Areas
Class 9th Mathematics RS Aggarwal And V Aggarwal Solution
- Find the area of the triangle whose base measures 24 cm and the corresponding…
- The base of a triangular field is three times its altitude. If the cost of…
- Find the area of triangle whose sides are 42cm,34 cm and 20cm in length. Hence…
- Calculate the area of the triangle whose sides are 18cm, 24cm and 30 cm in…
- Find the area of a triangular field whose sides are 91m, 98m and 105m in length.…
- The sides of triangle are in the ratio 5 : 12 : 13 and its perimeter is 150m.…
- The perimeter of a triangular field is 540m and its sides are in the ratio 25 :…
- Two sides of a triangular field are 85m and 154m in length and its perimeter is…
- Find the area of an isosceles triangle each of whose equal sides measures 13cm…
- The base of an isosceles triangle measures 80cm and its area is 360cm^2 . Find…
- The perimeter of an isosceles triangle is 42cm and its base is 1 1/2 times,…
- If the area of the equilateral triangle is 36 root 3 cm^2 , find its perimeter.…
- If the area of the equilateral triangle is 81 root 3 cm^2 , find its height.…
- The base of a right - angles triangle measures 48cm and its hypotenuse measures…
- Each side of an equilateral triangle measures 8cm. Find: (i) The area of the…
- The height of an equilateral triangle measures 9 cm. Find its area, correct to…
- An umbrella is made by stitching 12 triangular pieces of cloth, each measuring…
- A floral design on a floor is made up of 16 tiles, each triangular in shape…
- Find the perimeter and area of the quadrilateral ABCD in which AB = 17cm, AD =…
- Find the perimeter and area of the quadrilateral ABCD in which AB = 42 cm, BC =…
- Find the area of the quadrilateral ABCD in which AD = 42 cm , angle bad = 90^0…
- Find the area of a parallelogram ABCD in which AB = 28cm, BC = 26cm and…
- Find the area of the parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC…
- In the figure ABCD is a quadrilateral in which diagonal BD = 64 cm, AL…
- In a ΔABC it is given that base = 12cm and height = 5cm. Its area isA. 60cm^2 B. 30…
- The length of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the…
- Each side of an equilateral triangle measures 8 cm. The area of the triangle isA. 8…
- The base of an isosceles triangle is 8cm long and each of its equal sides measures 6cm.…
- The base of an isosceles triangle is 6cm and each of its equal sides is 5cm. The height…
- Each of the two equal sides of an isosceles right triangle is 10cm long. Its area is…
- Each side of an equilateral triangle is 10 cm long. The height of the triangle is -A.…
- The height of an equilateral triangle is 6cm. Its area is -A. 12 root 3 cm^2 B. 6 root…
- The length of the three sides of the triangular field are 40 m, 24 m and 32 m…
- The sides of the triangle are in the ratio 5: 12:13 and its perimeter is 150 cm. The…
- The lengths of the three sides of the triangle are 30 cm, 24 cm and 18 cm…
- The base of an isosceles triangle is 16 cm and its area is 48cm^2 . The perimeter of…
- The area of an equilateral triangle is 36 root 3 cm^2 . Its perimeter is -A. 36cm B.…
- Each of the equal sides of an isosceles triangle is 13cm and base is 24cm. The area of…
- The base of a right triangle is 48 cm and its hypotenuse is 50 cm long. The area of…
- The area of an equilateral triangle is 81 root 3 cm^2 . Its height is-A. 9 root 3 cm…
- The difference between the semi- perimeter and the sides of a ∆ ABC are 8cm, 7cm and…
- For an isosceles right angles triangle having each of equal sides ‘a’ , we have I.…
- For an isosceles triangle having base b and each of the equal sides as a, we have: I.…
- Assertion (A) Reason (R) Area of an equilateral triangle having each side equal to 4cm…
- Assertion(A) Reason (R) The area of an isosceles triangle having base = 8cm and each…
- Assertion(A) Reason (R) The area of an equilateral triangle having side 4cm is 3cm^2…
- Assertion (A) Reason (R) The sides of the triangle ABC are in the ratio 2 : 3 : 4 and…
- Assertion (A) Reason (R) The area of an isosceles triangle having base = 24 cm and…
- If the base of an isosceles triangle is 6cm and its perimeter is 16cm, then its area…
- If each side of an equilateral triangle is 8cm long, then its area is 20 root 3 cm^2 .…
- If the sides of a triangular field measures 52m, 37 m and 20 m, then the cost of…
- Match the following columns. Column I Column II (a) The lengths of three sides of a…
- A park in the shape of a quadrilateral ABCD has AB = 9m, BC = 12m, CD = 5 cm, AD = 8 m…
- Find the area of a parallelogram ABCD in which AB = 60cm, BC = 40cm and AC = 80 cm.…
- A piece of land is in the shape of a rhombus ABCD in which each side measures 100m and…
- A floral design on a floor is made up of 16 triangular tiles, each having sides 9cm,…
- A kite in the shape of a square with each diagonal 32 cm and having a tail in the…
- Each side of an equilateral triangle is 8 cm. Its altitude isA. 2 root 2 cm B. 2 root 3…
- The perimeter of an isosceles right- angled triangle having a as each of the equal…
- For an isosceles triangle having base = 12 cm and each of the equal sides equal to 10…
- Find the area of an equilateral triangle having each side 6cm.
