Chapter 11, Problem 30RP

To determine

To show: that the area of $\Delta CAT$ is equal to the area of $BRAT.$

Explanation of Solution

Given Information:

  $BRAT$ is a trapezoid

  $M$ is the mid-point of side $RB$

Formula used:

Areas of two congruent triangles are equal.

Proof:

As $BRAT$ is a trapezoid, $AR\parallel TB$

  $\therefore AR\parallel TC\mathrm{...}\left(T-B-C\right)$ (i)

In $\Delta ARM\text{and}MBC,$

  $\angle AMR=\angle CMB$ … (Vertically opposite angles)

  $\angle ARM=\angle CBM$ … (Alternate angles)

  $RM=BM$ … (Given)

  $\therefore \Delta ARM\cong \Delta CBM$ … (By ASA congruency criterion)

As we know that the areas of two congruent triangles are equal

  $\therefore ar\left(\Delta ARM\right)=ar\left(\Delta CBM\right)$ … (ii)

Now, $ar\left(\Delta CAT\right)=ar\left(\Delta CBM\right)+ar\left(▱BMAT\right)$

  $\therefore ar\left(\Delta CAT\right)=ar\left(\Delta ARM\right)+ar\left(▱BMAT\right)$ … From (ii)

  $\therefore ar\left(\Delta CAT\right)=ar\left(▱BRAT\right)$

Hence, area of $\Delta CAT$ is equal to the area of $BRAT.$