Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths

Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 1. If figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm, and CF = 10 cm, Find AD.

Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

Given that,

In parallelogram ABCD, CD = AB = 16 cm                                [∵ Opposite side of a parallelogramare equal]

We know that,

Area of parallelogram = Base × Corresponding altitude

Area of parallelogram ABCD = CD × AE = AD × CF

16 cm × cm = AD × 10 cm

Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics cm = 12.8 cm

Thus, The length of AD is 12.8 cm.

 

Q 2. In Q 1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, Find AB.

Solution:

We know that,

Area of a parallelogram ABCD = AD × CF ⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

Again area of parallelogram ABCD = CD × AE⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

Compare equation(1) and equation(2)

AD × CF   = CD × AE

⇒6×10 = D×8

⇒D = Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics  = 7.5 cm

∴ AB = DC = 7.5cm                                     [∵Opposite side of a parallelogramare equal]

 

Q 3. Let ABCD be a parallelogram of area 124 cm. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.

Solution:

Given,

Area of a parallelogram ABCD = 124 cm2

Construction: Draw AP⊥DC

Proof:-

Area of a parallelogram AFED = DF × AP  ⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

And  area of parallelogram EBCF = FC × AP⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

And DF = FC  ⋅⋅⋅⋅⋅⋅⋅⋅⋅(3)                                               [FisthemidpointofDC]

Compare equation (1), (2)  and (3)

Area of parallelogram AEFD = Area of parallelogram EBCF

∴ Area of parallelogram AEFD = Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsEx-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics  =  62cm2

 

Q 4. If ABCD is a parallelogram, then prove that

Ar (ΔABD)  = Ar (ΔBCD)  = Ar (ΔABC)  = Ar (ΔACD)  =  Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsAr(//gmABCD).

Solution:

Given:-

ABCD is a parallelogram,

To prove : – Ar (ΔABD)  = Ar (ΔBCD)  = Ar (ΔABC)  = Ar (ΔACD)  =  Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsAr(//gmABCD).

Proof:- We know that diagonal of a parallelogram divides it into two equilaterals .

Since, AC is the diagonal.

Then,  Ar (ΔABC)  = Ar (ΔACD)  =  Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsAr(//gmABCD) ⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

Since, BD is the diagonal.

Then ,  Ar (ΔABD)  = Ar (ΔBCD)  =  Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsAr(//gmABCD) ⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

Compare equation (1) and (2)

∴  Ar (ΔABC)  = Ar (ΔACD)  = Ar (ΔABD)  = Ar (ΔBCD)  =  Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsAr(//gmABCD)..

The document Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-15.2, Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. How do you find the area of a parallelogram?
Ans. To find the area of a parallelogram, we multiply the base (b) with the corresponding height (h) perpendicular to it. The formula for the area of a parallelogram is A = b × h.
2. How can we determine if two triangles are congruent?
Ans. Two triangles are said to be congruent if their corresponding sides and angles are equal. This can be determined using various congruence criteria such as SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), etc.
3. Can the area of a triangle be negative?
Ans. No, the area of a triangle cannot be negative. Area is a measure of space, and it cannot have a negative value. The area of a triangle is always positive or zero.
4. How do you find the area of a triangle when only the side lengths are given?
Ans. If only the side lengths of a triangle are given, we can find the area using Heron's formula. Heron's formula states that the area (A) of a triangle with side lengths a, b, and c can be calculated using the formula: A = √(s(s-a)(s-b)(s-c)) where s is the semi-perimeter of the triangle, given by s = (a + b + c)/2.
5. Is it possible for a parallelogram to have a right angle?
Ans. Yes, it is possible for a parallelogram to have a right angle. A parallelogram with a right angle is called a rectangle. In a rectangle, opposite sides are equal in length and all angles are right angles.
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