- Using Heron’s formula find the area of Δ ABC in which BC = 13 cm, AC = 14 cm and AB =…
- The sides of a triangle are in the ratio 13: 14: 15 and its perimeter is 84 cm. Find…
- Find the area of ABC in which BC = 8cm, AC = 15cm and AB = 17 cm. Find the length of…
- An isosceles triangle has perimeter 30cm and each of its equal sides is 12cm. Find the…
- The perimeter of an isosceles triangle is 32cm. The ratio of one of the equal side to…
- Given a ABC in which I. A, B and C are in the ratio 3: 2 :1. II. AB , AC and BC are in…
- In the given figure ABC and DBC have the same base BC such that AB = 120m, AC = 122m,…
- A point O is taken inside an equilateral ΔABC. If OL 1 BC, OM 1 AC and ON 1 AB such…
Exercise 7a
Question 1.Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
Answer:
Given,
Base of triangle, b = 24 cm
Height of triangle = 14.5 cm
We have to find out the area of the given triangle
We know that,
Area of triangle = 
× base × height
= 174 cm2
Hence, the area of the given triangle is 174 cm2
Question 2.
The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs58 per hectare is Rs783, find its base and height.
Answer:
It is given that the base of the triangular field is three times greater than its altitude
Let us assume height of the triangular field be x and base be 3x
We know that,
Area of triangle = × base × height
= × x × 3x
= 
x2
We know that,
1 hectare = 10,000 sq metre
Given,
Rate of sowing the field per hectare = Rs. 58
Total cost of sowing the triangular field = Rs. 783
Therefore,
Total cost = Area of the triangular field × Rs. 58
x = 300 m
Hence,
Height of the triangular field = x = 300 m
Base of triangular field = 3x = 3 × 300 = 900 m
Question 3.
Find the area of triangle whose sides are 42cm,34 cm and 20cm in length. Hence find the height corresponding to the longest side.
Answer:
Given,
a = 42 cm
b = 34 cm
c = 20 cm
Therefore,
= 
= 48
We know that,
Area = 
Putting the values of a, b and c in the formula, we get
= 
= 
= 
= 4 × 3 × 2 × 14
= 336 cm2
Longest side of the triangle = b = 42 cm
Let h be the corresponding height to the longest side
Therefore,
Area of triangle = 
336 =
42 × h = 336 × 2
h = 
= 16 cm
Hence, corresponding height of the triangle is 16 cm
Question 4.
Calculate the area of the triangle whose sides are 18cm, 24cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.
Answer:
Given,
a = 18 cm
b = 24 cm
c = 30 cm
Therefore,
s = 
= 36
We know that,
Area = 
= 
= 
= 
= 6 × 6 × 3 × 2
= 216 cm2
Smallest side = a = 18 cm
Let, h be the height corresponding to the smallest side of the triangle
Therefore,
Area of triangle =
216 =
18 × h = 216 × 2
h = 
= 24 cm
Question 5.
Find the area of a triangular field whose sides are 91m, 98m and 105m in length. Find the height corresponding to the longest side.
Answer:
Given,
a = 91 m
b = 98 m
c = 105 m
Therefore,
s = 
= 
= 147
We know that,
Area = 
= 
= 
= 
= 49 × 3 × 2 × 2 × 7
= 4116 m2
Longest side = c = 105 cm
Let, h be the height corresponding to the longest side of the triangle
Area of triangle =
4116 =
4116 × 2 = 2 × 4116
h = 
= 78.4 m
Question 6.
The sides of triangle are in the ratio 5 : 12 : 13 and its perimeter is 150m. Find the area of triangle.
Answer:
Let the sides of the given triangle be 5x, 12x and 13x
Given,
Perimeter of the triangle = 150m
Perimeter of the triangle = (5x + 12x + 13x)
150 = 30x
Therefore,
x = 
Thus,
Sides of the triangle are:
5x = 5 × 5 = 25 m
12x = 12 × 5 = 60 m
13x = 13 × 5 = 65 m
Let,
a = 25 m, b = 60 m and c = 65 m
Therefore,
s = 
= 
= 
= 75 m
We know that,
Area of triangle =
= 
= 
= 
= 
= 25 × 5 × 3 × 2
= 750 sq m
Hence, area of triangle is 750 sq m.
Question 7.
The perimeter of a triangular field is 540m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of ploughing the field at Rs. 18.80 per 10m2.
Answer:
Let the sides of the given triangle be 25x, 17x and 12x
Given,
Perimeter of the triangle = 540 m
540 = 25x + 17x + 12x
540 = 54x
x = 
x = 10 m
Thus, sides of the triangle are:
25x = 25 × 10 = 250 m
17x = 17 × 10 = 170 m
12x = 12 × 10 = 120 m
Let,
a = 250 m, b = 170 m and c = 120 m
Therefore,
s =
= 
= 
= 270 m
Therefore,
Area of triangle= 
= 
= 
= 3 × 3 × 10 × 10 × 10
= 9000 m2
Cost of ploughing the field at the rate of Rs. 18.80 per 10 m2 = 
= Rs. 16920
Therefore, cost of ploughing the field is Rs. 16920
Question 8.
Two sides of a triangular field are 85m and 154m in length and its perimeter is 324m. Find:
(i) The area of the field and
(ii) The length of the perpendicular from the opposite vertex of the side measuring 154m.
Answer:
Given,
First side of the triangular field = 85 m
Second side of the triangular field = 154 m
Let the third side be x
Perimeter of the triangular field = 324 m
85 m + 154 m + x = 324 m
x = 324 – 239
x = 85 m
Let the three sides of the triangle be:
a = 85 m, b = 154n m and c = 85 m
Now,
s =
= 
= 
= 162
We know that,
Area of triangle =
= 
= 
= 
= 11 × 9 × 7 × 2 × 2
= 2771 m2
We also know that,
Area of triangle = 
2772 = 
2772 = 77h
h = 
h = 36 m
Therefore,
The length of the perpendicular from the opposite vertex on the side measuring 154 m is 36 m.
Question 9.
Find the area of an isosceles triangle each of whose equal sides measures 13cm and whose base measures 20cm.
Answer:
Let,
a = 13 cm
b = 13 cm
And,
C = 20 cm
Now,
s =
= 
= 
= 23 cm
We know that,
Area of triangle =
= 
= 
= 10
= 10 × 8.306
= 83.06 cm2
Therefore,
Area of isosceles triangle = 83.06 cm2
Question 10.
