BC=10 cm , BD=7 cm . Area of∆ABC in 50 sq.cm .Find the area of ∆ABD and ∆ADC
Answers
Answer:
The value of DC is 6 cm.
Solution:
According to the question,
Given Data:
Area of triangle ABD is 10 square cm and area of triangle ADC is 12 sq.cm
To Find:
(a) Find BD: DC
(b) If BD=5cm,
Step-by-step Explanation:
Step 1:
Area of triangle ADB =h \times \frac{b_{1}}{2}=h×2b1
10 =h \times \frac{5}{2}=h×25
To find the value of h:
Step 2:
h=4 cm
The value of h is 4 cm
Area of ADC =h \times \frac{b_{2}}{2}=h×2b2
Step 3:
12= 4 \times \frac{b_{2}}{2}4×2b2
Step 4:
8frac{C D}{b_{2}}=\frac{24}{4}=6 \mathrm{cm}
BD: DC=5:6
Step 5:
Result:
The value of DC is 6 cm.
Given : BC=10 cm , BD=7 cm . Area of ∆ABC in 50 sq.cm
To Find : the area of ∆ABD and ∆ADC
Solution:
Draw AE⊥ BC , BD , CD
Area of Triangle = (1/2) * Base * height
Area of ∆ABC = (1/2) * BC * AE = 50
=> Area of ∆ABC = (1/2) *10 * AE = 50
=> (1/2) * AE = 5
Area of Δ ABD = (1/2) BD * AE
= 5 * BD
= 5 * 7
= 35 cm²
CD = BC - BD = 10 - 7 = 3 cm
Area of Δ ACD = (1/2) CD * AE
= 5 * CD
= 5 * 3
= 15 cm²
Area of Δ ABD = 35 cm²
Area of Δ ACD = 15 cm²
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