Rs Aggarwal 2019 2020 Solutions for Class 9 Math Chapter 14 Areas Of Triangles And Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Areas Of Triangles And Quadrilaterals are extremely popular among Class 9 students for Math Areas Of Triangles And Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 9 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Question 1:

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

We have:
Base = 24 cm
Height = 14.5 cm

Now,

#### 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.

Let the height of the triangle be h m.
∴ Base = 3h m
Now,
Area of the triangle =
We have:

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.

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:

#### 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.

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:

#### 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.

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.

#### 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.

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$×$5 = 25 m
12$×$5 = 60 m
13$×$5 = 65 m

Now,

#### 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.

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$×$10 = 250 m
17$×$10 = 170 m
12$×$10 = 120 m

Now,

Cost of ploughing 1 m2 field = Rs 5
Cost of ploughing 9000 m2 field = .

#### 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.

(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:

#### Question 9:

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

#### Question 10:

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

Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,

Also,

∴ 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.

The ratio of the equal side to its base is 3 : 2.
$⇒$Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.

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

Disclaimer: The answer does not match with the answer given in the book.

#### 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.

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

∴ By Heron's formula,

Hence, the area of the triangle is .

#### 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?

The sides of the triangle are of length 13 m, 14 m and 15 m.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

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.

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

So, equal sides of the triangle are 12 cm each.
Also,
Base = $\frac{3}{2}$a =
(ii)

(iii)

#### Question 15:

If the area of an equilateral triangle is , find its perimeter.

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

#### Question 16:

If the area of an equilateral triangle is , find its height.

Now, we have:

#### 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 $\sqrt{3}=1.732$.

Side of the equilateral triangle = 8 cm

(i)

(ii)

#### Question 18:

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

Height of the equilateral triangle = 9 cm
Thus, we have:

Also,

#### Question 19:

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

Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,

#### Question 20:

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

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

Area of ∆ADB = $\frac{1}{2}×DB×AD$
= $\frac{1}{2}×16×12$
= 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

∴ By Heron's formula,

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, $\sqrt{6}=2.45$).

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 = $\frac{1}{2}×AB×BC$
= $\frac{1}{2}×6×8$
= 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

∴ By Heron's formula,

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°.

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

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

#### 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.

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

Area of ∆ABC = $\frac{1}{2}×AB×AC$
= $\frac{1}{2}×21×20$
= 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

∴ By Heron's formula,

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.)

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

∴

Also, we know that $△$BDC is an equilateral triangle.

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.

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) =

#### Question 26:

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

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) =

#### Question 27:

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

#### 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.

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 =
$⇒475=\frac{1}{2}×\left(x+x+4\right)×19\phantom{\rule{0ex}{0ex}}⇒475×2=19\left(2x+4\right)\phantom{\rule{0ex}{0ex}}⇒950=38x+76\phantom{\rule{0ex}{0ex}}⇒38x=950-76\phantom{\rule{0ex}{0ex}}⇒38x=874\phantom{\rule{0ex}{0ex}}⇒x=\frac{874}{38}\phantom{\rule{0ex}{0ex}}⇒x=23\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

∴ 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.

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

∴ By Heron's formula,

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

Also,
Area of parallelogram DBCE = base × height

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.

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 =
= $\frac{1}{2}×\left(90+30\right)×80$
= 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.

#### 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.

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.

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

∴ By Heron's formula,

The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

∴ 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.

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

Adding all three equations, we get
3s − (a + b + c) = 8 + 7 + 5
⇒ 3s − (a + b + c) = 20
⇒ 3s − 2s = 20
s = 20 cm                  ...(2)

∴ By Heron's formula,

Hence, the area of the triangle is .

#### 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.

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

Now,
Area of 16 triangular-shaped tiles =
Cost of polishing tiles of area 1 cm2 = Rs 1
Cost of polishing tiles of area 1536 cm2 =

#### 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.

We know that the triangle is an isosceles triangle.
Thus, we can find out the area of one triangular piece of cloth.

