# Important Questions for CBSE Class 9 Maths Chapter 12 - Heron's Formula

## CBSE Class 9 Maths Chapter-12 Important Questions - Free PDF Download

Free PDF download of Important Questions with solutions for CBSE Class 9 Maths Chapter 12 - Heron"s Formula prepared by expert Mathematics teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your examination. You can also download NCERT Solutions Class 9 Maths to help you to revise complete syllabus and score more marks in your examinations. Students can also avail of NCERT Solutions Class 9 Science from our website. Besides, find NCERT Book Solutions to get more understanding of various subjects.

1 Marks Quetions

1. The measure of each side of an equilateral triangle whose area is cm2 is

(A) 8 cm

(B) 2 cm

(C) 4 cm

(D) 16 cm

Ans. (B) 2 cm

2. Measure of each side of an equilateral triangle is 12cm. Its area is given by

(A)

(B)

(C)

(D)

Ans. D)

3. Two adjacent side of a parallelogram are 74cm and 40cm one of Its diagonals is 102cm. area of the ||gram is

(A) 612 sq m

(B) 1224 sq m

(C) 2448 sq m

(D) 4896 sq m

Ans. (C) 2448 sq m

4.

(B) 8 cm

(C) 6 cm

(D) 5 cm

Ans. (A) 10 cm

5. The perimeter of a triangle is 60cm. If its sides are in the ratio 1:3:2, then its smallest side is

(a) 15

(b) 5

(c) 10

(d) none of these.

Ans. (c) 10

6. The perimeter of a triangle is 36cm. If its sides are in the ratio 1:3:2, then its largest side is

(a) 6

(b) 12

(c) 18

(d) none of these

Ans. (c) 18

7. If the perimeter of a rhombus is 20cm and one of the diagonals is 8cm. The area of the rhombus is

(a) 24 sq cm

(b) 48 sq cm

(c) 50 sq cm

(d) 30 sq cm

Ans. (a) 24 sq cm

8. One of the diagonals of a rhombus is 12cm and area is 54 sq cm. the perimeter of the rhombus is

(a) 72 cm

(b)

(c)

(d)

Ans. (d)

9. The side of a triangle is 12 cm, 16 cm, and 20 cm. Its area is

(A) 100sq cm

(B) 90sq cm

(C) 96sq cm

(D) 120sq cm.

Ans. (C) 96sq cm

10. The side of an equilateral triangle is. Its area is.

(A) sq cm

(B) sq cm

(C) sq cm

(D) sq cm.

Ans. (A) sq cm

11. It the perimeter of a rhombus is 20sq cm and one of the diagonals is 8 cm. then the area of the rhombus is

(A) 40sq cm

(B) 24sq cm

(C) 20sq cm

(D) 13sq cm.

Ans. (B) 24sq cm

12. One of the diagonals of a rhombus is 12 cm and Its area is 54sq cm. the perimeter of the rhombus is.

(A) 10 cm

(B) 8 cm

(C) 6 cm

(D) cm.

Ans. (D) cm.

13. The lengths of the side of a triangular park are 90m, 70m and 40m, find Its area.

(A) 1340 sq m

(B) 1344 sq m

(C) 1440 sq m

(D) 1444 sq m

Ans. (B) 1344 sq m

14. An equilateral triangle has a side 50cm long. Find the area of the triangles.

(A) 625sq cm

(B) 625 sq m

(C) 256sq m

(D) 625sq m

Ans. (A) 625sq cm

15. The area of an isosceles triangle is 12 sq cm. If one of the equal side is 5 cm, then the length of the base is

(A) 4 cm

(B) 5 cm

(C) 6 cm

(D) 8 cm

Ans. (C) 6 cm

16. Find the area of triangle whose side is 6 cm, 10 cm and 15cm.

(A) 404.9 sq cm

(B) 405.9 sq cm

(C) 402.9 sq cm

(D) 410sq cm

Ans. (A) 404.9 sq cm

17. If side of equilateral triangle is 25 m. Its area is

(a)

(b)

(c)

(d)

Ans. (a)

18. The perimeter of an equilateral triangle is 48 cm. Its area is

(a)

(b)

(c)

(d)

Ans. (c)

19. If area of isosceles triangle is 48 sq cm and length of one of its equal sides is 10 m, then what is the base?

(a) 16 cm or 12 cm

(b) 12 cm or 14 cm

(c) 14 cm or 16 cm

(d) 16 cm or 18 cm

Ans. (a) 16 cm or 12 cm

20. If AB = 14 cm, BC = 13 cm, CD = 17 cm, DA = 8 cm and AC = 15 cm then area of quadrilateral ABCD is

(a) 150 sq cm

(b) 144 sq cm

(c) 142 sq cm

(d) 140 sq cm

Ans. (b) 144 sq cm

2 Marks Quetions

1. There is slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”, (see figure). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Ans. Since, sides of coloured triangular wall are 15 m, 11 m and 6 m.

