RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

Class 7 : RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

Question 1:

Find the area in square centimetres of a triangle whose base and altitude are as under:
 (i) base = 18 cm, altitude = 3.5 cm
 (ii) base = 8 dm, altitude = 15 cm

Answer 1:

We know that the area of a triangle = 1/2×Base×Height
(i) Here, base = 18 cm and height = 3.5 cm
     ∴ Area of the triangle = (1/2×18×3.5)=31.5 cm2
(ii) Here, base = 8 dm = (8 x 10) cm = 80 cm     [Since 1 dm = 10 cm]
      and height = 3.5 cm
     ∴ Area of the triangle = (1/2×80×15)=600 cm2

 

Question 2:

Find the altitude of a triangle whose area is 42 cm2 and base is 12 cm.

Answer 2:

We have,
Altitude of a triangle = RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Here, base = 12 cm and area = 42 cm2
RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

 

Question 3:

The area of a triangle is 50 cm2. If the altitude is 8 cm, what is its base?

Answer 3:

We have,
Base of a triangle =   RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Here, altitude = 8 cm and area = 50 cm2

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

 

Question 4:

Find the area of a right angled triangle whose sides containing the right angle are of lengths 20.8 m and 14.7 m.

Answer 4:

In a right-angled triangle, the sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle 

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

 

Question 5:

The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.

Answer 5:

For the first triangle, we have,
Base = 15 cm and altitude = 7 cm
Thus, area of a triangle =  RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

 

It is given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 cm2
One side of the second triangle = 10.5 cm
Therefore,
The other side of the second triangle =    RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Hence, the other side of the second triangle will be 10 cm.

 

Question 6:

A rectangular field is 48 m long and 20 m wide. How many right triangular flower beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this field?

Answer 6:

We have,
Length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Area of the rectangular field = Length x Breadth = 48 m x 20 m = 960 m2

Area of one right triangular flower bed = 1/2×12 m×5 m = 30 m2
Therefore,

Required number of right triangular flower beds RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

 

Question 7:

In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ⊥ ACBM ⊥ ACDL = 16.5 cm and BM= 12 cm. Find the area of quadrilateral ABCD.

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Answer 7:

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Hence,
Area of quadrilateral ABCD = Area of Δ ADC  + Area of Δ ABC
                                                          = (693 + 504) cm2
                                                                             = 1197 cm

Question 8:

Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD measure 48 m and 32 m respectively and are perpendicular to each other.

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Answer 8:

We have,
Diagonal AC = 48 cm and diagonal BD = 32 m
∴ Area of a quadrilateral = 1/2 x Product of diagonals

 RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

 

Question 9:

In Fig 31, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such that EF ⊥ AD and EF= 14 cm. Calculate the area of the shaded region.

 

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Answer 9:

We have,
Area of the rectangle = AB x BC
                             = 32 m x 18 m
                             = 576 m2

 Area of the triangle  =  RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev
                           = 1/2x BC x FE       [Since AD = BC]
                           = 1/2x 18 m x 14 m
                           = 9 m x 14 m = 126 m2
∴ Area of the shaded region = Area of the rectangle − Area of the triangle
                                      =(576 − 126) m2
                                                  = 450 

 

Question 10:

In Fig. 32, ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm. If PQRS be the mid-points of the sides ABBCCD and DA respectively, find the area of the shaded region.

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Answer 10:

We have,
Join points PR and SQ.
These two lines bisect each other at point O.

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Here, ABDCSQ = 40 cm and AD = BC =RP = 25 cm

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

From the figure we observed that,
Area of Δ SPQ = Area of Δ SRQ
Hence, area of the shaded region  2 x (Area of Δ SPQ)

RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

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