Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

Question 11:

Find the area of a rhombus having each side equal to 15 cm and one of whose diagonals is 24 cm.

Answer 11:

Let ABCD be the rhombus where diagonals intersect at O.

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Then AB = 15 cm and AC = 24 cm.
The diagonals of a rhombus bisect each other at right angles.
Therefore, Δ AOB is a right-angled triangle, right angled at O such that

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

By Pythagoras theorem, we have,
(AB)2(OA)2 + (OB)2
⇒ (15)2 = (12)2 + (OB)2
⇒ (OB)= (15)− (12)2
⇒ (OB)= 225 − 144 = 81
⇒ (OB)2 = (9)2
⇒ OB = 9 cm
∴ BD = 2 x OB = 2 x 9 cm = 18 cm
Hence,
Area of the rhombus ABCDRD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Question 12:

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

Answer 12:

Let ABCD be the  rhombus whose diagonals intersect at O.

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Then AB = 20 cm and AC = 24 cm.
The diagonals of a rhombus bisect each other at right angles.
Therefore Δ AOB is a right-angled triangle, right angled at O such that
OA = 1/2AC = 12 cm and AB  = 20 cm
By Pythagoras theorem, we have,
(AB)2 = (OA)2 + (OB)2
⇒ (20)2 = (12)2 + (OB)2
⇒ (OB)= (20)− (12)2
⇒ (OB)= 400 − 144 = 256
⇒ (OB)2 = (16)2
 OB = 16 cm
∴ BD = 2 x OB = 2 x 16 cm = 32 cm
Hence,
Area of the rhombus ABCD =

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Question 13:

The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonals is 2 m?

Answer 13:

We have,
Side of a square = 4 m and one diagonal of a square = 2 m

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Area of the rhombus = Area of the square of side 4 m

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

⇒ AO=1/2AC = 8 m  and BO =1/2BD=1 m
By Pythagoras theorem, we have:   
    AO2BO2AB2
 AB2 = (8 m)2 + (1m)2 = 64 m2 + 1 m2 = 65 m2
⇒ Side of a rhombus = ABRD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Let DX be the altitude.
 Area of the rhombus = AB × DX
          16 m2 RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics x Dx

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Hence, the altitude of the rhombus will be  RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 

Question 14:

Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the sides of length 25 cm is 10 cm, find the altitude corresponding to the other pair of sides.

Answer 14:

We have,
ABCD is a parallelogram with longer side AB = 25 cm and  altitude AE = 10 cm.
As ABCD is a parallelogram .hence AB=CD    (opposite sides of parallelogram are equal)
The shorter side is AD =  20 cm and the corresponding altitude is CF.

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Area of a parallelogram = Base × Height
We have two altitudes and two corresponding bases.
So,

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Hence, the altitude corresponding to the other pair of the side AD is 12.5 cm.

 

Question 15:

The base and corresponding altitude of a parallelogram are 10 cm and 12 cm respectively. If the other altitude is 8 cm, find the length of the other pair of parallel sides.

Answer 15:

We have,
ABCD is a parallelogram with side ABCD = 10 cm (Opposite sides of parallelogram are equal) and corresponding altitude AM = 12 cm.
The other side is AD and the corresponding altitude is CN = 8 cm

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Area of a parallelogram = Base × Height
We have two altitudes and two corresponding bases.
So,

RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Hence, the length of the other pair of the parallel side = 15 cm.

 

Question 16:

A floral design on the floor of a building consists of 280 tiles. Each tile is in the shape of a parallelogram of altitude 3 cm and base 5 cm. Find the cost of polishing the design at the rate of 50 paise per cm2.

Answer 16:

We have,
Altitude of a tile = 3 cm
Base of a tile = 5 cm
Area of one tile = Altitude x Base = 5 cm x 3 cm = 15 cm2
Area of 280 tiles = 280 x 15 cm2 = 4200 cm2
Rate of polishing the tiles at 50 paise per cm2 = Rs. 0.5 per cm2
Thus,
Total cost of polishing the design = Rs. (4200 x 0.5) = Rs. 2100

 

The document RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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FAQs on RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math - RD Sharma Solutions for Class 7 Mathematics

1. What are the different shapes and figures covered in RD Sharma Solutions (Part - 2) for Class 7 Math?
Ans. RD Sharma Solutions (Part - 2) for Class 7 Math covers various shapes and figures such as rectangles, squares, triangles, circles, parallelograms, and trapeziums. It also includes concepts related to the area and perimeter of these shapes.
2. How can I find the area of a rectangle using RD Sharma Solutions (Part - 2) for Class 7 Math?
Ans. To find the area of a rectangle using RD Sharma Solutions (Part - 2) for Class 7 Math, you need to multiply the length and the breadth of the rectangle. The formula for the area of a rectangle is length × breadth.
3. What is the formula for finding the area of a triangle in RD Sharma Solutions (Part - 2) for Class 7 Math?
Ans. RD Sharma Solutions (Part - 2) for Class 7 Math provides the formula for finding the area of a triangle as 1/2 × base × height. The base is the length of the triangle's base, and the height is the perpendicular distance from the base to the opposite vertex.
4. How can I calculate the perimeter of a circle using RD Sharma Solutions (Part - 2) for Class 7 Math?
Ans. In RD Sharma Solutions (Part - 2) for Class 7 Math, to calculate the perimeter of a circle, you need to use the formula 2πr, where r is the radius of the circle. The value of π is approximately 3.14.
5. Can RD Sharma Solutions (Part - 2) for Class 7 Math help me solve problems related to irregular shapes?
Ans. Yes, RD Sharma Solutions (Part - 2) for Class 7 Math provides concepts and formulas that can be applied to solve problems related to irregular shapes as well. By breaking down the irregular shape into smaller regular shapes and finding their areas individually, you can determine the total area of the irregular shape.
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