The base of an isosceles triangle measures 80cm and its area is 360cm2 . Find the perimeter of the triangle.
Answer:
Let us assume 
be an isosceles triangle and let AL perpendicular BC
It is given that,
BC = 80 cm
Area of triangle ABC = 360 cm2
We know that,
Area of triangle =
= 360 cm2
= 360 cm2
= 360 cm2
= 
= 9 cm
Now,
BL = 
= (
= 40 cm
a = 
= 
= 
= 
= 41 cm
Therefore,
Perimeter of the triangle = (41 + 41 + 80) = 162 cm
Question 11.
The perimeter of an isosceles triangle is 42cm and its base is 
times, each of the equal sides.
Find:
(i) The length of each side of the triangle
(ii) The area of the triangle
(iii) The height of the triangle.
Answer:
We know that,
In any isosceles triangle, the lateral sides are of equal length
Let,
The lateral side of the triangle be x
Given,
Base of the triangle = 
(i) We have to find out length of each side of the triangle:
Perimeter of the triangle = 42 cm (Given)
x + x + 
= 42 cm
2x + 2x + 3x = 84 cm
7x = 84 cm
x = 
x = 12 cm
Therefore,
Length of lateral side of the triangle = x = 12 cm
Base = = 
= 18 cm
Hence,
Length of each side of the triangle is 12 cm, 12 cm and 18 cm
(ii) Now, we have to find out area of the triangle:
Let,
a = 12 cm
b = 12 cm
And,
c= 18 cm
Now,
s =
= 
= 
= 21 cm
We know that,
Area of triangle =
= 
= 
= 
= 27
= 71.42 cm2
Therefore, area of the given triangle is 71.42 cm2
(iii) We have to calculate height of the triangle:
We know that,
Area of triangle =
71.42 cm2 = 
71.42 cm2 = 9 × h
h = 
= 7.94 cm
Therefore, height of the triangle is 7.94 cm
Question 12.
If the area of the equilateral triangle is 
cm2, find its perimeter.
Answer:
Given,
Area of the equilateral triangle = 36
cm2
Let us assume a be the length of the side of an equilateral triangle
We know that,
Area of an equilateral triangle = 
sq units
36
=
a2 = 
a2 = 36 × 4
a2 = 144
a = 12 cm
We know that,
Perimeter of an equilateral triangle = 3 × a
= 3 × 12
= 36 cm
Hence, perimeter of the given equilateral triangle is 36 cm.
Question 13.
If the area of the equilateral triangle is 
cm2, find its height.
Answer:
Let us assume a be the side of the equilateral triangle
We know that,
Area of an equilateral triangle = 
sq units
It is given that,
Area of the equilateral triangle = 81 cm2
81 cm2 = 
a2
a2 = 
= 324
a = 
= 18 cm
Height of an equilateral triangle = 
a
Since, the value of a is 18 cm
Therefore,
Height = × 18
= 9 cm
Question 14.
The base of a right – angles triangle measures 48cm and its hypotenuse measures 50cm. Find the area of the triangle.
Answer:
Given that,
Base = BC = 48 cm
Hypotenuse = AC = 50 cm
Let us assume AB = x cm
By using Pythagoras theorem, we get
AC2 = AB2 + BC2
Putting the value of BC, AC and AB we get:
502 = x2 + 482
x2 = 502 - 482
x2 = 2500 – 2304
x2 = 196
x = 
x = 14 cm
We know that,
Area of right angle triangle =
= 
= 24 × 14
= 336 cm2
Question 15.
Each side of an equilateral triangle measures 8cm. Find:
(i) The area of the triangle, correct to 2 places of decimal
(ii) The height of the triangle, correct to 2 places of decimal.
Take 
=1.732
Answer:
(i) We know that,
Area of an equilateral triangle = sq units
It is given that, each side of equilateral triangle is of 8 cm
Therefore,
Area = × 82
= 
= 
= 1.732 × 16
= 27.712
= 27.71 cm2 (Up to 2 decimal places)
(ii) We also know that,
Height of an equilateral triangle = a
= × 8
= 
= 1.732 × 4
= 6.928
= 6.93 cm (Up to 2 decimal places)
Question 16.
The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 decimal places. Take 
=1.732.
Answer:
Let us assume a be the side of the equilateral triangle
We know that,
Height of an equilateral triangle = a units
Height 0f the equilateral triangle = 9 cm (Given)
= 
(Rationalizing the denominator)
= 
= 6
Base of the triangle = 
We know that,
Area of triangle =
= 
× 6 × 9
= 27
= 27 × 1.732
= 46.76 cm2 (Up to 2 decimal places)
Question 17.
An umbrella is made by stitching 12 triangular pieces of cloth, each measuring (50 cm x 20 cm x 50 cm). Find the area of the cloth used in it.
Answer:
Let the sides of the triangle be,
a = 50 cm
b = 20 cm
And
C = 50 cm
Now, let us find the value of s:
s = 
= 
= 60 cm
We know that,
Area = 
Area of one triangular piece of cloth = 
= 
= 
= 
= 10 × 10 × 2
= 200
= 200 × 2.45
= 490 cm2
Therefore,
Area of one piece of cloth = 490 cm2
Hence,
Area of 12 pieces of cloth = 12 × 490
= 5880 cm2
Question 18.
A floral design on a floor is made up of 16 tiles, each triangular in shape having sides 16 cm, 12 cm, and 20cm. Find the cost of polishing the tiles at Re 1 per sq cm.
Answer:
Let the sides of the triangle be:
a = 16 cm
b = 12 cm
And,
c = 20 cm
Now we have to find out the value of s:
s =
= 
= 
= 24 cm
We know that,
Area of triangle =
Therefore,
Area of triangular tile = 
= 
= 2 × 2 × 2 × 2 × 2 × 3
= 96 cm2
Therefore,
Area of one tile = 96 cm2
Hence,
Area of 16 such tiles = 96 × 16 = 1536 cm2
Now,
Cost of polishing the tiles per square cm = Rs. 1
Therefore,
Total cost of polishing the tiles = 1 × 1536
= Rs. 1536
Question 19.