Semi-perimeter of coloured triangular wall

= = = 16 m

Now, Using Heron’s formula,

Area of coloured triangular wall

=

=

= =

Hence area painted in blue colour =

2. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Ans. Given: = 18 cm, = 10 cm.

Since Perimeter = 42 cm

=42

18 + 10 + = 42

= 42 – 28 = 14 cm

Semi-perimeter of triangle

= = 21 cm

Area of triangle =

=

=

=

=

3. Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.

Ans. Let the sides of the triangle be and .

Therefore, = 540

= 10

The sides are 120 cm, 170 cm and 250 cm.

Semi-perimeter of triangle = 270 cm

Now, Area of triangle =

=

=

4. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Ans. Given: = 12 cm, = 12 cm

Since Perimeter = 30 cm =30

12 + 12 + = 30

= 30 – 24 = 6 cm

Semi-perimeter of triangle = = 15 cm

Area of triangle =

=

=

=

=

5. A park, in the shape of a quadrilateral ABCD has C = AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Ans. Since BD divides quadrilateral ABCD in two triangles:

(i) Right triangle BCD and (ii)ABD.

In right triangle BCD, right angled at C,

Therefore, Base = CD = 5 m and Altitude = BC = 12 m

Area of BCD =

=

In ABD, AB = 9 m, AD = 8 m

And BD = [Using Pythagoras theorem]

BD =

= = = 13 m

Now, Semi=perimeter of ABD = = 15 m

Using Heron’s formula,

Area of ABD =

=

=

= =

(approx.)

Area of quadrilateral ABCD = Area of BCD + Area of ABD

= 30 + 35.4

6. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Ans. In quadrilateral ABCE, diagonal AC divides it in two triangles, ABC and ADC.

In ABC, Semi-perimeter of ABC = = 6 cm

Using Heron’s formula,

Area of ABC =

=

=

Using Heron’s formula, Area of ABC =

=

= = 2

(approx.)

Now area of quadrilateral ABCD = Area of ABC + Area of ADC

= 6 + 9.2

7. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 29 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Ans. Semi-perimeter of triangle = = 42 cm

Using Heron’s formula,

Area of triangle =

=

=

According to question, Area of parallelogram = Area of triangle

Base Corresponding height = 336

28 Height = 336

Height = 12 cm

8. A kite is in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure.

How much paper of each side has been used in it?

Ans. Let ABCD is a square of side cm and diagonals AC = BD = 32 cm

In right triangle ABC, [Using Pythagoras theorem]

= 512

Area of square [Area of square =]

Diagonal BD divides the square in two equal triangular parts I and II.

Area of shaded part I = Area of shaded part II =

Now, semi-perimeter of shaded part III = 10 cm

Area of shaded part III =

=

= =

9. An umbrella is made by stitching 10 triangles pieces of cloth of two different colour, each piece measuring 20 cm 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Ans. a=20cm, b=50cm

cloth required for each colour

Area of one triangle piece

Thus, sq cm cloth of each colour is required

10. The perimeter of a rhombus ABCD is 40cm. find the area of rhombus of Its diagonals BD measures 12cm

Ans.

AB=10cm, BD =12cm and DA=10cm

Area of

=96 sq cm

11. Find area of triangle with two sides as 18cm & 10cm and the perimeter is 42cm.

Ans. Let a=18 cm, b=10 cm

Perimeter =42cm

So, C=14 cm

= 21 cm

new area of triangles =

12. Find the area of in isosceles triangle, the measure of one of Its equals side being ‘b’ and the third side ‘a’.

Ans. Here

sq units

13. Find the cost of leveling the ground in the form of a triangle having its sides are 40 m, 70 m and 90 m at Rs 8 per square meter. [use = 2.24]

Ans. Here S = = 100 m

Area of a triangular ground =

=

=

=

= 1344 sq m

Cost of leveling the ground = Rs

=Rs 10752

14. The triangular side’s walls of a flyover have been used for advertisements. The sides of the walls are 122m, 22m and 120m. The advertisement yield on earning of Rs 5000 per m2 per year. A company hired one of its walls for 4 months. How much rent did it pay?

