Question

4 the area of $\triangle abc$ in the figure on the right is $40\\ \mathrm{cm}^2$ and $bd = 4\\ \mathrm{cm}$, $dc = 6\\ \mathrm{cm}$. at this time, the area of $\triangle abd$ was obtained as follows. fill the appropriate numbers in the blanks.
if the base of $\triangle abd$ is $bd$ and the base of $\triangle adc$ is $dc$, the height of the two triangles is equal.
at this time, the ratio of the area of the triangle is equal to the ratio of the base.
$\triangle abd : \triangle adc = \square : \square$
$\triangle abd : \triangle abc = \square : \square$
$\triangle abd = 40 \times \dfrac{\square}{\square} = \square\\ (\mathrm{cm}^2)$
(figure: a triangle $abc$ with $d$ on $bc$, $bd = 4\\ \mathrm{cm}$, $dc = 6\\ \mathrm{cm}$)

5 the area of $\triangle abc$ in the figure on the right is $24\\ \mathrm{cm}^2$. given point $d$ is the midpoint of side $ac$ and $be : ec = 1 : 2$, find the area of the triangles below.
(1) find the area of $\triangle bdc$.
\boxed{\text{p 96}}
\langle \text{ans.} \
angle \underline{\quad\quad\quad\quad\quad\quad}
(2) find the area of $\triangle bde$.
\boxed{\text{p 96}}
\langle \text{ans.} \
angle \underline{\quad\quad\quad\quad\quad\quad}
(3) find the area of $\triangle dec$.
\langle \text{ans.} \
angle \underline{\quad\quad\quad\quad\quad\quad}
(figure: a triangle $abc$ with $d$ the midpoint of $ac$, $e$ on $bc$ with $be:ec = 1:2$)

Explanation:

Step1: Find area of ΔBDC

ΔBDC shares height with ΔABD; D is midpoint of AC, so base DC = AD. Area ratio ΔBDC:ΔABD = 1:1.
Area of ΔBDC = $\frac{1}{2} \times$ Area of ΔABC = $\frac{1}{2} \times 24 = 12$ cm².

Step2: Find area of ΔBDE

BE:EC = 1:2, so BE:BC = 1:3. ΔBDE and ΔBDC share height from D to BC.
Area of ΔBDE = $\frac{1}{3} \times$ Area of ΔBDC = $\frac{1}{3} \times 12 = 4$ cm².

Step3: Find area of ΔDEC

ΔDEC and ΔBDC share height from D to BC; EC:BC = 2:3.
Area of ΔDEC = $\frac{2}{3} \times$ Area of ΔBDC = $\frac{2}{3} \times 12 = 8$ cm².

Answer:

12
4
8