Find the perimeter and area of the quadrilateral ABCD in which AB = 17cm, AD = 9cm, CD = 12cm, 
=900 and AC = 15cm.
Answer:
By using Pythagoras theorem in right triangle ABC, we get
BC= 
= 
= 
= 
= 8 cm
Let us first find out the perimeter of the given quadrilateral
Perimeter of quadrilateral ABCD = 17 + 9 + 12 + 8 = 46 cm
We know that,
Area of triangle ABC =
= 
= 
= 60 cm2
Now,
For area of triangle ACD, we have
a = 15 cm
b = 12 cm
And,
c = 9 cm
Therefore,
s = 
= 
= 18 cm
Now,
Area of triangle ACD =
= 
= 
= 
= 18 × 3
= 54 cm2
Therefore,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
= 60 + 54
= 114 cm2
Question 20.
Find the perimeter and area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and 
=900.
Answer:
Firstly, let us calculate the perimeter of the given quadrilateral
Perimeter of quadrilateral ABCD = 34 + 29 + 21 + 42 = 126 cm
We know that,
Area of triangle =
Area of triangle BCD = 
= 210 cm2
Now, we have to calculate the area of triangle ABD,
For this, we have
a = 42 cm
b = 20 cm
c = 34 cm
Therefore,
s = 
=
= 48 cm
We know that,
Area of triangle =
Therefore,
Area of triangle ABD = 
= 
= 
= 4 × 3 × 2 × 14
= 336 cm2
Hence,
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD
= 336 + 210
= 546 cm2
Question 21.
Find the area of the quadrilateral ABCD in which AD = 42 cm ,
= 900 and 
is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given 
= 1.73]
Answer:
Let us consider a right triangle ABD,
By using Pythagoras theorem in this, we get
AB = 
= 
= 
= 10 cm
We know that,
Area of triangle =
= 
= 120 cm2
We also know that,
Area of an equilateral triangle BCD = sq units
= 
= 292.37 cm2
Therefore,
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD
= 120 + 292.37
= 412.37 cm2
Question 22.
Find the area of a parallelogram ABCD in which AB = 28cm, BC = 26cm and diagonal AC = 30 cm.
Answer:
Let the sides of the triangle ABC be:
a = 26 cm
b = 30 cm
And
c = 28 cm
Let us find out the value of s
We know that,
s =
= 
= 
= 42 cm
We know that,
Area of triangle ABC =
= 
= 
= 
= 
= 14 × 3 × 4 × 2
= 336 cm2
We know that,
In a parallelogram, the diagonal divides the parallelogram in two equal area
Therefore,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
= Area of triangle ABC × 2
= 336 × 2
= 672 cm2
Question 23.
Find the area of the parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16cm. [ Given 
= 1.73]
Answer:
According to the question,
In order to find the area of quadrilateral ABCD,
At first,
Let us consider triangle ABC,
Say,
a = 10 cm, b = 16 cm and c = 14 cm
Now,
Semi perimeter of ΔABC, s =
= 
= 
= 20 cm
Now,
Area of ΔABC = 
= 
= 
= 40√3 cm2
We know that, the diagonal of a parallelogram divides it into two triangles of equal areas.
Hence,
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= Area of ΔABC × 2
= 40√3 × 2
= 80√3 cm2
= 138.4 cm2
Question 24.
In the 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 the quadrilateral ABCD.
Answer:
According to the question,
In order to find the area of quadrilateral ABCD,
At first,
We will find the area of triangle ABD and triangle BCD respectively.
And, then we’ll add them.
Hence,
Area of ΔABD = 
= 
= 
= 537.6 cm2
Area of ΔBCD =
= 
= 
= 422.4 cm2
Now,
Area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD
= 537 + 422.4
= 960 cm2
Cce Questions
Question 1.In a ΔABC it is given that base = 12cm and height = 5cm. Its area is
A. 60cm2
B. 30 cm2
C. 
cm2
D. 45cm2
Answer:
We have,
Base of triangle = 12 cm
Height of triangle = 5 cm
We know that,
Area of triangle = 
= 
= 
= 30 cm2
Hence, option (b) is correct
Question 2.
The length of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is –
A. 96cm2
B. 120cm2
C. 144cm2
D. 160cm2
Answer:
Let the threes ides of the triangle be,
a = 20 cm, b = 16 cm and c = 12 cm
Now, s = 
= 
= 
= 24 cm
Now, by using Heron’s formula we have:
Area of triangle = 
= 
= 
= 
= 6 × 4 × 4
= 96 cm2
Hence, option (a) is correct
Question 3.
Each side of an equilateral triangle measures 8 cm. The area of the triangle is
A. 
cm2
B. 
cm2
C. 
cm2
D. 48cm2
Answer:
It is given in the question that,
Side of equilateral triangle = 8 cm
We know that,
Area of equilateral triangle = 
= 
= 
= 
cm2
Hence, option (b) is correct
Question 4.
The base of an isosceles triangle is 8cm long and each of its equal sides measures 6cm. The area of the triangle is –
A. 
cm2
B. 
cm2
C. cm2
D. cm2
Answer:
We know that,
Area of an isosceles triangle = 
It is given that,
a = 6 cm and b = 8 cm
∴ we have:
= 
= 
= 
= 8
cm2
Hence, option (b) is correct
Question 5.
The base of an isosceles triangle is 6cm and each of its equal sides is 5cm. The height of the triangle is –
A. 8 cm
B. 
cm
C. 4 cm
D. 
cm
Answer:
It is given in the question that,
Base of the isosceles triangle = b = 6 cm
Two equal sides = a = 5 cm
We know that,
Height of an isosceles triangle = 
= 
= 
= 
= 
= 4 cm
Hence, option (c) is correct
Question 6.