Ans. The lengths of the sides of the walls are 122m, 22m and 120m.

As,

= 14400 +484

= 14884

=

Walls are in the form of right triangles

Area of one wall =

=

= 1320 sq m.

Rent = Rs 5000/sq m per year

Rent for 4 month

= Rs

= Rs 22,00,000

15. Find the perimeter and area of a triangle whose sides are of length 2cm, 5cm and 5cm.

Ans. Here, a = 2cm, b = 5cm and c = 5cm

Perimeter = a+ b+ c = (2 + 5 + 5) = 12 cm

S = semi perimeter

= = 6 cm

using Heron’s formula,

Area of triangle =

=

=

= 4.9 sq cm

16. There is a slide in a park. One of its sides wall has been painted in some colour with a message “KEEF THE CITY GREEN AND CLEAN”.

If the sides of the wall are 15m, 11m and 6m. Find the area painted in colour.

Ans. The sides of the wall is in the triangular from with sides,

A = 15 m, b = 6 m and c = 11 m

S = m

= 16 m

Area to be painted in colour = Area of the side wall

=

=

=

=

17. Find the area of isosceles triangle whose side is 14 m, 12 m, 14m?

Ans.

Area of isosceles triangle=

= 75.6

18. The perimeter of a rhombus ABCD is 60 cm. find the area of the rhombus of Its diagonal BD measures 16 cm?

Ans. As side of rhombus are equal.

So,

Area of,

=101.6 sq cm

Area of rhombus =

19. Find the cost of leveling the ground in the from of a triangle having Its side as 70 cm, 50 cm, and 60 cm, at Rs 7 per square meter.

Ans.

=1469.7 sq m

cost of levelling the ground=

=10287.9

20. Find the area of a triangle two side of the triangle are 18 cm, and 12 cm. and the perimeter is 40 cm.

Ans.

So, a+b+c=40 cm

18+12+C=40

C=(40-30)cn =10cm

area of triangle=

=56.56 sq cm

21. Find the area of triangle whose side is 42m, 56m and 70m?

Ans. S = =

Area of ABC =

=

= 4228 sq m

= 1176sq m

22. Find the area of an isosceles triangle, the measure of one of Its equal side being b and the third side a.

Ans. S = =

Area of triangle = units

=

=

23. Find the area of equilateral triangle whose side is 12 cm using Heron’s formula.

Ans. S =

= = 18 cm

Area of equilateral =

=

=

=

24. Find the area of isosceles triangle whose equal side is 6 cm, 6 cm and 8 cm.

Ans. S =

=

Area of isosceles triangle =

=

= 17.8 sq cm

25. Find the area of an isosceles triangles, the measure of one of its equal sides being 10 cm and the third side is 6 cm.

Ans. S =

Area if tringle=sq cm

26. Find the area of equilateral triangle the length of one of its sides being 24 cm.

Ans. a = b = c = 24 cm

=36 cm

Area of triangle=sq cm

=246.12 sq cm

27. Find the perimeter and area of a triangle whose sides are 3 cm, 4 cm and 10 cm?

Ans. Perimeter = 3+4+5

= 12 cm

Or = 6cm

Area of triangle =sq cm

= 6 sq cm

28. Using Heron’s formula, find area of triangle whose sides are 6 cm, 8 cm and 10 cm?

Ans.

=12 cm

Area of triangle= sq cm

=24 sq cm

3 Marks Quetions

1. A traffic signal board, indicating ‘SCHOOLAHEAD’ is an equilateral triangle with side  "a". Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Ans. Let the Traffic signal board is ABC.
According to question, Semi-perimeter of ABC =
Using Heron’s Formula, Area of triangle ABC =

=

= =

=

Now, Perimeter of this triangle = 180 cm

Side of triangle = = 60 cm

Semi-perimeter of this triangle = = 90 cm

Using Heron’s Formula, Area of this triangle =

=

=

=

=

2. The triangular side walls of a flyover has been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisement yieldan earning of Rs. 5000 per per year. A company hired one of its walls for 3 months, how much rent did it pay?