Now,
Area of 1 triangular piece of cloth = 490 cm2
Area of 12 triangular pieces of cloth =

#### 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?

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 =
Area of yellow portion =
Area of green portion =

Now, in ∆AEF,
The sides of the triangle are of length 20 cm, 20 cm and 14 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

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.

#### 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.

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.

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

Area of trapezium =
⇒ 640 = $\frac{1}{2}×\left(10+6\right)×h$
⇒ 640 = $8×h$
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.

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

∴ By Heron's formula,

Also,
Area of ∆BCE = $\frac{1}{2}×\mathrm{Base}×\mathrm{Height}$

∴ 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.

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

Area of trapezium =
⇒ 312 = $\frac{1}{2}×\left(x+x-8\right)×24$
⇒ 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.

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

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.

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

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.

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

∴ By Heron's formula,

In ∆ACD,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

∴ 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.

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

(i) Area of the rhombus = $\frac{1}{2}×{d}_{1}×{d}_{2}$

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.

#### Question 1:

In a ABC it is given that base = 12 cm and height = 5 cm. Its area is
(a) 60 cm2
(b) 30 cm2
(c)
(d) 45 cm2

(b) 30 cm2

#### Question 2:

The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
(a) 96 cm2
(b) 120 cm2
(c) 144 cm2
(d) 160 cm2

(a) 96 cm2

#### Question 3:

Each side of an equilateral triangle measure 8 cm. The area of the triangle is
(a)
(b)
(c)
(d) 48 cm2

(b)

#### Question 4:

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
(a)
(b)
(c)
(d)

(b)

#### Question 5:

The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
(a) 8 cm
(b)
(c) 4 cm
(d)

(c) 4 cm

#### Question 6:

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
(a)
(b) 50 cm2
(c)
(d)
75 cm2

(b) 50 cm2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:

#### Question 7:

Each side of an equilateral triangle is 10 cm long. The height of the triangle is
(a)
(b)
(c)
(d)
5 cm

(b)

#### Question 8:

The height of an equilateral triangle is 6 cm. Its area is
(a)
(b)
(c)
(d)
18 cm2

(a)

#### Question 9:

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
(a) 480 m2
(b) 320 m2
(c) 384 m2
(d) 360 m2

(c) 384 m2

#### Question 10:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is
(a) 375 cm2
(b) 750 cm2
(c) 250 cm2
(d)
500 cm2

(b) 750 cm2

Let the sides of the triangle be 5x cm, 12x cm and 13x cm.
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, the sides of the triangle are 5$×$5 cm, 12$×$5 cm and 13$×$5 cm, i.e., 25 cm, 60 cm and 65 cm.

Now,

#### Question 11:

The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is
(a) 24 cm
(b) 18 cm
(c) 30 cm
(d)
12 cm

(a) 24 cm

The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:

#### Question 12:

The base of an isosceles triangle is 16 cm and its area is 48 cm2. The perimeter of the triangle is
(a) 41 cm
(b) 36 cm
(c) 48 cm
(d)
324 cm

(b) 36 cm

Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,

∴ Perimeter = (10 + 10 + 16) cm = 36 cm

#### Question 13:

The area of an equilateral triangle is . Its perimeter is
(a) 36 cm
(b)
(c) 24 cm
(d) 30
cm

(a) 36 cm

Now,
Perimeter
= 3 × Side = 3 × 12 = 36 cm

#### Question 14:

Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
(a) 156 cm2
(b) 78 cm2
(c) 60 cm2
(d) 120
cm2

(c) 60 cm2

#### Question 15:

The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is
(a) 168 cm2
(b) 252 cm2
(c) 336 cm2
(d) 504
cm2

(c) 336 cm2

Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,

#### Question 16:

The area of an equilateral triangle is . Its height is
(a)
(b)
(c)
(d) 9
cm