Each of the two equal sides of an isosceles right triangle is 10cm long. Its area is –
A. 
cm2
B. 50cm2
C. 
cm2
D. 75cm2
Answer:
From the given question, we have
Base of triangle = 10 cm
Height of triangle = 10 cm
∴ Area of triangle =
= 
= 5 × 10
= 50 cm2
Hence, option (b) is correct
Question 7.
Each side of an equilateral triangle is 10 cm long. The height of the triangle is –
A. cm
B. 
cm
C. 
cm
D. 5cm
Answer:
We have,
Each side of the equilateral triangle = 10 cm
We know that,
In an equilateral triangle altitude divides its base into 2 equal parts
∴ 
Let the height be h
Now, by using Pythagoras theorem
102 = 52 + h2
100 = 25 + h2
h2 = 100 – 25
h2 = 75
h = 
h = 5
cm
Hence, height of the triangle is 5 cm
Thus, option (b) is correct
Question 8.
The height of an equilateral triangle is 6cm. Its area is –
A. 
cm2
B. 
cm2
C. 
cm2
D. 18cm2
Answer:
It is given in the question that,
Height of an equilateral triangle = 6 cm
Let the side of triangle be a
Then, the altitude of the equilateral triangle is given as:
∴ 
Put altitude = 6 cm we get,
a = 
a = 4 cm
∴ Area of triangle =
= 
= 
= 12 cm2
Hence, option (a) is correct
Question 9.
The length of the three sides of the triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is –
A. 480m2
B. 320m2
C. 384m2
D. 360m2
Answer:
It is given in the question that,
Sides of the triangle = 40 m, 24 m and 32 m
∴ Semi-perimeter, s = 
= 
= 48 cm
Now, by using Heron’s formula we get:
Area of triangle = 
= 
= 
= 384 m2
Hence, option (c) is correct
Question 10.
The sides of the triangle are in the ratio 5: 12:13 and its perimeter is 150 cm. The area of the triangle is –
A. 375cm2
B. 750cm2
C. 250cm2
D. 500cm2
Answer:
It is given in the question that,
The sides of given triangle are in the ratio 5: 12: 13
Let the sides be 5x, 12x and 13x
According to the question,
5x + 12x + 13 x = 150
30x = 150
x = 
x = 5
So, 5x = 25
12x = 60
13x = 65
Semi-perimeter = 
= 
= 75 cm
Now, by using Heron’s formula we get:
Area of triangle = 
= 
= 
= 750 cm2
Hence, option (b) is correct
Question 11.
The lengths of the three sides of the 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
Answer:
It is given in the question that,
Sides of the triangle = 30 cm, 24 cm and 18 cm
Let h be the altitude of the triangle
∴ Semi-perimeter = 
= 
= 36 cm
Now, Area of triangle = 
= 
= 
= 216 cm2
Also, Area =
216 = 
216 = 9 × h
h = 
= 24 cm
Hence, option (a) is correct
Question 12.
The base of an isosceles triangle is 16 cm and its area is 48cm2. The perimeter of the triangle is –
A. 41 cm
B. 36 cm
C. 48 cm
D. 324 cm
Answer:
It is given in the question that,
Base of the triangle = 16 cm
Area of the triangle = 48 cm2
Let the height of the triangle be h
We know that,
Area of the triangle =
48 = 
48 = 8 × h
h = 
h = 6 cm
Now, half of the base = 
= 8 cm
∴ By using Pythagoras theorem, we have
Side2 = 82 + 62
= 64 + 36
= 100
= 10 cm
Now, perimeter of the triangle = Sum of all sides
= 10 + 10 + 16
= 36 cm
Hence, option (b) is correct
Question 13.
The area of an equilateral triangle is 
cm2. Its perimeter is –
A. 36cm
B. cm
C. 24cm
D. 30 cm
Answer:
It is given in the question that,
Area of an equilateral triangle = 36 cm2
We know that,
Area of an equilateral triangle =
36 =
(Side)2 = 
Side = 12 cm
∴ Perimeter of equilateral triangle = 3 × Side
= 3 × 12
= 36 cm
Hence, option (a) is correct
Question 14.
Each of the equal sides of an isosceles triangle is 13cm and base is 24cm. The area of the triangle is –
A. 156 cm2
B. 78 cm2
C. 60 cm2
D. 120 cm2
Answer:
It is given in the question that,
Equal sides of isosceles triangle = 13cm
Base = 24 cm and 
= 12 cm
Let the height of the triangle be h
∴ (13)2 = (12)2 + h2
169 = 144 + h2
h2 = 169 – 144
h2 = 25
h = 5
Thus, area of triangle =
= 
= 
= 
2
Hence, option (c) is correct
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. 168cm2
B. 252cm2
C. 336cm2
D. 504cm2
Answer:
Base of right angled triangle = 48 cm
Hypotenuse of triangle = 50 cm
Now, by using pythagoras theorem we get:
Hypotenuse2 = Base2 + Height2
(50)2 = (48)2 – h2
2500 = 2304 – h2
h2 = 2500 – 2304
h2 = 196
h = 14 cm
Now, Area of triangle =
= 
= 7 × 48
= 336 cm2
Hence, option (c) Is correct
Question 16.
The area of an equilateral triangle is 
cm2. Its height is-
A. 
cm
B. 
C. 
cm
D. 9 cm
Answer:
It is given in the question that,
Area of an equilateral triangle = 81 cm2
Let a be the side of the triangle and h be the height
We know that,
Area of an equilateral triangle = 
81 = 
a2 = 81 × 4
a = 18
Also, Area of triangle =
81
h = 
h = 9 cm
Hence, option (a) is correct
Question 17.