Ans. Given: = 122 m, = 22 m and = 120 m

Semi-perimeter of triangle = = = 132 m

Using Heron’s Formula,

Area of triangle =

=

=

=

Rent for advertisement on wall for 1 year = Rs. 5000 per

= Rs. 1650000

Hence rent paid by company = Rs. 16,50,000

3. Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

Ans. Area of triangular part I: Here, Semi-perimeter = 5.5 cm

Therefore, Area =

=

= =

=

Area of triangular part II =

=

Area of triangular part III (trapezium):

= (AB + DC)

= (1 + 2)

=

=

=

Area of triangular parts IV & V:

=

Total area = 2.4825 + 6.2 + 1.299 + 9

=

4. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, grass of how much area of grass field will each cow be getting?

Ans. Here, AB = BC = CD = DA = 30 m and Diagonal AC = 48 m which divides the rhombus ABCD in two congruent triangle.

Area of ABC = Area of ACD

Now, Semi-perimeter of ABC = = 54 m

Now Area of rhombus ABCD = Area of ABC + Area of ACD

= 2 x Area of ABC [ Area of ABC = Area of ACD]

= [ Using Heron’s formula]

=

= =

Field available for 18 cows to graze the grass

Field available for 1 cow to graze the grass =

5. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Ans. Here, sides of each of 10 triangular pieces of two different colours are 20 cm, 50 cm and 50 cm.

Semi-perimeter of each triangle = = 60 cm

Now, Area of each triangle =

=

= =

According to question, there are 5 pieces of red colour and 5 pieces of green colour.

Cloth required for 5 red pieces = =

And Cloth required to 5 green pieces = =

6. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 paise per cm2.

Ans. Here, Sides of a triangular shaped tile area 9 cm, 28 cm and 35 cm.

Semi-perimeter of tile = = 36 cm

Area of triangular shaped tile =

=

= =

= (approx.)

Area of 16 such tiles (Approx.)

Cost of polishing of tile = Rs. 0.50

Cost of polishing of tile = Rs. = Rs. 705.60 (Approx.)

7. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Ans. Let ABCD be a field in the shape of trapezium and parallel side AB = 10 m & CD = 25 m

And Non-parallel sides AD = 13 m and BC = 14 m

Draw BM DC and BE AD so that ABED is a parallelogram.

BE = AD = 13 m and DE = AB = 10 m

Now in BEC, Semi-perimeter

= 21 m

Area of BEC =

=

= =

And Area of BEC =

= 84

= 84

BM = = 11.2 m

Now area of trapezium ABCD =

=

8. From a point in the interior of an equilateral triangle perpendiculars drawn to the three sides are 8 cm, 10 cm and 11 cm respectively. Find the area of the triangle to the nearest cm. (use )

Ans. Let each side of the equilateral ABC measure be x cm.

Let OD= 11cm, OE =8cm and OF=10cm

Join OA, OB and OC

Sq cm

But area of equilateral, the measure of whose each side of x

From (i) and (ii)

sq cm

sq cm

9. A parallelogram, the length of whose side is 60 m and 25 m has one diagonal 65 m long. Find the area of the parallelogram.

Ans. AB=DC=60cm, BC=AD= 25m and AC=65m

Area of parallelogram ABCD= Area of ABC + area of ACD

= 2 Area of ABC [ar ABC =ar ABD]

Now

=sq m

=750 sq m ….(II)

From (i) and (ii), we get

Area of

=15000 sq m.

10. A parallelogram, the measures of whose adjacent sides are 28 cm and 42 cm, has one diagonals 38cm. Find Its altitude on the side 42cm.

Ans. AB=DC=42cm=C

And BD=38cm=a

Let A be the area of ABD

Now,

A=

whare h cm is altitude

=21 h sq cm

From (i) and (ii), we get

Thus, required altitude =

12. Find the area of a quadrilateral ABCD in which AB=3 cm, BC=4 cm, CD= 4 cm, DA=5cm and AC= 5 cm.

Ans. Area of quadrilateral ABC=area of ABC + area of ACD(i)

For ABC,

= 6sq cm …(i)

For ACD, S=

13. Find the base of an isosceles triangles whose area is 12cm and the length of one of the equal side is 5cm.

Ans. Area of an isosceles triangle =

= 12

or = 12

Squaring both sides, we have

= 14416

= 2304

or = 0

or= 0

Either a2 = 64 i.e. a = 8

ora2 = 36 i.e. a = 6

Required base = 8 cm or 6 cm

14. The perimeter of a triangle is 450m and its sides are in the ratio of 13:12:5. Find the area of the triangle.

Ans. Let the sides of the triangle be 13x, 12x and 5x

Perimeter of a triangle = 450 m

13x + 12x + 5x = 450 m

or 30x = 450

x = 15

The sides are 1315, 1215, and 515

I.e. 195 m, 180 m and 75 m

S = = = 225 m

Area of the triangle =

=

=

= (1515235) sq m

= 6750 sq m.

