Question
In the given figure ABCD and ABFE are parallelograms such that ar(quadrilateral EABC) = 17cm2 and ar(||gm ABCD) = 25cm2. Then, $\text{ar}(\triangle\text{BCF})=?$
- 4 cm2
- 4.8cm2
- 6cm2
- 8cm2
✓
Answer
- 8cm2
Solution:
From the given figure, we have ||gm ABCD and ||gm ABFE are lying on the same base and between the same parallels.
$\Rightarrow\ \text{ar}(||\text{gm ABCD})=\text{ar}(||\text{gm ABFE})=25\text{cm}^2$
$\Rightarrow\ \text{ar}(\triangle\text{BCF})=\text{ar}(||\text{gm ABFE})-\text{ar(quadrilateral EABC)}$
$=25-17$
$=8\text{cm}^2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In Fig , $a+b=$
→If x + y + z = 9 and xy + yz + zx = 23, the value of (x3 + y3 + z3 - 3xyz) = ?
A _______ is an exact location in space.
Which of the following is a false statement?
→- If the diagonals of a rhombus are 18cm and 14cm, then its area is 126cm2.
- Area of a ||gm $=\frac{1}{2}\times$ base × corresponding height.
- A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
- If the area of a ||gm with one side 24cm and corresponding height h cm is 192cm2, then h = 8cm.
The product of a nonzero rational number with an irrational number is always a/ an.
If x140 + 2x151 + k divisible by x + 1, then the value of k is:
The number of planks of dimension (4m × 5m × 2m) that can be stored in a pit which is 40m long, 12m wide and 16m deep, is:
- 190
- 192
- 184
- 180
