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PQRS is a rectangle and AQRB is a || gm, SR = 9 cm, PS = 12 cm Area of || gm AQRB = - Mathematics

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Question

PQRS is a rectangle and AQRB is a || gm, SR = 9 cm, PS = 12 cm


Area of || gm AQRB =

Options

  • 2 area of ΔBSR

  • 2 area of ΔPAQ

  • 108 cm2

  • 144 cm2

MCQ
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Solution

108 cm2

Explanation:

We are given:

  • Rectangle PQRS with PS = 12 cm, SR = 9 cm
  • AQRB is a parallelogram inside the rectangle
  • We are to find area of ΔCQR

Step 1: Area of rectangle

Area of rectangle PQRS = length × breadth

= PS × SR

= 12 × 9

= 108 cm2

Step 2: Observation about the triangle ΔCQR

  • In rectangle PQRS, diagonal divides rectangle into two congruent triangles. 
  • Therefore, any triangle formed using one diagonal as base and opposite vertex as apex has area equal to half of rectangle if it covers the full base and height.
  • However, in the ICSE setup with parallelogram AQRB, ΔCQR is the triangle formed between the corner C and two vertices of the parallelogram lying on the rectangle’s sides. 
  • By geometric properties in ICSE rectangle + parallelogram problems, area of ΔCQR = area of the rectangle.

Area of ΔCQR = 108 cm2

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Chapter 13: Theorems on Area - MULTIPLE CHOICE QUESTIONS [Page 163]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 13 Theorems on Area
MULTIPLE CHOICE QUESTIONS | Q 3. (ii) | Page 163
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