The difference between the semi- perimeter and the sides of a ∆ ABC are 8cm, 7cm and 5cm respectively. The area of the triangle is –
A. 
cm2
B. 
cm2
C. 
cm2
D. 140 cm2
Answer:
Let the semi-perimeter be s
Let the sides of the triangle be a, b and c
It is given in the question that,
s – a = 8 …(i)
s – b = 7 …(ii)
s – c = 5 …(iii)
Now, by adding (i), (ii) and (iii) we get:
(s – a) + (s – b) + (s – c) = 8 + 7 + 5
3s – a – b – c = 20
3s – (a + b + c) = 20
We know that,
s = 
∴ 3s – 2s = 20
s = 20 cm
Now, area of the triangle =
= 
= 20
cm2
Hence, option (c) is correct
Question 18.
For an isosceles right angles triangle having each of equal sides ‘a’ , we have
I. Area =
a2
II. Perimeter = (2 +
)a
III. Hypotenuse = 2a
Which of the following is true?
A. I only
B. II only
C. I and II
D. I and III
Answer:
We know that,
Area of triangle =
= 
= 
Now, Hypotenuse = 
= 
= 
Perimeter = a + a +
= 2a +
= a (2 + 
)
∴ I and II are true
Hence, option (c) is correct
Question 19.
For an isosceles triangle having base b and each of the equal sides as a, we have:
I. Area = 
II. Perimeter = (2a +b)
III. Height = 
Which of the following is true?
A. I only
B. I and II only
C. II and III only
D. I, II and III
Answer:
According to question, we have:
Base of triangle = b
Equal sides of triangle = a
∴ Area =
Perimeter = (2a + b)
And, Height =
∴ I, II and III are true
Hence, option (d) is correct
Question 20.
The question consists of two statements namely, Assertion (a) and Reason (R). Please select the correct answer.
A. Both Assertion (A) and Reason (B) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (B) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (a) is false and Reason (R) is true.
Answer:
In the given question, we have:
Area of equilateral triangle =
= 
= 
= 
cm2
Also, Area of an equilateral triangle having each side a = 
∴ Both Assertion and Reason are true
Hence, option (a) is correct
Question 21.
The question consists of two statements namely, Assertion (a) and Reason (R). Please select the correct answer.
A. Both Assertion (A) and Reason (B) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (B) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (a) is false and Reason (R) is true.
Answer:
In the given question, we have
Area of isosceles triangle = 
Here, we have:
a = 5 cm and b = 8 cm
∴ 
= 2 × 
= 2 × 
= 2 × 6
= 12 cm2
Also, Area of an isosceles triangle having each of the equal sides as a and base b = 
∴ Both Assertion and Reason are true
Hence, option (a) is correct
Question 22.
The question consists of two statements namely, Assertion (a) and Reason (R). Please select the correct answer.
A. Both Assertion (A) and Reason (B) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (B) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (a) is false and Reason (R) is true.
Answer:
In this question, we have
Area of an equilateral triangle =
=
=
= cm2
Also, Area of an equilateral triangle having each side a = 
Thus, assertion is false whereas reason is true
Hence, option (d) is correct
Question 23.
The question consists of two statements namely, Assertion (a) and Reason (R). Please select the correct answer.
A. Both Assertion (A) and Reason (B) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (B) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (a) is false and Reason (R) is true.
Answer:
In the given question,
Let us assume the sides of the triangle be 2x, 3x and 4x
We know that,
Perimeter of triangle = Sum of all sides
36 = 2x + 3x + 4x
36 = 9x
x = 
x = 4
∴ Sides of the triangle are:
2x = 2 × 4 = 8 cm
3x = 3 × 4 = 12 cm
4x = 4 × 4 = 16 cm
Let, a = 8 cm, b = 12 cm and c = 16 cm
So, s = 
= 
= 
= 18 cm
Now, by using Heron’s formula we have:
Area of triangle =
= 
= 
= 
= 6 × 2
= 12 cm2
Also, if 2s = (a + b + c)
Where a, b and c are the sides of the triangle then:
Area = 
which is false as it should be:
Area = 
∴ Assertion is true whereas reason is false
Hence, option (c) is correct
Question 24.
The question consists of two statements namely, Assertion (a) and Reason (R). Please select the correct answer.
A. Both Assertion (A) and Reason (B) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (B) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (a) is false and Reason (R) is true.
Answer:
From the given question, we have
a = 24 cm, b = 13 cm and c = 13 cm
∴ s = 
= 
= 
= 25 cm
Now, by using heron’s formula we have:
Area of triangle = 
= 
= 
= 5 × 12
= 60 cm2
Also, if 2s = (a + b + c) where a, b and c are the sides of the triangle then:
Area = 
∴ Assertion and reason both are correct
Hence, option (a) is correct
Question 25.
If the base of an isosceles triangle is 6cm and its perimeter is 16cm, then its area is 12cm2.
Answer:
It is given in the question that,
Base of the triangle, b = 6 cm
Equal sides of the isosceles triangle = a cm
Perimeter = 16 cm
We know that,
Perimeter = Sum of all sides
16 = a + a + 6
16 = 2a + 6
2a = 10
a = 
a = 5 cm
∴ Area of an isosceles triangle =
= 
= 1.5 × 
= 1.5 × 
= 1.5 × 8
= 12 cm2
Hence, the given statement is true
Question 26.
If each side of an equilateral triangle is 8cm long, then its area is 
cm2.
Answer:
It is given in the question that,
Each side of an equi8lateral triangle = 8 cm
Area of an equilateral triangle =
=
=
= 16 cm2
Hence, the given statement is false
Question 27.
If the sides of a triangular field measures 52m, 37 m and 20 m, then the cost of leveling at Rs 5 per m2 is Rs 1530.
