RD Sharma Chapter 15 Class 9 Maths Exercise 15.2 Solutions

RD Sharma Chapter 15 Class 9 Maths Exercise 15.2 Solutions is based on the topics like- Rectangular region, Polygon region, Area Axioms, Triangular region, etc., which is concerned under the Area of Parallelograms and Triangles. Students will learn about ‘how to calculate Areas of Parallelograms and Triangles’ through the questions mentioned in the attached PDF.

The RD Sharma Chapter 15 Class 9 Maths Exercise 15.2 Solutions PDF is prepared by our subject experts, provided the varieties of questions based on the rectangular region, polygon region, etc., with detailed explanation and example. Go down to the article and learn about the topics and subtopics concerned under this exercise of Chapter 15.

Learn about RD Sharma Class 9 Chapter 15 (Areas Of And Triangles)

Download RD Sharma Chapter 15 Class 9 Maths Exercise 15.2 Solutions PDF

Solutions for Class 9 Maths Chapter 15 Area of Parallelogram and Triangles Exercise 15.2

Important Definitions RD Sharma Chapter 15 Class 9 Maths Exercise 15.2 Solutions

The topics and subtopics of Area of Parallelograms and Triangles are mentioned below-

1. Polygon Regions
2. Triangular region
3. Rectangular region
4. Polygon region
5. Area Axiom
1. Area Axiom
2. Congruent area axiom
3. Area monotone axiom

Examples of RD Sharma Chapter 15 Class 9 Maths Exercise 15.2 Solutions

Ques 1- If PQRS is a parallelogram, then prove that-

ar(Δ PQS) = ar(Δ QRS) = ar(Δ PQR)=ar(Δ PRS) = 1/2 ar(parallelogram PQRS)

Solution-

PQRS is a parallelogram.

When we join the diagonal of the parallelogram, it divides it into two quadrilaterals.

Step 1: Let PR is diagonal, then, Area (ΔPQR) = Area (ΔPRS) = 1/2(Area of parallelogram PQRS)

Step 2: Let QS be another diagonal

Area (ΔPQS) = Area (ΔQRS) = 1/2( Area of parallelogram PQRS)

Now,

From Step 1 and step 2, we have

Area (ΔPQR) = Area (ΔPQS) = Area (ΔPQS) = Area (ΔQRS) = 1/2(Area of parallelogram PQRS)

Hence Proved.

Ques 2- Let PQRS be a parallelogram of area 124 cm2. If T and U are the mid-points of sides PQ and RS, respectively, then find the area of parallelogram PTUS.

Solution-

PQRS be a parallelogram.

Area of parallelogram = 124 cm2 (Given)

Consider a point A and join PA, which is perpendicular to SR.

Now, Area of parallelogram TQRU = UR x PA and

Area of parallelogram PUTS = SU x PA

Since F is the mid-point of SR, so SU = UR

From the above results, we have

Area of parallelogram PTUS = Area of parallelogram TQRU = 1/2 (Area of parallelogram PQRS)

= 124/2

= 62

The area of parallelogram PTUS is 62 cm2 (Hence Proved).

Frequently Asked Questions (FAQs) of RD Sharma Chapter 15 Class 9 Maths Exercise 15.2 Solutions

Ques 1- What is the area of parallelogram and triangle?

Ans- The area of a triangle is ½ x base x height. Practice questions in this theorem from areas of parallelograms and triangles If a triangle and parallelogram are on the corresponding base and within the same parallels, then an area of the triangle is equivalent to half the area of a parallelogram.

Ques 2- What is the relationship between a parallelogram and a triangle?

Ans- Each triangle brings up specifically one-half of a parallelogram. From this, the area of a triangle is one-half the area of a parallelogram or an area of a parallelogram is two times (twice) the area of a triangle.