In the given figure, area of AABC = 27 cm and the area of trapezium BDEC =165cm". If
BC =6 cm, then find (i) DE (ii) BD, if AB =9 cm.
Answers
Answer:
Step-by-step explanation:
Area of trapezium ABCD = Area of ∆ ABD + Area of ∆ CBD
= 1/2 × a × h + 1/2 × b × h
= 1/2 × h × (a + b)
= 1/2 (sum of parallel sides) × (perpendicular distance between them)
(i) DE= 16cm
(ii) BD= 15cm
Step-by-step explanation:
Given:
A(ΔABC)=27 sq.cm
Area of trapezium BDEC= 165sq.cm
BC= 6cm, AB= 9cm
To find:
(i) DE
(ii) BD
Solution:
A(ΔADE)= A(ΔABC)+ A trapezium(BDEC)
A(ΔADE)= 27+165
A(ΔADE)= 192 sq.cm
In ΔABC & ΔADE,
side BC= side DE.......(the sides of a trapezium are parallel)
Thus, the angles formed by these parallel lines will be congruent
∠ABC = ∠ADE
∠ACB = ∠AED
∴ ΔABC ~ ∆XYZ
According to the theorem of similar triangles
The ratio of the heights and bases and areas of similar triangles will be equal
Symbolically, A(ΔABC) = BC²
A(ΔADE) DE²
∴ 27 = (6)²
192 DE²
∴ 27 = 36
192 DE²
∴ 27 X DE² = 36 X 192
∴ 27 X DE² =6912
∴DE² =6912/27
∴ DE² = 256
∴ DE = √256
∴ DE = ±16 cm
But length cannot be negative, hence DE= 16cm
The triangles ABC & ADE are similar
So, the ratio of corresponding sides will be equal
∴ AB = BC
AD DE
∴ 9 = 6
AD 16
∴ 9 X 16= AD X 6
∴ 144 = AD X 6
∴ 144/6 = AD
∴ AD = 24cm
Now, side AD= side AB+ side BD.......(A-B-D)
24= 9+ side BD
24-9= side BD
side BD= 15cm
The length of sides (i) DE= 16cm, (ii) BD= 15cm
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