KnowledgeBoat Logo
|

Mathematics

Assertion (A): In the given figure, square ABCD and △APB are equal in area.

In the given figure, square ABCD and △APB are equal in area. Assertion Reasoning, Concise Mathematics Solutions ICSE Class 9.

Reason (R): Square ABCD and △APB are on the same base (AB) and between the same parallels (AB//DP).

⇒ Area of △APB = 12×Area of square ABCD\dfrac{1}{2}\times \text{Area of square ABCD}

  1. A is true, R is false.
  2. A is false, R is true.
  3. Both A and R are true.
  4. Both A and R are false.

Theorems on Area

1 Like

Answer

A is false, R is true.

Explanation

Given,

  • ABCD is a square.

  • AB is parallel to CD.

  • △APB shares the base AB with square ABCD.

  • The height of △APB is equal to the side of the square (h).

Let BC = h (height of square)

Height of triangle = h (height of square)

Area of (Δ ABP) = 12\dfrac{1}{2} x base x height

= 12\dfrac{1}{2} x AB x h

= 12\dfrac{1}{2} x h x h (∴ As all sides of square are equal)

= 12\dfrac{1}{2} x h2 …………..(1)

Area of square ABCD = (side)2

= h2…………..(2)

From (1) and (2),

Area of (Δ ABP) = 12\dfrac{1}{2} x Area of square ABCD

∴ Assertion (A) is false.

As proved above,

Area of (Δ ABP) = 12\dfrac{1}{2} x Area of square ABCD

∴ Reason (R) is true.

Hence, Assertion (A) is false, Reason (R) is true.

Answered By

3 Likes


Related Questions