Question

Match the two columns for the figure shown below.
(4 marks)

$\begin{array}{cc}\text{Column - I}& \text{Column - II}\\ \text{(A) Area of}\square DABC& \text{(p) 41}\\ \text{(B) Area of}\Delta ADC& \text{(q) 126}\\ \text{(C) Area of}\Delta ABC& \text{(r) 306}\\ \text{(D) AC}& \text{(s) 180}\end{array}$

Answer 1

Using Pythagorus theorem in $\Delta ABC$

$AC=\sqrt{A{B}^2+B{C}^2}$

$=\sqrt{9^2+{40}^2}$

$=\sqrt{81+1600}=41m$

(A) Area of \(\square DABC = area~(\Delta ABC) + area~(\Delta ADC)\

$area\left(\Delta ABC\right)=\frac{1}{2}\left(a\right)\left(40\right)=180m^{2}for\Delta ADC$

$S=\frac{28+15+41}{2}=42m$

$area\left(\Delta ADC\right)=\sqrt{42(42-28)(42-15)(42-41)}$

$=\sqrt{42\left(14\right)\left(27\right)}$

$=\sqrt{2\times 3\times 7\times 2\times 7\times 3\times 3\times 3}$

= 2 x 3 x 3 x 7 = 126$m^2$

$\therefore area\square DABC=180+126=306m^2$ (1 mark)

(B) $Areaof\Delta ABC=126m^2$ (1 mark)

(C) $Areaof\Delta ABC=180m^2$ (1 mark)

(D) $AC=41m$ (1 mark)

$\left(A\right)\to r;\left(B\right)\to q;\left(C\right)\to s;\left(D\right)\to p$