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. $\triangle abd : \triangle adc = \square : \square$ $\triangle abd : \triangle abc = \square : \square$ $\triangle abd = 40 \times \dfrac{\square}{\square} = \square\\ (\mathrm{cm}^2)$ 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 $\triangle bdc$. $\langle$ans.$\
angle$ \underline{\qquad\qquad\qquad\qquad} (2) find the area of $\triangle bde$. $\langle$ans.$\
angle$ \underline{\qquad\qquad\qquad\qquad} (3) find the area of $\triangle dec$. $\langle$ans.$\
angle$ \underline{\qquad\qquad\qquad\qquad}

Explanation:

Step1: Solve problem 4

Substep1: Ratio of ΔABD to ΔADC

ΔABD and ΔADC share the same height from A to BC. Their bases are BD=4cm and DC=6cm. Area ratio = base ratio: 4:6 = 2:3.

Substep2: Ratio of ΔABD to ΔABC

ΔABC area = ΔABD + ΔADC. Total base BC=4+6=10cm. ΔABD base ratio: 4:10 = 2:5.

Substep3: Calculate area of ΔABD

ΔABD area = 40 × (2/5) = 16 cm².

Step2: Solve problem 5(1)

D is midpoint of AC, so ΔBDC area = ½ ΔABC area = ½×24=12 cm².

Step3: Solve problem5(2)

BE:EC=1:2, so ΔBDE area = (1/(1+2))×ΔBDC area = (1/3)×12=4 cm².

Step4: Solve problem5(3)

ΔDEC area = ΔBDC area - ΔBDE area =12-4=8 cm².

Answer:

Problem4:
ΔABD:ΔADC=2:3
ΔABD:ΔABC=2:5
ΔABD=40×2/5=16 cm²

Problem5:
(1)12 cm²
(2)4 cm²
(3)8 cm²