Question
Construct a parallelogram ABCD in which BC = 5cm, $\angle\text{BCD}=120^\circ$ and CD = 4.8cm.
✓
Answer
Steps of construction:Step 1: Draw BC = 5cm.
Step 2: Make an $\angle\text{BCD}=120^\circ.$
Step 2: With C as centre draw an arc of 4.8cm, name that point as D.
Step 3: With D as centre draw an arc 5cm, name that point as A.
Step 4: With B as centre draw another arc 4.8cm cutting the previous arc at A.
Step 5: Join AD and AB.
Then, ABCD is a required parallelogram.
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
Plot the given points on a graph sheet and check if they lie on a straight line. If not, name the shape they form when joined in the given order.
J(4, 3), K(6, 1), L(6, 5) and M(4, 7)
J(4, 3), K(6, 1), L(6, 5) and M(4, 7)
Simplify:
$\left(2 x^2+3 x-5\right)\left(3 x^2-5 x+4\right)$
→$\left(2 x^2+3 x-5\right)\left(3 x^2-5 x+4\right)$
If the diagonals of a rhombus are 12cm and 16cm, find the length of each side.
Jyoti and Meena run a ready-made garment shop. They mark the garments at such a price that even after allowing a discount a of 12.5%, they make a profit of 10%. Find the marked price of a suit which costs them Rs. 1470.
Verify the division algorithm i.e., Dividend = Divisor × Quotient + Remainder, in the following. Also write the quotient and remainder.
Dividend: $6 y^5-28 y^2+30 y-9$
Divisor:$2 y^2-6$
→Dividend: $6 y^5-28 y^2+30 y-9$
Divisor:$2 y^2-6$
Ratna obtained a loan of Rs. 25000 from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after 2 years to discharge her debt?
A die is thrown at random. Find the probability of getting:
- 2.
- A number less than 3.
- A composite number.
- A number not less than 4.
Evaluate $\left(3.2 x^6 y^3\right) \times\left(2.1 x^2 y^2\right)$ when $x=1$ and $y=0.5$
→Verify the following:
$\frac{-13}{24}\times\Big(\frac{-12}{5}\times\frac{35}{36}\Big)=\Big(\frac{-13}{24}\times\frac{-12}{5}\Big)\times\frac{35}{36}$
→$\frac{-13}{24}\times\Big(\frac{-12}{5}\times\frac{35}{36}\Big)=\Big(\frac{-13}{24}\times\frac{-12}{5}\Big)\times\frac{35}{36}$
Solve the following equation and also check your result in case:
$\frac{(1-2\text{x})}{7}-\frac{(2-3\text{x})}{8}=\frac{3}{2}+\frac{\text{x}}{4}$
→$\frac{(1-2\text{x})}{7}-\frac{(2-3\text{x})}{8}=\frac{3}{2}+\frac{\text{x}}{4}$