15. The sides of a triangle are 39cm, 42cm and 45cm. A parallelogram stands on the greatest side of the triangle and has the same area as that of the triangle. Find the height of the parallelogram.

Ans. To find the area of ABC

S = cm

= 63 cm

Area of ABC =

=

= 97232 sq cm

= 756 sq cm

Let h be the height of the parallelogram

Now,

Area of parallelogram BCDE = Area of ABC

hBC = 756

or 45h = 756

h =

h = 16.8 cm

Hence, height of the parallelogram = 16.8 cm

16. The students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA whiles the other group through the lanes AC, CD and DA [fig1.1]. Then they cleaned the area enclosed within their lanes. If AB=9m, BC=40m, CD=15m, DA=28m and =90o, which group cleaned more area and by how much? Find also the total area cleaned by the students.

Ans. We have, right angle ABC,

AC = 41

The first group has to clean the area of ABC which is right angled triangle

Now,

Area of ABC = 40 m 9 m

= 180 sq m

The second group has to clean the area ofACD which has AD=28 m,

DC=15m and AC=41

Hence,

S =

= 42 m

Area of ACD =

=

=

= 126 sq m

First group cleaned more = (180 – 126) sq m

= 54 sq m

Total area cleaned by students = (180 + 126) sq m

= 306 sq m

17. A traffic signal board indicating ‘school ahead’ is an equilateral triangle with side ‘a’ find the area of the signal board using heron’s. Its perimeter is 180 cm, what will be Its area?

Ans.

Area of triangle=

Now, perimeter = 180 cm

18. A parallelogram the length of whose sides are 80m, and 40m has one diagonal 75m long. Find the area of the parallelogram?

Ans.

and AC = 75 cm

=97.5 cm

Area of triangle ABC=

=

sq m

=1485.75 sq m

Area of parallelogram ABCD=

=2971.4 sq m

19. The side of a triangular field is 52m, 56m, and 60m find the cost of leveling the field Rs 18 per meter, if a space of 4cm is to be left for entry gate.

Ans. Side of the field are 52m, 56m and 60m.

Sq m

sq m

sq m

=1344 sq m

Total cost of levelling the field

=Rs 24192

20. A floral design of a floor is made up of 16 tiles which are triangular. The side of the triangle being 9 cm, 28 cm, and 35 cm. find the cost of polishing the tiles, at RS 50 paisa/sq cm.

Ans. For each triangular tile, we have

sq cm

21. The measure of one side of a right triangle is 42m. If the difference in lengths of Its hypotenuse and other side is 14 cm, find the measure of two unknown side?

Ans. Let AB = y and AC = x and BC = 42 cm

By the given condition,

x - y = 14 (i)

By Pythagoras theorem,

(x + y) (x – y) = 1764

14 (x + y) = 1764 using (ii)

x + y =

Adding (ii) and (iii), we get

2x = 140

i.e. x = 70

y = 126-x

y = 126 – 70

= 56

22. The perimeter of a rhombus ABCD is 80 cm. find the area of rhombus if Its diagonal BD measures 12 cm.

Ans. AB = BC = CD = DA =

Now in ABD,

S = =26

so,

Area of ABD =

= 114.4 sq cm

Area of rhombus =

= 2 114.4 sq cm

= 228.8 sq cm

23. Find area of quadrilateral ABCD in which AB= 5 cm, BC= 6 cm, CD 6 cm, DA =7cm, And AC=7 cm.

Ans. Area of quadrilateral ABCD = Area of ABC + Area of ACD (i)

In ABC,

S = = 9 cm

Area of ABC =

=

= =14.4 sq cm

In ACD,

S = = 10 cm

Area of ACD =

=

= 18.9 sq cm

Area of quadrilateral ABCD = (14.4 + 18.9) sq cm

= 33.3 sq cm.

24. Shashi Kant has a vegetable garden in the shape of a rhombus. The length of each side of garden is 35 m And Its diagonal is 42 m long. After growing the vegetables in it. He wants to divide it in seven equal parts And look after each part once a week. Find the area of the garden which he has to look after daily.

Ans. Let ABCD be garden

DC = 35 m . . . . . . (Given)

DB = 42 m . . . . . . (Given)

Draw CEDB

We know that the diagonals of a rhombus bisect each other at right angles.

