41:

A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure 120 m and 44 m. If one of the sides of the parallelogram measures 66 m, find its corresponding altitude.

Answer 41:

Diagonals d₁ and d₂ of the rhombus measure 120 m and 44 m, respectively.

Base of the parallelogram = 66 m

Now,
Area of the rhombus = Area of the parallelogram
$$\begin{gathered} \Rightarrow \frac{1}{2}\times d_1\times d_2=\mathrm{Base}\times \mathrm{Height} \\ \Rightarrow \frac{1}{2}\times 120\times 44=66\times \mathrm{Height} \\ \Rightarrow 60\times 44=66\times \mathrm{Height} \\ \Rightarrow 2640=66\times \mathrm{Height} \\ \Rightarrow \mathrm{Height}=\frac{2640}{66} \\ \Rightarrow \mathrm{Height}=40 \mathrm{m} \end{gathered}$$

Hence, the measure of the altitude of the parallelogram is 40 m.

Question 42:

A parallelogram and a square have the same area. If the sides of the square measure 40 m and altitude of the parallelogram measures 25 m, find the length of the corresponding base of the parallelogram.

Answer 42:

It is given that,
Sides of the square = 40 m
Altitude of the parallelogram = 25 m

Now,
Area of the parallelogram = Area of the square
$$\begin{gathered} \Rightarrow \mathrm{Base}\times \mathrm{Height}={\left(\mathrm{side}\right)}^2 \\ \Rightarrow \mathrm{Base}\times 25={\left(40\right)}^2 \\ \Rightarrow \mathrm{Base}\times 25=1600 \\ \Rightarrow \mathrm{Base}=\frac{1600}{25} \\ \Rightarrow \mathrm{Base}=64 \mathrm{m} \end{gathered}$$

Hence, the length of the corresponding base of the parallelogram is 64 m.

Question 43:

Find the area of a rhombus one side of which measures 20 cm and one of whose diagonals is 24 cm.

Answer 43:

It is given that,
The sides of rhombus = 20 cm.
One of the diagonal = 24 cm.

 

In ∆ABC,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+24}{2}=\frac{64}{2}=32 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{32\left(32-20\right)\left(32-20\right)\left(32-24\right)} \\ =\sqrt{32\left(12\right)\left(12\right)\left(8\right)} \\ =192 {\mathrm{cm}}^2 ...\left(1\right) \end{gathered}$$

In ∆ACD,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+24}{2}=\frac{64}{2}=32 \mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{32\left(32-20\right)\left(32-20\right)\left(32-24\right)} \\ =\sqrt{32\left(12\right)\left(12\right)\left(8\right)} \\ =192 {\mathrm{cm}}^2 ...\left(2\right) \end{gathered}$$


∴ Area of the rhombus = Area of ∆ABC + Area of ∆ACD
                                      = 192 + 192
                                      = 384 cm²

Hence, the area of a rhombus is 384 cm².

Question 44:

The area of a rhombus is 480 cm², and one of its diagonals measures 48 cm. Find (i) the length of the other diagonal, (ii) the length of each of its sides, and (iii) its perimeter.

Answer 44:

It is given that,
Area of rhombus = 480 cm².
One of the diagonal = 48 cm.

(i) Area of the rhombus = $\frac{1}{2}\times d_1\times d_2$
$$\begin{gathered} \Rightarrow 480=\frac{1}{2}\times 48\times d_2 \\ \Rightarrow 480=24\times d_2 \\ \Rightarrow d_2=\frac{480}{24} \\ \Rightarrow d_2=20 \mathrm{cm} \end{gathered}$$

Hence, the length of the other diagonal is 20 cm.

(ii) We know that the diagonals of the rhombus bisect each other at right angles.

 

In right angled ∆ABO,
AB² = AO² + OB²          (Pythagoras Theorem)
⇒ AB² = 24² + 10²
⇒ AB² = 576 + 100
⇒ AB² = 676
⇒ AB = 26 cm

Hence, the length of each of the sides of the rhombus is 26 cm.

(iii) Perimeter of the rhombus = 4 × side
                                                = 4 × 26
                                                = 104 cm

Hence, the perimeter of the rhombus is 104 cm.