Question

In fig seg DH ⊥ seg EF and seg GK ⊥ seg EF.
If DH = 12cm, GK= 20 cm
and A(∆DEF) = 300 cm² then find
(i) EF
(ii) A(∆GEF)
(iii) A(⎕DFGE)

Answer 1

(i) We know:
Area of a triangle = $\frac{1}{2}$ × Base × Height
Thus, we get:
A(ΔDEF) = $\frac{1}{2}$ × EF × DH
$\Rightarrow$300 = $\frac{1}{2}$ × EF × 12
$\Rightarrow$300 = EF × 6
$\Rightarrow$EF = 50 cm

(ii)
From the figure, we observe that ΔDEF and ΔGEF have a common base EF.
We know that the ratio of the areas of two triangles with the common base is equal to the ratio of their corresponding heights.
Thus, we get:
$$\begin{gathered} \frac{\mathrm{A}\left(∆\mathrm{DEF}\right)}{\mathrm{A}\left(∆\mathrm{GEF}\right)}=\frac{\mathrm{DH}}{\mathrm{GK}} \\ \Rightarrow \frac{300}{\mathrm{A}\left(∆\mathrm{GEF}\right)}=\frac{12}{20} \\ \Rightarrow \mathrm{A}\left(∆\mathrm{GEF}\right)=\frac{300\times 20}{12}=500{\mathrm{cm}}^2 \end{gathered}$$

(iii)
From the figure, we observe that the area of quadrilateral DFGE is equal to the sum of the areas of ΔDEF and ΔGEF.
Thus, we get:
A(DFGE) = A(ΔDEF) + A(ΔGEF)
= 300 + 500
= 800 cm²