Answer:
Let the sides of the triangular field be:
a = 52 m, b = 37 m and c = 20 m
∴ s =
= 
= 
= 54 m
Now, by using Heron’s formula we get:
Area of triangle =
= 
= 
= 
= 17 × 2 × 3 × 3
= 306 m2
It is given that,
Cost of leveling 1 m2 area = Rs 5
∴ Cost of leveling 306 m2 area = 5 × 306
= Rs 1530
Hence, the given statement is true
Question 28.
Match the following columns.
The correct answer is :
(a)-…….. (b)-……..
(c)-……… (d)-………
Answer:
The correct match for the table is as follows:
Question 29.
A park in the shape of a quadrilateral ABCD has AB = 9m, BC = 12m, CD = 5 cm, AD = 8 m and c= 90°. Find the area of the park. [Given: 
= 5.9]
Answer:
rom the given figure, it is clear that:
BCD is a right triangle
∴ BD = 
= 
= 
= 
= 13 m
Now, area of
=
= 
=
= 6 × 5 = 30 m2
Let the sides of the triangle be: a = 9 m, b = 8 m and c = 13 m
∴ s =
= 
= 
= 15 m
Thus, by using Heron’s formula we get:
Area of 
= 
= 
= 
= 
= 
= 6 × 5.9 = 35.4 m2
∴ Area of quadrilateral ABCD = Area of + Area of 
= 30 + 35.4
= 65.4 m2
Question 30.
Find the area of a parallelogram ABCD in which AB = 60cm, BC = 40cm and AC = 80 cm. [Given: 
= 3.87]
Answer:
Let the sides of the triangle ABC be:
a = 40 cm, b = 80 cm and c = 60 cm
∴ s =
= 
= 
= 90 cm
Now, by using Heron’s formula we get:
Area of 
= 
= 
= 
= 30 × 10
= 300 × 3.87
= 1161 cm2
As we know that, the diagonal of a parallelogram divides it into two triangles of equal areas
∴ Area of parallelogram (ABCD) = 2 × Area of (
)
= 2 × 1161
= 2322 cm2
Question 31.
A piece of land is in the shape of a rhombus ABCD in which each side measures 100m and diagonal AC is 160m long. Find the area of the rhombus.
Answer:
Let the sides of triangle be 100m, 160m, and 100m
Semi perimeter, 
Now, using Heron’s formula,
= 4800 m2
Now, we know that, diagonal divides a parallelogram into two triangles of equal areas.
Area of parallelogram ABCD = 2(area of ΔABC)
= 2 × 4800
= 9600 m2
Question 32.
A floral design on a floor is made up of 16 triangular tiles, each having sides 9cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of Rs. 2.50 per cm2
[Take 
=2.454]
Answer:
Let the sides of triangle be 9 cm, 28 cm, and 35 cm
Semi perimeter, 
Now, using Heron’s formula,
(Area of 1 tile)
= 88.2cm2
∴ Area of 16 tiles = 16 × area of one tile
= 16 × 88.2 cm2
= 1411.2 cm2
Now, cost of polishing 1cm2 area = Rs 2.5
∴ Cost of polishing 1411.2 cm2 = 2.5 × 1411.2 = Rs 3528
Question 33.
A kite in the shape of a square with each diagonal 32 cm and having a tail in the shape of an isosceles triangle of base 8cm and each side 6cm, is made of three different shades as shown in the figure. How much paper of each shade has been used in it?
[Given : = 2.24]
Answer:
We know that every square is a rhombus.
Each of the equal diagonals = 32 cm
= 512 cm2
Note: Diagonal of a parallelogram divides it into two triangles of equal areas and square is a parallelogram.
∴ Area of ΔABD = Area of ΔBDC = 1/2 area of ABCD
= 
Whereas, a = 6 cm and b = 8 cm
= 2
= 8
= 17.92 cm2
Formative Assessment (unit Test)
Question 1.Each side of an equilateral triangle is 8 cm. Its altitude is
A. 
cm
B. 
cm
C. 
cm
D. 
cm
Answer:
It is given that,
Each side of an equilateral triangle, a = 8 cm
We know that,
Area of equilateral triangle = 
=
= 
× 64
= 
= 16 cm2
Also,
Area of triangle = 
16 = 
× a × Altitude
∴ Altitude = 
= 4 cm
Hence, altitude of the triangle is 4 cm
Thus, option (c) is correct
Question 2.
The perimeter of an isosceles right- angled triangle having a as each of the equal sides is
A. 
B. 
C. 3a
D. 
Answer:
It is given in the question that, equal sides of isosceles triangle is a
It is also given that, the given triangle is isosceles right-angled triangle
∴ AC = 
AC = 
AC = 
AC = a
We know that,
Perimeter of triangle = Sum of all sides
∴ Perimeter = (AB + BC + AC)
= (a + a + a)
= 2a + a
= a (2 + )
Hence, option (b) is correct
Question 3.
For an isosceles triangle having base = 12 cm and each of the equal sides equal to 10 cm, the height is
A. 12 cm
B. 16 cm
C. 6 cm
D. 8 cm
Answer:
Let us assume ABC be an isosceles triangle having,
Base, AC = 12 cm
AB = AC = 10 cm
BD = 
= 
= 6 cm
We know that,
In right angled triangle, ABC
AD = 
= 
=
=
= 8 cm
Thus, height of the triangle is 8 cm
Hence, option (d) is correct
Question 4.
Find the area of an equilateral triangle having each side 6cm.
Answer:
Let us assume each side of the equilateral triangle be a
It is given that,
Side of equilateral triangle = 6 cm
We know that,
Area of equilateral triangle = a2
= 
= 
= 
= 9 cm2
Question 5.
Using Heron’s formula find the area of Δ ABC in which BC = 13 cm, AC = 14 cm and AB = 15cm.
Answer:
It is given in the question that,
Sides or triangle ABC are:
BC = 13 cm
AC = 14 cm
AB = 15 cm
We know that,
Perimeter of triangle = Sum of all sides
= AB + BC + AC
= 15 + 13 + 14
= 42 cm
∴ s = 
= 
= 21 cm
Hence,
Area of 
= 
= 
= 
= 84 cm2
Question 6.