DE = DB =

Now

= 784

CE = 28 m

Now area of DBC, =

=

= 588 sq m

Area of the garden ABCD = 2588 sq m

= 1176 sq m

Area of the garden he has to look after, daily

= sq m

= 168 sq m

25. The perimeter of a triangle is 480 meters and its sides are in the ratio of 1:2:3. Find the area of triangle?

Ans. Let the sides of the triangle be x, 2x, 3x

Perimeter of the triangle = 480 m

x + 2x + 3x = 480 m

6x = 480 m

x = 80 m

The sides are 80 m, 160 m, 240 m

So,

And,

= 0 sq m

Triangle doesn’t exist with the ratio 1:2:3 whose perimeter is 480 m.

26. Find the cost of leveling the ground in the form of equilateral triangle whose side is 12 m at Rs 5 per square meter.

Ans. Here, sides are 12 m, 12 m, 12 m,

= 18 cm

And,

Area of equilateral triangle =

Cost of leveling ground =

= Rs 311.4 m

27. A kite in the shape of a square with diagonal 32 cm and an isosceles triangle of base 8 cm and side 6 cm each is to be made of three different shades. How much paper of each shade has been used in it? ( use = 2.24)

Ans.
Let ABCD be the square and CEF be an isosceles triangle.

Let the diagonals bisect each other at O.

Then,
AO =cm
= 16 cm
Area of shaded portion I =
= 256 sq cm
And,
Area of portion III =

Thus, the papers of three shades required are 256 sq cm, 256 sq cm and 17.92 sq cm.

28. The sides of a quadrangular field, taken in order are 29 m, 36 m, 7m and 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Ans.
In ADC, which are right angles?

= 84 sq m

For ABC,

From (I), (II) and (III) we get area of quadrilateral ABCD = (84+360) sq m = 444 sq m

29. Find the area of a triangle whose perimeter is 180 cm and two of its sides are 80 cm and 18 cm. hence calculate the altitude of the triangle taking the longest side as base.

Ans. Let a= 80 cm and b=18 cm, perimeter = 180 cm

c = 82cm
Now,

= 720 sq cm
The longest side of triangle is 82 cm
Let h cm be length of altitude corresponding to longest side.

4 Marks Quetions

1. A field in the shape of a trapezium whose parallel side are 25m and 10m. The non- parallel side are 14 m and 13 m. Find the area of the field.

Ans. AB=25m, CD =10m

EB=AB-AE= (25-10) m= 15m

In BCE, a=15m, b=14m and C=13 m

=21 m

=84 sq m ….(i)

Also area of

[using (i)]

Now, area of parallelogram

….(ii)

Thus area of trapezium ABCD = area of parallelogram + area of

=112 sq m +84 sq m [using (i) and (ii)]

=196 sq m

2. The perimeter of a right triangle is 24 cm. If its hypotenuse is 10 cm, find the other two sides. Find its area by using the formula area of a right triangle. Verify your result by using Heron’s formula.

Ans. Let x and y be the two lines of the right.

AB = x cm, BC = y cm and AC = 10 cm

By the given condition,

Perimeter = 24 cm

Or x + y = 14 (1)

By Pythagoras theorem,

= (10)2

= 100 (2)

From (1), = (14)2

Or + 2xy = 196

100 + 2xy =196 from (2)

xy =

= 48 sq cm (3)

Area of ABC = sq cm

= sq cm

=24 sq cm (4)

again,

Or x-y = 2

(i) When, x-y = 2 and x+y = 14, then 2x = 16

or x = 8, y = 6

(ii) When, X – y = -2 and x + y = 14, then 2x = 12

Or x = 6, y = 8

Verification by using Heron’s formula:

Sides are 6 cm, 8 cm and 10 cm

S = = 12 cm

Area of ABC =

= sq cm

= 24 sq cm

Which is same as found in (4)

Thus, the result is verified.

3. Radha made a picture of an aero plane with colored paper as shown in fig. find the total area of the paper used.

Ans. Area (1) = area of isosceles triangle with a =1 cm and b = 5 cm

(approx)

Area (ii) = area of rectangle with

L = 6.5 cm and b=1 cm

Area (iii) = Area of trapezium

Area of equilateral

= 1.3 sq cm (approx.)

Area of (IV + V) =

total area of the paper used = Area (I+II+III+IV+V)

= (2.5+6.5+1.3+9) sq cm

=19.3 sq cm