The base of a triangle is $14\ cm$ and its height is $8\ cm.$ The area of the triangle is:
- A$112 \mathrm{~cm}^2$
- ✓$56 \mathrm{~cm}^2$
- C$122 \mathrm{~cm}^2$
- D$66 \mathrm{~cm}^2$
Answer: B.
View full solution →Question types
Area of a Trapezium and a Polygon question types
45 questions across 4 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
45
Questions
4
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5
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Sample Questions
Area of a Trapezium and a Polygon questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
The parallel sides of a trapezium are $54 \ cm$ and $26\ cm$ and the distance between them is $15\ cm.$ The area of the trapezium is:
- A$ 702 \mathrm{~cm}^2 $
- B$ 810 \mathrm{~cm}^2 $
- C$ 405 \mathrm{~cm}^2 $
- ✓$ 600 \mathrm{~cm}^2 $
Answer: D.
View full solution →Tick the correct answer in the following: The lengths of the parallel sides of a trapezium are $19\ cm$ and $19\ cm$ and its area is $128\ cm^2$. The distance between the parallel sides is:
- A$9\ cm$
- B$7\ cm$
- ✓$8\ cm$
- D$12.5\ cm$
Answer: C.
View full solution →Tick the correct answer in the following: In the given figure, $AB \| DC$ and $\text{AB}\perp\text{DC}$ If $DC = 7\ cm, BC = 10\ cm, AB = 13\ cm$ and $\text{CL}\perp\text{AB},$ the area of trap. $\text{ABCD}$ is:
- A$ 84 \mathrm{~cm}^2 $
- B$ 72 \mathrm{~cm}^2 $
- ✓$ 80 \mathrm{~cm}^2 $
- D$ 91 \mathrm{~cm}^2 $
Answer: C.
View full solution →The area of a rhombus is $120\ cm^2$ and one of its diagonals is $24\ cm. $ Each side of the rhombus is:
- A$10\ cm$
- ✓$13\ cm$
- C$12\ cm$
- D$15\ cm$
Answer: B.
View full solution →The length of the fence of a trapezium-shaped field $ABCD$ is $130m$ and side $AB$ is perpendicular to each of the parallel sides $AD$ and $BC$. If $BC = 54\ m, CD = 19\ m$ and $AD = 42\ m$, find the area of the field.
View full solution →The shape of the top surface of a table is trapezium. Its parallel sides are $1\ m$ and $1.4\ m$ and the perpendicular distance between them is $0.9\ m$. Find its area.
View full solution →Find the area of a trapezium whose parallel sides are $24\ cm$ and $20\ cm$ and the distance between them is $15\ cm$.
View full solution →Find the area enclosed by the given figure $ABCDEF$ as per dimensions given herewith.
View full solution →The area of a trapezium is $1080cm^2$. If the lengths of its parallel sides be $55\ cm$ and $35\ cm$, find the distance between them.
View full solution →The area of a trapezium is $216m^2$ and its height is $12m.$ If one of the parallel sides is $14m$ less than the other, find the length of each of the parallel sides.
View full solution →The area of a trapezium is $405\ cm^2$. Its parallel sides are in the ratio $4 : 5$ and the distance between them is $18\ cm.$ Find the length of each of the parallel sides.
View full solution →The base of a triangular field is three times its height and its area is $1350m^2$. Find the base and height of the field.
View full solution →Find the area of pentagon $ABCDE$ in which $\text{BL}\perp\text{AC},\text{cm}\perp\text{AD}$ and $\text{EN}\perp\text{AD}$ such that $AC = 10\ cm, AD = 12\ cm, BL = 3\ cm, \ cm = 7\ cm$ and $EN = 5\ cm.$
View full solution →Find the area of a quadrilateral one of whose diagonals is $40\ cm$ and the lengths of the perpendiculars drawn from the opposite vertices on the diagonal are $16\ cm$ and $12\ cm.$
View full solution →Fill in the blanks.
The parallel sides of a trapezium are $14\ cm$ and $18\ cm$ and the distance between them is $8\ cm.$ The area of the trapezium is _______ $cm^2$.
View full solution →The parallel sides of a trapezium are $14\ cm$ and $18\ cm$ and the distance between them is $8\ cm.$ The area of the trapezium is _______ $cm^2$.
Fill in the blanks. Area of a parallelogram $=\frac{1}{2}\times(\ \underline{\ \ \ \ \ \ \ }\ )\times(\ \underline{\ \ \ \ \ \ \ }\ ).$
View full solution →Fill in the blanks. Area of triangle $=\frac{1}{2}\times(\ \underline{\ \ \ \ \ \ \ }\ )\times(\ \underline{\ \ \ \ \ \ \ }\ ).$
View full solution →Fill in the blanks. Area of a trapezium $=\frac{1}{2}\times(\ \underline{\ \ \ \ \ \ \ }\ )\times(\ \underline{\ \ \ \ \ \ \ }\ ).$
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