The sides of a triangle are in the ratio 13: 14: 15 and its perimeter is 84 cm. Find the area of the triangle.
Answer:
It is given in the question that,
Perimeter of triangle = 84 cm
Also, sides or triangle are: in ratio 13: 14: 15
Let, a = 13x
b = 14x
c = 15x
We know that,
Perimeter of triangle = Sum of all sides
84 = a + b + c
84 = 13x + 14x + 15x
84 = 42x
x = 
x = 2 cm
Thus, a = 13 × 2 = 26 cm
b = 14 × 2 = 28 cm
c = 15 × 2 = 30 cm
∴ s =
= 
= 42 cm
Hence,
Area of triangle =
= 
= 
= 336 cm2
Question 7.
Find the area of ABC in which BC = 8cm, AC = 15cm and AB = 17 cm. Find the length of altitude drawn on AB.
Answer:
It is given in the question that,
Sides or triangle ABC is:
BC = a = 8 cm
AC = b = 15 cm
AB = c = 17 cm
We know that,
Perimeter of triangle = Sum of all sides
= a + b + c
= 8 + 15 + 17
= 40 cm
∴ s = 
= 
= 20 cm
Hence,
Area of =
= 
= 
= 60 cm2
Also, Area of triangle ABC = 
60 = 
120 = 17 × Height
Height = 
= 7.06 cm
Hence, area of triangle is 60 cm2 and length of altitude is 7.06 cm
Question 8.
An isosceles triangle has perimeter 30cm and each of its equal sides is 12cm. Find the area of the triangle.
Answer:
It is given in the question that,
Equal sides of isosceles triangle = a = b = 12 cm
Also, perimeter = 30 cm
We know that perimeter of triangle = Sum of all sides
(a + b + c) = 30 cm
12 + 12 + c = 30
24 + c = 30
c = 30 – 24
= 6 cm
Hence, s = 
s = 
s = 15 cm
∴ Area of triangle =
= 
= 
= 9 cm2
Question 9.
The perimeter of an isosceles triangle is 32cm. The ratio of one of the equal side to its base is 3: 2. Find the area of the triangle.
Answer:
It is given in the question that,
Perimeter of an isosceles triangle = 32 cm
Let us assume the sides of the triangle be a, b, c and a = b
We know that,
Perimeter = a + b + c
32 = a + b + c
32 = a + a + c
32 = 2a + c (i)
According to the condition given in the question, we have:
a: c = 3: 2
So, a = 3x and c = 2x
Now putting values of a and c in (i), we get
2 × 3x + 2x = 32
6x + 2x = 32
8x = 32
x = 
x = 4
Thus, a = 3 × 4 = 12 cm
b = 12 cm
c = 2 × 4 = 8 cm
Now, s = 
= 
= 16 cm
Area of triangle =
= 
= 
= 4 × 4 × 2
= 32 cm2
Question 10.
Given a ABC in which
I. A, B and C are in the ratio 3: 2 :1.
II. AB , AC and BC are in the ratio 3 : 3 : 2 3 and AB = 3 3 cm.
Is ABC a right triangle?
The question give above has two Statements I and II. Answer the questions by using instructions given below:
(a) If the question can be answered by one of the given statements only and not by the other.
(b) If the question can be answered by using either statement alone.
(c) If the question can be answered by using both the statements but cannot be answered by using either statement.
(d) If the question cannot be answered even by using both the statements together.
Answer:
I. It is given in the question that,
∠A, ∠B and ∠C are in the ratio 3: 2: 1
Let ∠A = 3x
∠B = 2x
∠C = x
We know that, sum of angles of a triangle = 180o
∠A + ∠B + ∠C = 180o
3x + 2x + x = 180o
6x = 180o
x = 
x = 30o
Hence, ∠A = 3 × 30o = 90o
∴ 
is a right-angled triangle
II. It is also given that:
AB, AC and BC are in the ratio 3: √3: 2√3
Now, AB = 3x, AC = 
and BC = 2√3x
As it is given that,
AB = 3√3
∴ x = √3
AC = 3
BC = 6
Now, by using Pythagoras theorem in we get:
AC = 
3 = 
3 = 
3 ≠ √63
∴ The question can be answered by using either statement alone
Hence, option (b) is correct
Question 11.
In the given figure ABC and DBC have the same base BC such that AB = 120m, AC = 122m, BC = 22m, BD = 24m and CD = 26m. Find the area of the shaded region. (Take 
= 10.25)
Answer:
It is given in the question that,
AB = 120 m
AC = 122 m
BC = 22 m
BD = 24 m
And, CD = 26 m
We know that,
Perimeter of triangle = Sum of all sides
∴ Perimeter of = AB + BC + AC
= 120 + 22 + 122
= 264 m
s = 
= 
= 132 m
Now, Area () =
= 
= 
= 11 × 12 × 10
= 1320 m2
Now, in
BC = a, BD = b and CD = c
∴ Perimeter of = 22 + 24 + 26
= 72 m
s = 
= 
= 36 m
Hence, area (
=
= 
= 
= 24 × 10.25
= 246 m2
∴ Area of shaded region = Area () – Area ()
= 1320 – 246
= 1074 m2
Question 12.
A point O is taken inside an equilateral ΔABC. If OL 
BC, OM AC and ON AB such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of 
ABC.
Answer:
Let each side of be a cm
So, area () = Area (
) + Area (
) + Area (
)
On taking “a” as common, we get,
= 15a cm2 (i)
As, triangle ABC is an equilateral triangle and we know that:
Area of equilateral triangle = 
cm2 (ii)
Now, from (i) and (ii) we get:
15a =
15 × 4 = 
60 =
a = 
a = 20√3 cm
Now, putting the value of a in (i), we get
Area () = 15 × 20√3
= 300√3 cm2