Question : In the given figure, the area of isosceles triangle $\mathrm{ABE}$ is $72\;\mathrm{cm^2}$ and $\mathrm{BE = AB}$ and $\mathrm{AB = 2 AD}$, $\mathrm{AE \parallel DC}$, then what is the area (in$\;\mathrm{cm^2}$) of the trapezium $\mathrm{ABCD}$?
Option 1: 108
Option 2: 124
Option 3: 136
Option 4: 144
Correct Answer: 144
Solution : Given that the area of triangle $\mathrm{ABE}$ is $72\;\mathrm{cm^2}$. The triangle is an isosceles. $\mathrm{AB} = \mathrm{BE} = \sqrt{2 \times 72} = 12\;\mathrm{cm}$ $\mathrm{AB = 2 AD}$ ⇒ $\mathrm{AD} = \frac{\mathrm{AB}}{2} = 6\;\mathrm{cm}$ Since $\mathrm{AE} \parallel \mathrm{DC}$ $\mathrm{CE} = \mathrm{AD} = 6\;\mathrm{cm}$, and $\mathrm{BC} = \mathrm{BE} + \mathrm{CE} = 12 + 6 = 18\;\mathrm{cm}$ The area of trapezium $\mathrm{ABCD}$ $=\frac{1}{2} \times (\mathrm{sum\; of\; parallel\; sides}) \times \mathrm{height} = \frac{1}{2} \times (18 + 6) \times 12 = 144\;\mathrm{cm^2}$ Hence, the correct answer is 144.
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Question : Parallel sides of a trapezium are $26\;\mathrm{cm}$ and $40\;\mathrm{cm}$ and the area is $792\;\mathrm{cm^2}$. What is the value of the distance (in $\mathrm{cm}$) between parallel sides?
Option 1: $24\;\mathrm{cm}$
Option 2: $48\;\mathrm{cm}$
Option 3: $12\;\mathrm{cm}$
Option 4: $36\;\mathrm{cm}$
Question : In $\Delta\mathrm{ ABC,AD}$ and $\mathrm{AE}$ are bisectors of $\angle \mathrm{BAC}$ and $\angle \mathrm{BAD}$ respectively. If $\angle \mathrm{BAE}=30^{\circ}, \mathrm{AE}=9\;\mathrm{cm}$ and $\mathrm{EC}=15\;\mathrm{cm}$, what is the area (in$\;\mathrm{cm^2}$ ) of $\Delta \mathrm{CAE}$?
Option 1: $36$
Option 2: $54$
Option 3: $72$
Option 4: $216$
Question : In $\Delta \mathrm{ABC}$, a line parallel to side $\mathrm{BC}$ cuts the sides $\mathrm{AB}$ and $\mathrm{AC}$ at points $\mathrm{D}$ and $\mathrm{E}$ respectively and also point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio of $\mathrm{1 : 4}$. If the area of $\Delta \mathrm{ABC}$ is $200\;\mathrm{cm^2}$, then what is the area (in $\mathrm{cm^2}$) of quadrilateral $\mathrm{DECB}$?
Option 1: 192
Option 2: 50
Option 3: 120
Option 4: 96
Question : In the given figure, $\mathrm{DE} \| \mathrm{BC}$. If $\mathrm{AD}=5 \mathrm{~cm}, \mathrm{DB}=10 \mathrm{~cm}$, and $\mathrm{AE}=8 \mathrm{~cm}$, then $\mathrm{AC}$ is:
Option 1: 24 cm
Option 2: 32 cm
Option 3: 8 cm
Option 4: 16 cm
Question : For a triangle ABC, D and E are two points on AB and AC such that $\mathrm{AD}=\frac{1}{6} \mathrm{AB}$, $\mathrm{AE}=\frac{1}{6} \mathrm{AC}$. If BC = 22 cm, then DE is _______. (Consider up to two decimals)
Option 1: 1.33 cm
Option 2: 1.67 cm
Option 3: 3.67 cm
Option 4: 3.33 cm
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